INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
3 −1 Ç 2
3 −1 Ç 2
3 −1
Ç2
È6
Ù
È
Ù
È
5
0 Ù ~ È 0 −4
3Ù ~ È 0 −4
3ÙÙ
È
ÈÉ 2 −5
7 ÙÚ ÈÉ 0 −8
8ÙÚ ÈÉ 0
0
2ÙÚ
The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in
common.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
Ç2
È0
È
ÈÉ 2
4
1
3
4
−2
0
4 Ç2
−2 ÙÙ ~ ÈÈ 0
0 ÙÚ ÈÉ 0
4
1
−1
4
−2
−4
4 Ç2
−2 ÙÙ ~ ÈÈ 0
−4ÙÚ ÈÉ 0
4
1
0
4
−2
−6
4
−2ÙÙ
−6ÙÚ
The system is consistent, and using the argument from Example 2, there is only one solution. So the
three planes have only one point in common.
4
h
Ç 1 h 4 Ç1
19. È
~È
Ù
Ù Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no
É3 6 8Ú É 0 6 − 3h −4 Ú
solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
h −5 Ç 1
h
−5
Ç1
Write c for −8 − 2h. If c = 0, that is, if h = –4, then the system
20. È
~È
Ù
6 Ú É 0 −8 − 2h 16 ÚÙ
É 2 −8
has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.
−2
4
Ç 1 4 −2 Ç 1
21. È
~È
Ù
Ù Write c for h − 12 . Then the second equation cx2 = 0 has a solution
É3 h −6 Ú É 0 h − 12 0 Ú
for every value of c. So the system is consistent for all h.
Ç −4
h È
~
−3ÙÚ È 0
É
and only if h = 6.
Ç −4
22. È
É 2
12
−6
12
0
h
h
h ÙÙ The system is consistent if and only if −3 + = 0, that is, if
2
−3 +
2Ú
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description applies to a single solution. The solution set consists of all possible
solutions. Only in special cases does the solution set consist of exactly one solution. Mark a
statement True only if the statement is always true.
d. True. See the box before Example 2.
24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
c. False. The definition of equivalent systems is in the second paragraph after equation (2).
d. True. By definition, a consistent system has at least one solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
3 −1 Ç 2
3 −1 Ç 2
3 −1
Ç2
È6
Ù
È
Ù
È
5
0 Ù ~ È 0 −4
3Ù ~ È 0 −4
3ÙÙ
È
ÈÉ 2 −5
7 ÙÚ ÈÉ 0 −8
8ÙÚ ÈÉ 0
0
2ÙÚ
The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in
common.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
Ç2
È0
È
ÈÉ 2
4
1
3
4
−2
0
4 Ç2
−2 ÙÙ ~ ÈÈ 0
0 ÙÚ ÈÉ 0
4
1
−1
4
−2
−4
4 Ç2
−2 ÙÙ ~ ÈÈ 0
−4ÙÚ ÈÉ 0
4
1
0
4
−2
−6
4
−2ÙÙ
−6ÙÚ
The system is consistent, and using the argument from Example 2, there is only one solution. So the
three planes have only one point in common.
4
h
Ç 1 h 4 Ç1
19. È
~È
Ù
Ù Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no
É3 6 8Ú É 0 6 − 3h −4 Ú
solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
h −5 Ç 1
h
−5
Ç1
Write c for −8 − 2h. If c = 0, that is, if h = –4, then the system
20. È
~È
Ù
6 Ú É 0 −8 − 2h 16 ÚÙ
É 2 −8
has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.
−2
4
Ç 1 4 −2 Ç 1
21. È
~È
Ù
Ù Write c for h − 12 . Then the second equation cx2 = 0 has a solution
É3 h −6 Ú É 0 h − 12 0 Ú
for every value of c. So the system is consistent for all h.
Ç −4
h È
~
−3ÙÚ È 0
É
and only if h = 6.
Ç −4
22. È
É 2
12
−6
12
0
h
h
h ÙÙ The system is consistent if and only if −3 + = 0, that is, if
2
−3 +
2Ú
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description applies to a single solution. The solution set consists of all possible
solutions. Only in special cases does the solution set consist of exactly one solution. Mark a
statement True only if the statement is always true.
d. True. See the box before Example 2.
24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
c. False. The definition of equivalent systems is in the second paragraph after equation (2).
d. True. By definition, a consistent system has at least one solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
6
CHAPTER 1
• Linear Equations in Linear Algebra
7 g
7
g
7
g
Ç 1 −4
Ç1 −4
Ç1 −4
È
Ù
È
Ù
È
Ù
h Ù ~ È0
h
3 −5 h Ù ~ È 0
3 −5
3 −5
25. È 0
Ù
ÈÉ −2
5 −9 k ÙÚ ÈÉ0 −3
5 k + 2 g ÙÚ ÈÉ 0
0
0 k + 2 g + h ÙÚ
Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix
above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution
if and only if k + 2g + h = 0.
26. Row reduce the augmented matrix for the given system:
Ç2
Èc
É
4
d
Ç1
~È
Ù
g Ú Éc
f
2
Ç1
~È
Ù
g Ú É0
f /2
d
2
f /2
d − 2c
g − c ( f / 2) ÙÚ
This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f
and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus
d 2c.
27. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible
since a is nonzero. Then replace R2 by R2 + (–c)R1.
Ça
È
Éc
b
d
Ç1
~È
Ù
g Ú Éc
f
Ç1
~È
Ù
g Ú É0
b/a
f /a
d
b/a
f /a
g − c( f / a ) ÚÙ
d − c(b / a )
The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity
g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be
written as ad – bc ≠ 0, or ad ≠ bc.
28. A basic principle of this section is that row operations do not affect the solution set of a linear
system. Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and
then perform any elementary row operations to produce other augmented matrices. Here are three
examples. The fact that they are all row equivalent proves that they all have the solution set (3, –2, –
1).
Ç1
È0
È
ÉÈ0
0
1
0
0
0
1
3 Ç1
−2ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 0
0
1
0
0
0
1
3 Ç1
4ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 2
0
1
0
0
0
1
3
4ÙÙ
5ÚÙ
29. Swap R1 and R3; swap R1 and R3.
30. Multiply R3 by –1/5; multiply R3 by –5.
31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.
32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2.
33. The first equation was given. The others are:
T2 = (T1 + 20 + 40 + T3 )/4,
or
4T2 − T1 − T3 = 60
T3 = (T4 + T2 + 40 + 30)/4,
or
4T3 − T4 − T2 = 70
T4 = (10 + T1 + T3 + 30)/4,
or
4T4 − T1 − T3 = 40
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
3 −1 Ç 2
3 −1 Ç 2
3 −1
Ç2
È6
Ù
È
Ù
È
5
0 Ù ~ È 0 −4
3Ù ~ È 0 −4
3ÙÙ
È
ÈÉ 2 −5
7 ÙÚ ÈÉ 0 −8
8ÙÚ ÈÉ 0
0
2ÙÚ
The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in
common.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
Ç2
È0
È
ÈÉ 2
4
1
3
4
−2
0
4 Ç2
−2 ÙÙ ~ ÈÈ 0
0 ÙÚ ÈÉ 0
4
1
−1
4
−2
−4
4 Ç2
−2 ÙÙ ~ ÈÈ 0
−4ÙÚ ÈÉ 0
4
1
0
4
−2
−6
4
−2ÙÙ
−6ÙÚ
The system is consistent, and using the argument from Example 2, there is only one solution. So the
three planes have only one point in common.
4
h
Ç 1 h 4 Ç1
19. È
~È
Ù
Ù Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no
É3 6 8Ú É 0 6 − 3h −4 Ú
solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
h −5 Ç 1
h
−5
Ç1
Write c for −8 − 2h. If c = 0, that is, if h = –4, then the system
20. È
~È
Ù
6 Ú É 0 −8 − 2h 16 ÚÙ
É 2 −8
has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.
−2
4
Ç 1 4 −2 Ç 1
21. È
~È
Ù
Ù Write c for h − 12 . Then the second equation cx2 = 0 has a solution
É3 h −6 Ú É 0 h − 12 0 Ú
for every value of c. So the system is consistent for all h.
Ç −4
h È
~
−3ÙÚ È 0
É
and only if h = 6.
Ç −4
22. È
É 2
12
−6
12
0
h
h
h ÙÙ The system is consistent if and only if −3 + = 0, that is, if
2
−3 +
2Ú
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description applies to a single solution. The solution set consists of all possible
solutions. Only in special cases does the solution set consist of exactly one solution. Mark a
statement True only if the statement is always true.
d. True. See the box before Example 2.
24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
c. False. The definition of equivalent systems is in the second paragraph after equation (2).
d. True. By definition, a consistent system has at least one solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
6
CHAPTER 1
• Linear Equations in Linear Algebra
7 g
7
g
7
g
Ç 1 −4
Ç1 −4
Ç1 −4
È
Ù
È
Ù
È
Ù
h Ù ~ È0
h
3 −5 h Ù ~ È 0
3 −5
3 −5
25. È 0
Ù
ÈÉ −2
5 −9 k ÙÚ ÈÉ0 −3
5 k + 2 g ÙÚ ÈÉ 0
0
0 k + 2 g + h ÙÚ
Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix
above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution
if and only if k + 2g + h = 0.
26. Row reduce the augmented matrix for the given system:
Ç2
Èc
É
4
d
Ç1
~È
Ù
g Ú Éc
f
2
Ç1
~È
Ù
g Ú É0
f /2
d
2
f /2
d − 2c
g − c ( f / 2) ÙÚ
This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f
and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus
d 2c.
27. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible
since a is nonzero. Then replace R2 by R2 + (–c)R1.
Ça
È
Éc
b
d
Ç1
~È
Ù
g Ú Éc
f
Ç1
~È
Ù
g Ú É0
b/a
f /a
d
b/a
f /a
g − c( f / a ) ÚÙ
d − c(b / a )
The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity
g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be
written as ad – bc ≠ 0, or ad ≠ bc.
28. A basic principle of this section is that row operations do not affect the solution set of a linear
system. Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and
then perform any elementary row operations to produce other augmented matrices. Here are three
examples. The fact that they are all row equivalent proves that they all have the solution set (3, –2, –
1).
Ç1
È0
È
ÉÈ0
0
1
0
0
0
1
3 Ç1
−2ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 0
0
1
0
0
0
1
3 Ç1
4ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 2
0
1
0
0
0
1
3
4ÙÙ
5ÚÙ
29. Swap R1 and R3; swap R1 and R3.
30. Multiply R3 by –1/5; multiply R3 by –5.
31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.
32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2.
33. The first equation was given. The others are:
T2 = (T1 + 20 + 40 + T3 )/4,
or
4T2 − T1 − T3 = 60
T3 = (T4 + T2 + 40 + 30)/4,
or
4T3 − T4 − T2 = 70
T4 = (10 + T1 + T3 + 30)/4,
or
4T4 − T1 − T3 = 40
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
7
Rearranging,
4T1
−
T2
−T1
+
4T2
−T2
−
−T1
T4
= 30
− T3
+ 4T3
−
T4
= 60
= 70
−
+ 4T4
= 40
T3
34. Begin by interchanging R1 and R4, then create zeros in the first column:
Ç 4
È −1
È
È 0
È
É −1
−1
0
−1
4
−1
−1
4
0
−1
0
−1
4
Ç −1
Ù
60 È −1
Ù~È
70 Ù È 0
Ù È
40 Ú É 4
0
−1
4
4
−1
−1
4
0
−1
−1
0
−1
30
Ç −1
Ù
60 È 0
Ù~È
70 Ù È 0
Ù È
30 Ú É 0
40
0
−1
4
4
−1
0
4
−4
−1
−1
−4
15
40
20 Ù
Ù
70 Ù
Ù
190 Ú
Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
−1
0
4
−1
−1
−1
−4
15
Ç1
Ù
È
5
0
Ù~È
70 Ù È 0
Ù È
190 Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
−4
14
Ç1
Ù
È
5
0
Ù~È
75Ù È 0
Ù È
195Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
0
12
−40
5Ù
Ù
75Ù
Ù
270 Ú
Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
0
0
4
−1
−2
0
0
1
Ç1
Ù
5 È0
Ù~È
75Ù È0
Ù È
22.5Ú É0
−40
0
1
0
1
0
0
4
0
0
0
0
1
Ç1
Ù
27.5 È 0
Ù~È
120 Ù È0
Ù È
22.5Ú É0
50
0
1
0
1
0
0
1
0
0
0
0
1
50
27.5Ù
Ù
30 Ù
Ù
22.5Ú
The last step is to replace R1 by R1 + (–1)R3:
Ç1
È0
~È
È0
È
É0
0
0
0
1
0
0
1
0
0
0
0
1
20.0
27.5Ù
Ù . The solution is (20, 27.5, 30, 22.5).
30.0 Ù
Ù
22.5Ú
Notes: The Study Guide includes a “Mathematical Note” about statements, “If … , then … .”
This early in the course, students typically use single row operations to reduce a matrix. As a result,
even the small grid for Exercise 34 leads to about 80 multiplications or additions (not counting operations
with zero). This exercise should give students an appreciation for matrix programs such as MATLAB.
Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not
already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in
connection with an LU factorization.
For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB
notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/84+/89
calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section
1.1 describes how to access the data that is available for all numerical exercises in the text. This feature
has the ability to save students time if they regularly have their matrix program at hand when studying
linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale.
These commands are included in the text data sets, available from the text web site,
www.pearsonhighered.com/lay.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
3 −1 Ç 2
3 −1 Ç 2
3 −1
Ç2
È6
Ù
È
Ù
È
5
0 Ù ~ È 0 −4
3Ù ~ È 0 −4
3ÙÙ
È
ÈÉ 2 −5
7 ÙÚ ÈÉ 0 −8
8ÙÚ ÈÉ 0
0
2ÙÚ
The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in
common.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
Ç2
È0
È
ÈÉ 2
4
1
3
4
−2
0
4 Ç2
−2 ÙÙ ~ ÈÈ 0
0 ÙÚ ÈÉ 0
4
1
−1
4
−2
−4
4 Ç2
−2 ÙÙ ~ ÈÈ 0
−4ÙÚ ÈÉ 0
4
1
0
4
−2
−6
4
−2ÙÙ
−6ÙÚ
The system is consistent, and using the argument from Example 2, there is only one solution. So the
three planes have only one point in common.
4
h
Ç 1 h 4 Ç1
19. È
~È
Ù
Ù Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no
É3 6 8Ú É 0 6 − 3h −4 Ú
solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
h −5 Ç 1
h
−5
Ç1
Write c for −8 − 2h. If c = 0, that is, if h = –4, then the system
20. È
~È
Ù
6 Ú É 0 −8 − 2h 16 ÚÙ
É 2 −8
has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.
−2
4
Ç 1 4 −2 Ç 1
21. È
~È
Ù
Ù Write c for h − 12 . Then the second equation cx2 = 0 has a solution
É3 h −6 Ú É 0 h − 12 0 Ú
for every value of c. So the system is consistent for all h.
Ç −4
h È
~
−3ÙÚ È 0
É
and only if h = 6.
Ç −4
22. È
É 2
12
−6
12
0
h
h
h ÙÙ The system is consistent if and only if −3 + = 0, that is, if
2
−3 +
2Ú
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description applies to a single solution. The solution set consists of all possible
solutions. Only in special cases does the solution set consist of exactly one solution. Mark a
statement True only if the statement is always true.
d. True. See the box before Example 2.
24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
c. False. The definition of equivalent systems is in the second paragraph after equation (2).
d. True. By definition, a consistent system has at least one solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
6
CHAPTER 1
• Linear Equations in Linear Algebra
7 g
7
g
7
g
Ç 1 −4
Ç1 −4
Ç1 −4
È
Ù
È
Ù
È
Ù
h Ù ~ È0
h
3 −5 h Ù ~ È 0
3 −5
3 −5
25. È 0
Ù
ÈÉ −2
5 −9 k ÙÚ ÈÉ0 −3
5 k + 2 g ÙÚ ÈÉ 0
0
0 k + 2 g + h ÙÚ
Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix
above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution
if and only if k + 2g + h = 0.
26. Row reduce the augmented matrix for the given system:
Ç2
Èc
É
4
d
Ç1
~È
Ù
g Ú Éc
f
2
Ç1
~È
Ù
g Ú É0
f /2
d
2
f /2
d − 2c
g − c ( f / 2) ÙÚ
This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f
and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus
d 2c.
27. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible
since a is nonzero. Then replace R2 by R2 + (–c)R1.
Ça
È
Éc
b
d
Ç1
~È
Ù
g Ú Éc
f
Ç1
~È
Ù
g Ú É0
b/a
f /a
d
b/a
f /a
g − c( f / a ) ÚÙ
d − c(b / a )
The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity
g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be
written as ad – bc ≠ 0, or ad ≠ bc.
28. A basic principle of this section is that row operations do not affect the solution set of a linear
system. Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and
then perform any elementary row operations to produce other augmented matrices. Here are three
examples. The fact that they are all row equivalent proves that they all have the solution set (3, –2, –
1).
Ç1
È0
È
ÉÈ0
0
1
0
0
0
1
3 Ç1
−2ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 0
0
1
0
0
0
1
3 Ç1
4ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 2
0
1
0
0
0
1
3
4ÙÙ
5ÚÙ
29. Swap R1 and R3; swap R1 and R3.
30. Multiply R3 by –1/5; multiply R3 by –5.
31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.
32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2.
33. The first equation was given. The others are:
T2 = (T1 + 20 + 40 + T3 )/4,
or
4T2 − T1 − T3 = 60
T3 = (T4 + T2 + 40 + 30)/4,
or
4T3 − T4 − T2 = 70
T4 = (10 + T1 + T3 + 30)/4,
or
4T4 − T1 − T3 = 40
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
7
Rearranging,
4T1
−
T2
−T1
+
4T2
−T2
−
−T1
T4
= 30
− T3
+ 4T3
−
T4
= 60
= 70
−
+ 4T4
= 40
T3
34. Begin by interchanging R1 and R4, then create zeros in the first column:
Ç 4
È −1
È
È 0
È
É −1
−1
0
−1
4
−1
−1
4
0
−1
0
−1
4
Ç −1
Ù
60 È −1
Ù~È
70 Ù È 0
Ù È
40 Ú É 4
0
−1
4
4
−1
−1
4
0
−1
−1
0
−1
30
Ç −1
Ù
60 È 0
Ù~È
70 Ù È 0
Ù È
30 Ú É 0
40
0
−1
4
4
−1
0
4
−4
−1
−1
−4
15
40
20 Ù
Ù
70 Ù
Ù
190 Ú
Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
−1
0
4
−1
−1
−1
−4
15
Ç1
Ù
È
5
0
Ù~È
70 Ù È 0
Ù È
190 Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
−4
14
Ç1
Ù
È
5
0
Ù~È
75Ù È 0
Ù È
195Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
0
12
−40
5Ù
Ù
75Ù
Ù
270 Ú
Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
0
0
4
−1
−2
0
0
1
Ç1
Ù
5 È0
Ù~È
75Ù È0
Ù È
22.5Ú É0
−40
0
1
0
1
0
0
4
0
0
0
0
1
Ç1
Ù
27.5 È 0
Ù~È
120 Ù È0
Ù È
22.5Ú É0
50
0
1
0
1
0
0
1
0
0
0
0
1
50
27.5Ù
Ù
30 Ù
Ù
22.5Ú
The last step is to replace R1 by R1 + (–1)R3:
Ç1
È0
~È
È0
È
É0
0
0
0
1
0
0
1
0
0
0
0
1
20.0
27.5Ù
Ù . The solution is (20, 27.5, 30, 22.5).
30.0 Ù
Ù
22.5Ú
Notes: The Study Guide includes a “Mathematical Note” about statements, “If … , then … .”
This early in the course, students typically use single row operations to reduce a matrix. As a result,
even the small grid for Exercise 34 leads to about 80 multiplications or additions (not counting operations
with zero). This exercise should give students an appreciation for matrix programs such as MATLAB.
Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not
already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in
connection with an LU factorization.
For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB
notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/84+/89
calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section
1.1 describes how to access the data that is available for all numerical exercises in the text. This feature
has the ability to save students time if they regularly have their matrix program at hand when studying
linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale.
These commands are included in the text data sets, available from the text web site,
www.pearsonhighered.com/lay.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
8
CHAPTER 1
1.2
• Linear Equations in Linear Algebra
SOLUTIONS
Notes: The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises
7–14, in preparation for Section 1.5.)
1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c.
2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.
Ç1
3. ÈÈ 2
ÈÉ 3
2
4
6
4
6
9
Ç1
~ ÈÈ0
ÈÉ0
Ç1
È
4. È 2
ÉÈ 4
2
4
5
8 Ç1
8ÙÙ ~ ÈÈ0
12 ÙÚ ÈÉ0
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
4 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
2
0
0
5 Ç1
4 ÙÙ ~ ÈÈ0
2 ÚÙ ÉÈ0
2
0
−3
4
−3
−12
2
1
0
4
4
1
5 Ç1
6 ÙÙ ~ ÈÈ 0
2 ÙÚ ÈÉ0
̈
0 ÙÚ
Ç̈
5. È
É0
* Ç̈
,
̈ÙÚ ÈÉ 0
* Ç0
,
0 ÙÚ ÈÉ 0
Ç1
7. È
É3
3
9
7 Ç1
~
6 ÙÚ ÈÉ 0
4
7
4
−2
−3
4
1
0
2
0
0
4
5
4
2
0
0
3
0
2
1
0
8 Ç1
−8ÙÙ ~ ÈÈ0
−12 ÙÚ ÈÉ0
0
1
0
2
0
0
4
1
−3
−8
4ÙÙ . Pivot cols 1 and 3.
0ÙÚ
5
Ç1
Ù
È
−6 Ù ~ È0
−18ÚÙ ÉÈ0
4
0
1
2
−3
0
5 Ç1
−2 ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
7 Ç1
~
−15ÙÚ ÈÉ0
Corresponding system of equations:
x1
3
0
0
0
1
+ 3 x2
x3
2
4
6
4
6
9
1
Pivot cols
−2 ÙÙ .
1, 2, and 3
2 ÙÚ
* Ç̈
̈ÙÙ , ÈÈ 0
0 ÚÙ ÉÈ 0
4
1
Ç1
È2
È
ÈÉ 3
5
Ç1
Ù
È
− 18Ù ~ È 0
−6 ÚÙ ÉÈ 0
4
− 12
−3
0
1
0
Ç̈
6. ÈÈ 0
ÉÈ 0
4
−5
8
4ÙÙ
−12ÙÚ
* Ç0
0 ÙÙ , ÈÈ0
0ÚÙ ÉÈ0
7 Ç1
~
3ÙÚ ÈÉ 0
8
8ÙÙ
12 ÙÚ
2
1
0
4
4
−3
Ç1
È2
È
ÈÉ 4
2
4
5
5
6ÙÙ
−6ÚÙ
4
5
4
5
4 ÙÙ
2 ÙÚ
̈
0 ÙÙ
0 ÙÚ
3
0
0
1
−5
3ÙÚ
= −5
=
3
The basic variables (corresponding to the pivot positions) are x1 and x3. The remaining variable x2 is
free. Solve for the basic variables in terms of the free variable. The general solution is
Ê x1 = −5 − 3 x2
Í
Ë x2 is free
Íx = 3
Ì 3
Note: Exercise 7 is paired with Exercise 10.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
3 −1 Ç 2
3 −1 Ç 2
3 −1
Ç2
È6
Ù
È
Ù
È
5
0 Ù ~ È 0 −4
3Ù ~ È 0 −4
3ÙÙ
È
ÈÉ 2 −5
7 ÙÚ ÈÉ 0 −8
8ÙÚ ÈÉ 0
0
2ÙÚ
The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in
common.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
Ç2
È0
È
ÈÉ 2
4
1
3
4
−2
0
4 Ç2
−2 ÙÙ ~ ÈÈ 0
0 ÙÚ ÈÉ 0
4
1
−1
4
−2
−4
4 Ç2
−2 ÙÙ ~ ÈÈ 0
−4ÙÚ ÈÉ 0
4
1
0
4
−2
−6
4
−2ÙÙ
−6ÙÚ
The system is consistent, and using the argument from Example 2, there is only one solution. So the
three planes have only one point in common.
4
h
Ç 1 h 4 Ç1
19. È
~È
Ù
Ù Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no
É3 6 8Ú É 0 6 − 3h −4 Ú
solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
h −5 Ç 1
h
−5
Ç1
Write c for −8 − 2h. If c = 0, that is, if h = –4, then the system
20. È
~È
Ù
6 Ú É 0 −8 − 2h 16 ÚÙ
É 2 −8
has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.
−2
4
Ç 1 4 −2 Ç 1
21. È
~È
Ù
Ù Write c for h − 12 . Then the second equation cx2 = 0 has a solution
É3 h −6 Ú É 0 h − 12 0 Ú
for every value of c. So the system is consistent for all h.
Ç −4
h È
~
−3ÙÚ È 0
É
and only if h = 6.
Ç −4
22. È
É 2
12
−6
12
0
h
h
h ÙÙ The system is consistent if and only if −3 + = 0, that is, if
2
−3 +
2Ú
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description applies to a single solution. The solution set consists of all possible
solutions. Only in special cases does the solution set consist of exactly one solution. Mark a
statement True only if the statement is always true.
d. True. See the box before Example 2.
24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
c. False. The definition of equivalent systems is in the second paragraph after equation (2).
d. True. By definition, a consistent system has at least one solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
6
CHAPTER 1
• Linear Equations in Linear Algebra
7 g
7
g
7
g
Ç 1 −4
Ç1 −4
Ç1 −4
È
Ù
È
Ù
È
Ù
h Ù ~ È0
h
3 −5 h Ù ~ È 0
3 −5
3 −5
25. È 0
Ù
ÈÉ −2
5 −9 k ÙÚ ÈÉ0 −3
5 k + 2 g ÙÚ ÈÉ 0
0
0 k + 2 g + h ÙÚ
Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix
above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution
if and only if k + 2g + h = 0.
26. Row reduce the augmented matrix for the given system:
Ç2
Èc
É
4
d
Ç1
~È
Ù
g Ú Éc
f
2
Ç1
~È
Ù
g Ú É0
f /2
d
2
f /2
d − 2c
g − c ( f / 2) ÙÚ
This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f
and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus
d 2c.
27. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible
since a is nonzero. Then replace R2 by R2 + (–c)R1.
Ça
È
Éc
b
d
Ç1
~È
Ù
g Ú Éc
f
Ç1
~È
Ù
g Ú É0
b/a
f /a
d
b/a
f /a
g − c( f / a ) ÚÙ
d − c(b / a )
The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity
g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be
written as ad – bc ≠ 0, or ad ≠ bc.
28. A basic principle of this section is that row operations do not affect the solution set of a linear
system. Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and
then perform any elementary row operations to produce other augmented matrices. Here are three
examples. The fact that they are all row equivalent proves that they all have the solution set (3, –2, –
1).
Ç1
È0
È
ÉÈ0
0
1
0
0
0
1
3 Ç1
−2ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 0
0
1
0
0
0
1
3 Ç1
4ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 2
0
1
0
0
0
1
3
4ÙÙ
5ÚÙ
29. Swap R1 and R3; swap R1 and R3.
30. Multiply R3 by –1/5; multiply R3 by –5.
31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.
32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2.
33. The first equation was given. The others are:
T2 = (T1 + 20 + 40 + T3 )/4,
or
4T2 − T1 − T3 = 60
T3 = (T4 + T2 + 40 + 30)/4,
or
4T3 − T4 − T2 = 70
T4 = (10 + T1 + T3 + 30)/4,
or
4T4 − T1 − T3 = 40
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
7
Rearranging,
4T1
−
T2
−T1
+
4T2
−T2
−
−T1
T4
= 30
− T3
+ 4T3
−
T4
= 60
= 70
−
+ 4T4
= 40
T3
34. Begin by interchanging R1 and R4, then create zeros in the first column:
Ç 4
È −1
È
È 0
È
É −1
−1
0
−1
4
−1
−1
4
0
−1
0
−1
4
Ç −1
Ù
60 È −1
Ù~È
70 Ù È 0
Ù È
40 Ú É 4
0
−1
4
4
−1
−1
4
0
−1
−1
0
−1
30
Ç −1
Ù
60 È 0
Ù~È
70 Ù È 0
Ù È
30 Ú É 0
40
0
−1
4
4
−1
0
4
−4
−1
−1
−4
15
40
20 Ù
Ù
70 Ù
Ù
190 Ú
Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
−1
0
4
−1
−1
−1
−4
15
Ç1
Ù
È
5
0
Ù~È
70 Ù È 0
Ù È
190 Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
−4
14
Ç1
Ù
È
5
0
Ù~È
75Ù È 0
Ù È
195Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
0
12
−40
5Ù
Ù
75Ù
Ù
270 Ú
Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
0
0
4
−1
−2
0
0
1
Ç1
Ù
5 È0
Ù~È
75Ù È0
Ù È
22.5Ú É0
−40
0
1
0
1
0
0
4
0
0
0
0
1
Ç1
Ù
27.5 È 0
Ù~È
120 Ù È0
Ù È
22.5Ú É0
50
0
1
0
1
0
0
1
0
0
0
0
1
50
27.5Ù
Ù
30 Ù
Ù
22.5Ú
The last step is to replace R1 by R1 + (–1)R3:
Ç1
È0
~È
È0
È
É0
0
0
0
1
0
0
1
0
0
0
0
1
20.0
27.5Ù
Ù . The solution is (20, 27.5, 30, 22.5).
30.0 Ù
Ù
22.5Ú
Notes: The Study Guide includes a “Mathematical Note” about statements, “If … , then … .”
This early in the course, students typically use single row operations to reduce a matrix. As a result,
even the small grid for Exercise 34 leads to about 80 multiplications or additions (not counting operations
with zero). This exercise should give students an appreciation for matrix programs such as MATLAB.
Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not
already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in
connection with an LU factorization.
For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB
notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/84+/89
calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section
1.1 describes how to access the data that is available for all numerical exercises in the text. This feature
has the ability to save students time if they regularly have their matrix program at hand when studying
linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale.
These commands are included in the text data sets, available from the text web site,
www.pearsonhighered.com/lay.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
8
CHAPTER 1
1.2
• Linear Equations in Linear Algebra
SOLUTIONS
Notes: The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises
7–14, in preparation for Section 1.5.)
1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c.
2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.
Ç1
3. ÈÈ 2
ÈÉ 3
2
4
6
4
6
9
Ç1
~ ÈÈ0
ÈÉ0
Ç1
È
4. È 2
ÉÈ 4
2
4
5
8 Ç1
8ÙÙ ~ ÈÈ0
12 ÙÚ ÈÉ0
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
4 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
2
0
0
5 Ç1
4 ÙÙ ~ ÈÈ0
2 ÚÙ ÉÈ0
2
0
−3
4
−3
−12
2
1
0
4
4
1
5 Ç1
6 ÙÙ ~ ÈÈ 0
2 ÙÚ ÈÉ0
̈
0 ÙÚ
Ç̈
5. È
É0
* Ç̈
,
̈ÙÚ ÈÉ 0
* Ç0
,
0 ÙÚ ÈÉ 0
Ç1
7. È
É3
3
9
7 Ç1
~
6 ÙÚ ÈÉ 0
4
7
4
−2
−3
4
1
0
2
0
0
4
5
4
2
0
0
3
0
2
1
0
8 Ç1
−8ÙÙ ~ ÈÈ0
−12 ÙÚ ÈÉ0
0
1
0
2
0
0
4
1
−3
−8
4ÙÙ . Pivot cols 1 and 3.
0ÙÚ
5
Ç1
Ù
È
−6 Ù ~ È0
−18ÚÙ ÉÈ0
4
0
1
2
−3
0
5 Ç1
−2 ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
7 Ç1
~
−15ÙÚ ÈÉ0
Corresponding system of equations:
x1
3
0
0
0
1
+ 3 x2
x3
2
4
6
4
6
9
1
Pivot cols
−2 ÙÙ .
1, 2, and 3
2 ÙÚ
* Ç̈
̈ÙÙ , ÈÈ 0
0 ÚÙ ÉÈ 0
4
1
Ç1
È2
È
ÈÉ 3
5
Ç1
Ù
È
− 18Ù ~ È 0
−6 ÚÙ ÉÈ 0
4
− 12
−3
0
1
0
Ç̈
6. ÈÈ 0
ÉÈ 0
4
−5
8
4ÙÙ
−12ÙÚ
* Ç0
0 ÙÙ , ÈÈ0
0ÚÙ ÉÈ0
7 Ç1
~
3ÙÚ ÈÉ 0
8
8ÙÙ
12 ÙÚ
2
1
0
4
4
−3
Ç1
È2
È
ÈÉ 4
2
4
5
5
6ÙÙ
−6ÚÙ
4
5
4
5
4 ÙÙ
2 ÙÚ
̈
0 ÙÙ
0 ÙÚ
3
0
0
1
−5
3ÙÚ
= −5
=
3
The basic variables (corresponding to the pivot positions) are x1 and x3. The remaining variable x2 is
free. Solve for the basic variables in terms of the free variable. The general solution is
Ê x1 = −5 − 3 x2
Í
Ë x2 is free
Íx = 3
Ì 3
Note: Exercise 7 is paired with Exercise 10.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
8.
Ç 1
È −3
É
Ç 1 −3 0 −5 Ç 1
~È
~
Ù
−2 0 −6 Ú É 0
7 0
1 0
3ÙÚ ÈÉ 0
x
= 4
Corresponding system of equations: 1
x2
= 3
−3
0
Ç1
~È
Ù
9Ú É0
−5
−3
0
−5
0
0
1
0
• Solutions
9
4
3ÙÚ
The basic variables (corresponding to the pivot positions) are x1 and x2. The remaining variable x3 is
free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic
variables do not depend on the value of the free variable.
Ê x1 = 4
Í
General solution: Ë x2 = 3
Í x is free
Ì 3
Note: A common error in Exercise 8 is to assume that x3 is zero. To avoid this, identify the basic
variables first. Any remaining variables are free. (This type of computation will arise in Chapter 5.)
Ç0
9. È
É1
1
−2
−3
4
Ç1
~È
Ù
−6 Ú É 0
3
Corresponding system:
−3
4
1
−2
x1
x2
Ç1
~È
Ù
3Ú É 0
−6
− 2 x3
= 3
− 2 x3
= 3
0
−2
1
−2
3
3ÙÚ
Ê x1 = 3 + 2 x3
Í
Basic variables: x1, x2; free variable: x3. General solution: Ë x2 = 3 + 2 x3
Í x is free
Ì 3
Ç 1
10. È
É −2
−2
−1
4
−5
Ç1
~È
Ù
6 Ú É0
4
Corresponding system:
x1
−2
−1
0
−7
Ç1
~È
Ù
14 Ú É 0
− 2 x2
= 2
4
x3
−2
0
0
1
2
−2 ÙÚ
= −2
Ê x1 = 2 + 2 x2
Í
Basic variables: x1, x3; free variable: x2. General solution: Ë x2 is free
Í x = −2
Ì 3
Ç3
È
11. È9
ÈÉ6
−2
−6
−4
4
12
8
0 Ç3
0 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
−2
0
0
x1
Corresponding system:
−
4
0
0
0 Ç1
0ÙÙ ~ ÈÈ 0
0ÙÚ ÈÉ 0
2
x2
3
+
4
x3
3
0
0
−2 3
0
0
43
0
0
0
0ÙÙ
0ÙÚ
= 0
= 0
= 0
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
3 −1 Ç 2
3 −1 Ç 2
3 −1
Ç2
È6
Ù
È
Ù
È
5
0 Ù ~ È 0 −4
3Ù ~ È 0 −4
3ÙÙ
È
ÈÉ 2 −5
7 ÙÚ ÈÉ 0 −8
8ÙÚ ÈÉ 0
0
2ÙÚ
The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in
common.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
Ç2
È0
È
ÈÉ 2
4
1
3
4
−2
0
4 Ç2
−2 ÙÙ ~ ÈÈ 0
0 ÙÚ ÈÉ 0
4
1
−1
4
−2
−4
4 Ç2
−2 ÙÙ ~ ÈÈ 0
−4ÙÚ ÈÉ 0
4
1
0
4
−2
−6
4
−2ÙÙ
−6ÙÚ
The system is consistent, and using the argument from Example 2, there is only one solution. So the
three planes have only one point in common.
4
h
Ç 1 h 4 Ç1
19. È
~È
Ù
Ù Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no
É3 6 8Ú É 0 6 − 3h −4 Ú
solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
h −5 Ç 1
h
−5
Ç1
Write c for −8 − 2h. If c = 0, that is, if h = –4, then the system
20. È
~È
Ù
6 Ú É 0 −8 − 2h 16 ÚÙ
É 2 −8
has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.
−2
4
Ç 1 4 −2 Ç 1
21. È
~È
Ù
Ù Write c for h − 12 . Then the second equation cx2 = 0 has a solution
É3 h −6 Ú É 0 h − 12 0 Ú
for every value of c. So the system is consistent for all h.
Ç −4
h È
~
−3ÙÚ È 0
É
and only if h = 6.
Ç −4
22. È
É 2
12
−6
12
0
h
h
h ÙÙ The system is consistent if and only if −3 + = 0, that is, if
2
−3 +
2Ú
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description applies to a single solution. The solution set consists of all possible
solutions. Only in special cases does the solution set consist of exactly one solution. Mark a
statement True only if the statement is always true.
d. True. See the box before Example 2.
24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
c. False. The definition of equivalent systems is in the second paragraph after equation (2).
d. True. By definition, a consistent system has at least one solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
6
CHAPTER 1
• Linear Equations in Linear Algebra
7 g
7
g
7
g
Ç 1 −4
Ç1 −4
Ç1 −4
È
Ù
È
Ù
È
Ù
h Ù ~ È0
h
3 −5 h Ù ~ È 0
3 −5
3 −5
25. È 0
Ù
ÈÉ −2
5 −9 k ÙÚ ÈÉ0 −3
5 k + 2 g ÙÚ ÈÉ 0
0
0 k + 2 g + h ÙÚ
Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix
above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution
if and only if k + 2g + h = 0.
26. Row reduce the augmented matrix for the given system:
Ç2
Èc
É
4
d
Ç1
~È
Ù
g Ú Éc
f
2
Ç1
~È
Ù
g Ú É0
f /2
d
2
f /2
d − 2c
g − c ( f / 2) ÙÚ
This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f
and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus
d 2c.
27. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible
since a is nonzero. Then replace R2 by R2 + (–c)R1.
Ça
È
Éc
b
d
Ç1
~È
Ù
g Ú Éc
f
Ç1
~È
Ù
g Ú É0
b/a
f /a
d
b/a
f /a
g − c( f / a ) ÚÙ
d − c(b / a )
The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity
g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be
written as ad – bc ≠ 0, or ad ≠ bc.
28. A basic principle of this section is that row operations do not affect the solution set of a linear
system. Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and
then perform any elementary row operations to produce other augmented matrices. Here are three
examples. The fact that they are all row equivalent proves that they all have the solution set (3, –2, –
1).
Ç1
È0
È
ÉÈ0
0
1
0
0
0
1
3 Ç1
−2ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 0
0
1
0
0
0
1
3 Ç1
4ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 2
0
1
0
0
0
1
3
4ÙÙ
5ÚÙ
29. Swap R1 and R3; swap R1 and R3.
30. Multiply R3 by –1/5; multiply R3 by –5.
31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.
32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2.
33. The first equation was given. The others are:
T2 = (T1 + 20 + 40 + T3 )/4,
or
4T2 − T1 − T3 = 60
T3 = (T4 + T2 + 40 + 30)/4,
or
4T3 − T4 − T2 = 70
T4 = (10 + T1 + T3 + 30)/4,
or
4T4 − T1 − T3 = 40
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
7
Rearranging,
4T1
−
T2
−T1
+
4T2
−T2
−
−T1
T4
= 30
− T3
+ 4T3
−
T4
= 60
= 70
−
+ 4T4
= 40
T3
34. Begin by interchanging R1 and R4, then create zeros in the first column:
Ç 4
È −1
È
È 0
È
É −1
−1
0
−1
4
−1
−1
4
0
−1
0
−1
4
Ç −1
Ù
60 È −1
Ù~È
70 Ù È 0
Ù È
40 Ú É 4
0
−1
4
4
−1
−1
4
0
−1
−1
0
−1
30
Ç −1
Ù
60 È 0
Ù~È
70 Ù È 0
Ù È
30 Ú É 0
40
0
−1
4
4
−1
0
4
−4
−1
−1
−4
15
40
20 Ù
Ù
70 Ù
Ù
190 Ú
Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
−1
0
4
−1
−1
−1
−4
15
Ç1
Ù
È
5
0
Ù~È
70 Ù È 0
Ù È
190 Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
−4
14
Ç1
Ù
È
5
0
Ù~È
75Ù È 0
Ù È
195Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
0
12
−40
5Ù
Ù
75Ù
Ù
270 Ú
Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
0
0
4
−1
−2
0
0
1
Ç1
Ù
5 È0
Ù~È
75Ù È0
Ù È
22.5Ú É0
−40
0
1
0
1
0
0
4
0
0
0
0
1
Ç1
Ù
27.5 È 0
Ù~È
120 Ù È0
Ù È
22.5Ú É0
50
0
1
0
1
0
0
1
0
0
0
0
1
50
27.5Ù
Ù
30 Ù
Ù
22.5Ú
The last step is to replace R1 by R1 + (–1)R3:
Ç1
È0
~È
È0
È
É0
0
0
0
1
0
0
1
0
0
0
0
1
20.0
27.5Ù
Ù . The solution is (20, 27.5, 30, 22.5).
30.0 Ù
Ù
22.5Ú
Notes: The Study Guide includes a “Mathematical Note” about statements, “If … , then … .”
This early in the course, students typically use single row operations to reduce a matrix. As a result,
even the small grid for Exercise 34 leads to about 80 multiplications or additions (not counting operations
with zero). This exercise should give students an appreciation for matrix programs such as MATLAB.
Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not
already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in
connection with an LU factorization.
For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB
notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/84+/89
calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section
1.1 describes how to access the data that is available for all numerical exercises in the text. This feature
has the ability to save students time if they regularly have their matrix program at hand when studying
linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale.
These commands are included in the text data sets, available from the text web site,
www.pearsonhighered.com/lay.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
8
CHAPTER 1
1.2
• Linear Equations in Linear Algebra
SOLUTIONS
Notes: The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises
7–14, in preparation for Section 1.5.)
1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c.
2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.
Ç1
3. ÈÈ 2
ÈÉ 3
2
4
6
4
6
9
Ç1
~ ÈÈ0
ÈÉ0
Ç1
È
4. È 2
ÉÈ 4
2
4
5
8 Ç1
8ÙÙ ~ ÈÈ0
12 ÙÚ ÈÉ0
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
4 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
2
0
0
5 Ç1
4 ÙÙ ~ ÈÈ0
2 ÚÙ ÉÈ0
2
0
−3
4
−3
−12
2
1
0
4
4
1
5 Ç1
6 ÙÙ ~ ÈÈ 0
2 ÙÚ ÈÉ0
̈
0 ÙÚ
Ç̈
5. È
É0
* Ç̈
,
̈ÙÚ ÈÉ 0
* Ç0
,
0 ÙÚ ÈÉ 0
Ç1
7. È
É3
3
9
7 Ç1
~
6 ÙÚ ÈÉ 0
4
7
4
−2
−3
4
1
0
2
0
0
4
5
4
2
0
0
3
0
2
1
0
8 Ç1
−8ÙÙ ~ ÈÈ0
−12 ÙÚ ÈÉ0
0
1
0
2
0
0
4
1
−3
−8
4ÙÙ . Pivot cols 1 and 3.
0ÙÚ
5
Ç1
Ù
È
−6 Ù ~ È0
−18ÚÙ ÉÈ0
4
0
1
2
−3
0
5 Ç1
−2 ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
7 Ç1
~
−15ÙÚ ÈÉ0
Corresponding system of equations:
x1
3
0
0
0
1
+ 3 x2
x3
2
4
6
4
6
9
1
Pivot cols
−2 ÙÙ .
1, 2, and 3
2 ÙÚ
* Ç̈
̈ÙÙ , ÈÈ 0
0 ÚÙ ÉÈ 0
4
1
Ç1
È2
È
ÈÉ 3
5
Ç1
Ù
È
− 18Ù ~ È 0
−6 ÚÙ ÉÈ 0
4
− 12
−3
0
1
0
Ç̈
6. ÈÈ 0
ÉÈ 0
4
−5
8
4ÙÙ
−12ÙÚ
* Ç0
0 ÙÙ , ÈÈ0
0ÚÙ ÉÈ0
7 Ç1
~
3ÙÚ ÈÉ 0
8
8ÙÙ
12 ÙÚ
2
1
0
4
4
−3
Ç1
È2
È
ÈÉ 4
2
4
5
5
6ÙÙ
−6ÚÙ
4
5
4
5
4 ÙÙ
2 ÙÚ
̈
0 ÙÙ
0 ÙÚ
3
0
0
1
−5
3ÙÚ
= −5
=
3
The basic variables (corresponding to the pivot positions) are x1 and x3. The remaining variable x2 is
free. Solve for the basic variables in terms of the free variable. The general solution is
Ê x1 = −5 − 3 x2
Í
Ë x2 is free
Íx = 3
Ì 3
Note: Exercise 7 is paired with Exercise 10.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
8.
Ç 1
È −3
É
Ç 1 −3 0 −5 Ç 1
~È
~
Ù
−2 0 −6 Ú É 0
7 0
1 0
3ÙÚ ÈÉ 0
x
= 4
Corresponding system of equations: 1
x2
= 3
−3
0
Ç1
~È
Ù
9Ú É0
−5
−3
0
−5
0
0
1
0
• Solutions
9
4
3ÙÚ
The basic variables (corresponding to the pivot positions) are x1 and x2. The remaining variable x3 is
free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic
variables do not depend on the value of the free variable.
Ê x1 = 4
Í
General solution: Ë x2 = 3
Í x is free
Ì 3
Note: A common error in Exercise 8 is to assume that x3 is zero. To avoid this, identify the basic
variables first. Any remaining variables are free. (This type of computation will arise in Chapter 5.)
Ç0
9. È
É1
1
−2
−3
4
Ç1
~È
Ù
−6 Ú É 0
3
Corresponding system:
−3
4
1
−2
x1
x2
Ç1
~È
Ù
3Ú É 0
−6
− 2 x3
= 3
− 2 x3
= 3
0
−2
1
−2
3
3ÙÚ
Ê x1 = 3 + 2 x3
Í
Basic variables: x1, x2; free variable: x3. General solution: Ë x2 = 3 + 2 x3
Í x is free
Ì 3
Ç 1
10. È
É −2
−2
−1
4
−5
Ç1
~È
Ù
6 Ú É0
4
Corresponding system:
x1
−2
−1
0
−7
Ç1
~È
Ù
14 Ú É 0
− 2 x2
= 2
4
x3
−2
0
0
1
2
−2 ÙÚ
= −2
Ê x1 = 2 + 2 x2
Í
Basic variables: x1, x3; free variable: x2. General solution: Ë x2 is free
Í x = −2
Ì 3
Ç3
È
11. È9
ÈÉ6
−2
−6
−4
4
12
8
0 Ç3
0 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
−2
0
0
x1
Corresponding system:
−
4
0
0
0 Ç1
0ÙÙ ~ ÈÈ 0
0ÙÚ ÈÉ 0
2
x2
3
+
4
x3
3
0
0
−2 3
0
0
43
0
0
0
0ÙÙ
0ÙÚ
= 0
= 0
= 0
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
10
CHAPTER 1
• Linear Equations in Linear Algebra
2
4
Ê
Í x1 = 3 x2 − 3 x3
Í
Basic variable: x1; free variables x2, x3. General solution: Ë x2 is free
Í x is free
Í 3
Ì
12. Since the bottom row of the matrix is equivalent to the equation 0 = 1, the system has no solutions.
Ç1
È0
13. È
È0
È
É0
−3
0
−1
0
1
0
0
0
0
1
−4
9
0
0
0
0
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
−2
−3
0
0
9
1
0
0
0
0
1
−4
9
0
0
0
0
x1
x2
Corresponding system:
x4
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
2
− 3 x5
= 5
− 4 x5
=
0
0
0
−3
1
0
0
0
0
1
−4
9
0
0
0
0
5
1Ù
Ù
4Ù
Ù
0Ú
1
+ 9 x5 = 4
0 = 0
Ê x1 = 5 + 3x5
Í x = 1 + 4x
5
ÍÍ 2
Basic variables: x1, x2, x4; free variables: x3, x5. General solution: Ë x3 is free
Íx = 4 − 9x
5
Í 4
ÍÌ x5 is free
Note: The Study Guide discusses the common mistake x3 = 0.
Ç1
È0
14. È
È0
È
É0
0
1
−5
4
0
−1
−8
0
0
0
0
0
0
0
1
0
3 Ç1
6 ÙÙ ÈÈ0
~
0 Ù È0
Ù È
0 Ú É0
−5
4
0
−1
0
0
0
0
0
0
0
0
1
0
− 5 x3
x1
Corresponding system:
0
1
x2
+ 4 x3
3
6 ÙÙ
0Ù
Ù
0Ú
= 3
−
x4
= 6
x5 = 0
0 = 0
Ê x1 = 3 + 5 x3
Íx = 6 − 4x + x
3
4
ÍÍ 2
Basic variables: x1, x2, x5; free variables: x3, x4. General solution: Ë x3 is free
Í x is free
Í 4
ÍÌ x5 = 0
15. a. The system is consistent. There are many solutions because x3 is a free variable.
b. The system is consistent. There are many solutions because x1 is a free variable.
16. a. The system is inconsistent. (The rightmost column of the augmented matrix is a pivot column).
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
3 −1 Ç 2
3 −1 Ç 2
3 −1
Ç2
È6
Ù
È
Ù
È
5
0 Ù ~ È 0 −4
3Ù ~ È 0 −4
3ÙÙ
È
ÈÉ 2 −5
7 ÙÚ ÈÉ 0 −8
8ÙÚ ÈÉ 0
0
2ÙÚ
The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in
common.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
Ç2
È0
È
ÈÉ 2
4
1
3
4
−2
0
4 Ç2
−2 ÙÙ ~ ÈÈ 0
0 ÙÚ ÈÉ 0
4
1
−1
4
−2
−4
4 Ç2
−2 ÙÙ ~ ÈÈ 0
−4ÙÚ ÈÉ 0
4
1
0
4
−2
−6
4
−2ÙÙ
−6ÙÚ
The system is consistent, and using the argument from Example 2, there is only one solution. So the
three planes have only one point in common.
4
h
Ç 1 h 4 Ç1
19. È
~È
Ù
Ù Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no
É3 6 8Ú É 0 6 − 3h −4 Ú
solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
h −5 Ç 1
h
−5
Ç1
Write c for −8 − 2h. If c = 0, that is, if h = –4, then the system
20. È
~È
Ù
6 Ú É 0 −8 − 2h 16 ÚÙ
É 2 −8
has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.
−2
4
Ç 1 4 −2 Ç 1
21. È
~È
Ù
Ù Write c for h − 12 . Then the second equation cx2 = 0 has a solution
É3 h −6 Ú É 0 h − 12 0 Ú
for every value of c. So the system is consistent for all h.
Ç −4
h È
~
−3ÙÚ È 0
É
and only if h = 6.
Ç −4
22. È
É 2
12
−6
12
0
h
h
h ÙÙ The system is consistent if and only if −3 + = 0, that is, if
2
−3 +
2Ú
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description applies to a single solution. The solution set consists of all possible
solutions. Only in special cases does the solution set consist of exactly one solution. Mark a
statement True only if the statement is always true.
d. True. See the box before Example 2.
24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
c. False. The definition of equivalent systems is in the second paragraph after equation (2).
d. True. By definition, a consistent system has at least one solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
6
CHAPTER 1
• Linear Equations in Linear Algebra
7 g
7
g
7
g
Ç 1 −4
Ç1 −4
Ç1 −4
È
Ù
È
Ù
È
Ù
h Ù ~ È0
h
3 −5 h Ù ~ È 0
3 −5
3 −5
25. È 0
Ù
ÈÉ −2
5 −9 k ÙÚ ÈÉ0 −3
5 k + 2 g ÙÚ ÈÉ 0
0
0 k + 2 g + h ÙÚ
Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix
above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution
if and only if k + 2g + h = 0.
26. Row reduce the augmented matrix for the given system:
Ç2
Èc
É
4
d
Ç1
~È
Ù
g Ú Éc
f
2
Ç1
~È
Ù
g Ú É0
f /2
d
2
f /2
d − 2c
g − c ( f / 2) ÙÚ
This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f
and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus
d 2c.
27. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible
since a is nonzero. Then replace R2 by R2 + (–c)R1.
Ça
È
Éc
b
d
Ç1
~È
Ù
g Ú Éc
f
Ç1
~È
Ù
g Ú É0
b/a
f /a
d
b/a
f /a
g − c( f / a ) ÚÙ
d − c(b / a )
The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity
g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be
written as ad – bc ≠ 0, or ad ≠ bc.
28. A basic principle of this section is that row operations do not affect the solution set of a linear
system. Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and
then perform any elementary row operations to produce other augmented matrices. Here are three
examples. The fact that they are all row equivalent proves that they all have the solution set (3, –2, –
1).
Ç1
È0
È
ÉÈ0
0
1
0
0
0
1
3 Ç1
−2ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 0
0
1
0
0
0
1
3 Ç1
4ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 2
0
1
0
0
0
1
3
4ÙÙ
5ÚÙ
29. Swap R1 and R3; swap R1 and R3.
30. Multiply R3 by –1/5; multiply R3 by –5.
31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.
32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2.
33. The first equation was given. The others are:
T2 = (T1 + 20 + 40 + T3 )/4,
or
4T2 − T1 − T3 = 60
T3 = (T4 + T2 + 40 + 30)/4,
or
4T3 − T4 − T2 = 70
T4 = (10 + T1 + T3 + 30)/4,
or
4T4 − T1 − T3 = 40
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
7
Rearranging,
4T1
−
T2
−T1
+
4T2
−T2
−
−T1
T4
= 30
− T3
+ 4T3
−
T4
= 60
= 70
−
+ 4T4
= 40
T3
34. Begin by interchanging R1 and R4, then create zeros in the first column:
Ç 4
È −1
È
È 0
È
É −1
−1
0
−1
4
−1
−1
4
0
−1
0
−1
4
Ç −1
Ù
60 È −1
Ù~È
70 Ù È 0
Ù È
40 Ú É 4
0
−1
4
4
−1
−1
4
0
−1
−1
0
−1
30
Ç −1
Ù
60 È 0
Ù~È
70 Ù È 0
Ù È
30 Ú É 0
40
0
−1
4
4
−1
0
4
−4
−1
−1
−4
15
40
20 Ù
Ù
70 Ù
Ù
190 Ú
Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
−1
0
4
−1
−1
−1
−4
15
Ç1
Ù
È
5
0
Ù~È
70 Ù È 0
Ù È
190 Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
−4
14
Ç1
Ù
È
5
0
Ù~È
75Ù È 0
Ù È
195Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
0
12
−40
5Ù
Ù
75Ù
Ù
270 Ú
Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
0
0
4
−1
−2
0
0
1
Ç1
Ù
5 È0
Ù~È
75Ù È0
Ù È
22.5Ú É0
−40
0
1
0
1
0
0
4
0
0
0
0
1
Ç1
Ù
27.5 È 0
Ù~È
120 Ù È0
Ù È
22.5Ú É0
50
0
1
0
1
0
0
1
0
0
0
0
1
50
27.5Ù
Ù
30 Ù
Ù
22.5Ú
The last step is to replace R1 by R1 + (–1)R3:
Ç1
È0
~È
È0
È
É0
0
0
0
1
0
0
1
0
0
0
0
1
20.0
27.5Ù
Ù . The solution is (20, 27.5, 30, 22.5).
30.0 Ù
Ù
22.5Ú
Notes: The Study Guide includes a “Mathematical Note” about statements, “If … , then … .”
This early in the course, students typically use single row operations to reduce a matrix. As a result,
even the small grid for Exercise 34 leads to about 80 multiplications or additions (not counting operations
with zero). This exercise should give students an appreciation for matrix programs such as MATLAB.
Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not
already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in
connection with an LU factorization.
For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB
notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/84+/89
calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section
1.1 describes how to access the data that is available for all numerical exercises in the text. This feature
has the ability to save students time if they regularly have their matrix program at hand when studying
linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale.
These commands are included in the text data sets, available from the text web site,
www.pearsonhighered.com/lay.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
8
CHAPTER 1
1.2
• Linear Equations in Linear Algebra
SOLUTIONS
Notes: The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises
7–14, in preparation for Section 1.5.)
1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c.
2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.
Ç1
3. ÈÈ 2
ÈÉ 3
2
4
6
4
6
9
Ç1
~ ÈÈ0
ÈÉ0
Ç1
È
4. È 2
ÉÈ 4
2
4
5
8 Ç1
8ÙÙ ~ ÈÈ0
12 ÙÚ ÈÉ0
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
4 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
2
0
0
5 Ç1
4 ÙÙ ~ ÈÈ0
2 ÚÙ ÉÈ0
2
0
−3
4
−3
−12
2
1
0
4
4
1
5 Ç1
6 ÙÙ ~ ÈÈ 0
2 ÙÚ ÈÉ0
̈
0 ÙÚ
Ç̈
5. È
É0
* Ç̈
,
̈ÙÚ ÈÉ 0
* Ç0
,
0 ÙÚ ÈÉ 0
Ç1
7. È
É3
3
9
7 Ç1
~
6 ÙÚ ÈÉ 0
4
7
4
−2
−3
4
1
0
2
0
0
4
5
4
2
0
0
3
0
2
1
0
8 Ç1
−8ÙÙ ~ ÈÈ0
−12 ÙÚ ÈÉ0
0
1
0
2
0
0
4
1
−3
−8
4ÙÙ . Pivot cols 1 and 3.
0ÙÚ
5
Ç1
Ù
È
−6 Ù ~ È0
−18ÚÙ ÉÈ0
4
0
1
2
−3
0
5 Ç1
−2 ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
7 Ç1
~
−15ÙÚ ÈÉ0
Corresponding system of equations:
x1
3
0
0
0
1
+ 3 x2
x3
2
4
6
4
6
9
1
Pivot cols
−2 ÙÙ .
1, 2, and 3
2 ÙÚ
* Ç̈
̈ÙÙ , ÈÈ 0
0 ÚÙ ÉÈ 0
4
1
Ç1
È2
È
ÈÉ 3
5
Ç1
Ù
È
− 18Ù ~ È 0
−6 ÚÙ ÉÈ 0
4
− 12
−3
0
1
0
Ç̈
6. ÈÈ 0
ÉÈ 0
4
−5
8
4ÙÙ
−12ÙÚ
* Ç0
0 ÙÙ , ÈÈ0
0ÚÙ ÉÈ0
7 Ç1
~
3ÙÚ ÈÉ 0
8
8ÙÙ
12 ÙÚ
2
1
0
4
4
−3
Ç1
È2
È
ÈÉ 4
2
4
5
5
6ÙÙ
−6ÚÙ
4
5
4
5
4 ÙÙ
2 ÙÚ
̈
0 ÙÙ
0 ÙÚ
3
0
0
1
−5
3ÙÚ
= −5
=
3
The basic variables (corresponding to the pivot positions) are x1 and x3. The remaining variable x2 is
free. Solve for the basic variables in terms of the free variable. The general solution is
Ê x1 = −5 − 3 x2
Í
Ë x2 is free
Íx = 3
Ì 3
Note: Exercise 7 is paired with Exercise 10.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
8.
Ç 1
È −3
É
Ç 1 −3 0 −5 Ç 1
~È
~
Ù
−2 0 −6 Ú É 0
7 0
1 0
3ÙÚ ÈÉ 0
x
= 4
Corresponding system of equations: 1
x2
= 3
−3
0
Ç1
~È
Ù
9Ú É0
−5
−3
0
−5
0
0
1
0
• Solutions
9
4
3ÙÚ
The basic variables (corresponding to the pivot positions) are x1 and x2. The remaining variable x3 is
free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic
variables do not depend on the value of the free variable.
Ê x1 = 4
Í
General solution: Ë x2 = 3
Í x is free
Ì 3
Note: A common error in Exercise 8 is to assume that x3 is zero. To avoid this, identify the basic
variables first. Any remaining variables are free. (This type of computation will arise in Chapter 5.)
Ç0
9. È
É1
1
−2
−3
4
Ç1
~È
Ù
−6 Ú É 0
3
Corresponding system:
−3
4
1
−2
x1
x2
Ç1
~È
Ù
3Ú É 0
−6
− 2 x3
= 3
− 2 x3
= 3
0
−2
1
−2
3
3ÙÚ
Ê x1 = 3 + 2 x3
Í
Basic variables: x1, x2; free variable: x3. General solution: Ë x2 = 3 + 2 x3
Í x is free
Ì 3
Ç 1
10. È
É −2
−2
−1
4
−5
Ç1
~È
Ù
6 Ú É0
4
Corresponding system:
x1
−2
−1
0
−7
Ç1
~È
Ù
14 Ú É 0
− 2 x2
= 2
4
x3
−2
0
0
1
2
−2 ÙÚ
= −2
Ê x1 = 2 + 2 x2
Í
Basic variables: x1, x3; free variable: x2. General solution: Ë x2 is free
Í x = −2
Ì 3
Ç3
È
11. È9
ÈÉ6
−2
−6
−4
4
12
8
0 Ç3
0 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
−2
0
0
x1
Corresponding system:
−
4
0
0
0 Ç1
0ÙÙ ~ ÈÈ 0
0ÙÚ ÈÉ 0
2
x2
3
+
4
x3
3
0
0
−2 3
0
0
43
0
0
0
0ÙÙ
0ÙÚ
= 0
= 0
= 0
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
10
CHAPTER 1
• Linear Equations in Linear Algebra
2
4
Ê
Í x1 = 3 x2 − 3 x3
Í
Basic variable: x1; free variables x2, x3. General solution: Ë x2 is free
Í x is free
Í 3
Ì
12. Since the bottom row of the matrix is equivalent to the equation 0 = 1, the system has no solutions.
Ç1
È0
13. È
È0
È
É0
−3
0
−1
0
1
0
0
0
0
1
−4
9
0
0
0
0
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
−2
−3
0
0
9
1
0
0
0
0
1
−4
9
0
0
0
0
x1
x2
Corresponding system:
x4
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
2
− 3 x5
= 5
− 4 x5
=
0
0
0
−3
1
0
0
0
0
1
−4
9
0
0
0
0
5
1Ù
Ù
4Ù
Ù
0Ú
1
+ 9 x5 = 4
0 = 0
Ê x1 = 5 + 3x5
Í x = 1 + 4x
5
ÍÍ 2
Basic variables: x1, x2, x4; free variables: x3, x5. General solution: Ë x3 is free
Íx = 4 − 9x
5
Í 4
ÍÌ x5 is free
Note: The Study Guide discusses the common mistake x3 = 0.
Ç1
È0
14. È
È0
È
É0
0
1
−5
4
0
−1
−8
0
0
0
0
0
0
0
1
0
3 Ç1
6 ÙÙ ÈÈ0
~
0 Ù È0
Ù È
0 Ú É0
−5
4
0
−1
0
0
0
0
0
0
0
0
1
0
− 5 x3
x1
Corresponding system:
0
1
x2
+ 4 x3
3
6 ÙÙ
0Ù
Ù
0Ú
= 3
−
x4
= 6
x5 = 0
0 = 0
Ê x1 = 3 + 5 x3
Íx = 6 − 4x + x
3
4
ÍÍ 2
Basic variables: x1, x2, x5; free variables: x3, x4. General solution: Ë x3 is free
Í x is free
Í 4
ÍÌ x5 = 0
15. a. The system is consistent. There are many solutions because x3 is a free variable.
b. The system is consistent. There are many solutions because x1 is a free variable.
16. a. The system is inconsistent. (The rightmost column of the augmented matrix is a pivot column).
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
• Solutions
11
b. The system is consistent. There are many solutions because x2 is a free variable.
4
Ç 1 −1 4 Ç 1 −1
17. È
The system has a solution for all values of h since the augmented
~È
Ù
3 h Ú É0 1 h + 8ÙÚ
É −2
column cannot be a pivot column.
1 Ç1
1
−3
Ç 1 −3
18. È
If 3h + 6 is zero, that is, if h = –2, then the system has a
~È
Ù
6 −2 Ú É 0 3h + 6 − h − 2 ÙÚ
Éh
solution, because 0 equals 0. When h ≠ −2, the system has a solution since the augmented column
cannot be a pivot column. Thus the system has a solution for all values of h.
2
h
Ç 1 h 2 Ç1
19. È
~È
Ù
Ù
É 4 8 k Ú É 0 8 − 4 h k − 8Ú
a. When h = 2 and k ≠ 8, the augmented column is a pivot column, and the system is inconsistent.
b. When h ≠ 2, the system is consistent and has a unique solution. There are no free variables.
c. When h = 2 and k = 8, the system is consistent and has many solutions.
−3
1
Ç 1 −3 1 Ç 1
20. È
~È
Ù
h k Ú É 0 h + 6 k − 2 ÙÚ
É2
a. When h = –6 and k ≠ 2, the system is inconsistent, because the augmented column is a pivot
column.
b. When h ≠ −6, the system is consistent and has a unique solution. There are no free variables.
c. When h = –6 and k = 2, the system is consistent and has many solutions.
21. a.
b.
c.
d.
e.
False. See Theorem 1.
False. See the second paragraph of the section.
True. Basic variables are defined after equation (4).
True. This statement is at the beginning of Parametric Descriptions of Solution Sets.
False. The row shown corresponds to the equation 5x4 = 0, which does not by itself lead to a
contradiction. So the system might be consistent or it might be inconsistent.
22. a. True. See Theorem 1.
b. False. See Theorem 2.
c. False. See the beginning of the subsection Pivot Positions. The pivot positions in a matrix are
determined completely by the positions of the leading entries in the nonzero rows of any echelon
form obtained from the matrix.
d. True. See the paragraph just before Example 4.
e. False. The existence of at least one solution is not related to the presence or absence of free
variables. If the system is inconsistent, the solution set is empty. See the solution of Practice
Problem 2.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
3 −1 Ç 2
3 −1 Ç 2
3 −1
Ç2
È6
Ù
È
Ù
È
5
0 Ù ~ È 0 −4
3Ù ~ È 0 −4
3ÙÙ
È
ÈÉ 2 −5
7 ÙÚ ÈÉ 0 −8
8ÙÚ ÈÉ 0
0
2ÙÚ
The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in
common.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
Ç2
È0
È
ÈÉ 2
4
1
3
4
−2
0
4 Ç2
−2 ÙÙ ~ ÈÈ 0
0 ÙÚ ÈÉ 0
4
1
−1
4
−2
−4
4 Ç2
−2 ÙÙ ~ ÈÈ 0
−4ÙÚ ÈÉ 0
4
1
0
4
−2
−6
4
−2ÙÙ
−6ÙÚ
The system is consistent, and using the argument from Example 2, there is only one solution. So the
three planes have only one point in common.
4
h
Ç 1 h 4 Ç1
19. È
~È
Ù
Ù Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no
É3 6 8Ú É 0 6 − 3h −4 Ú
solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
h −5 Ç 1
h
−5
Ç1
Write c for −8 − 2h. If c = 0, that is, if h = –4, then the system
20. È
~È
Ù
6 Ú É 0 −8 − 2h 16 ÚÙ
É 2 −8
has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.
−2
4
Ç 1 4 −2 Ç 1
21. È
~È
Ù
Ù Write c for h − 12 . Then the second equation cx2 = 0 has a solution
É3 h −6 Ú É 0 h − 12 0 Ú
for every value of c. So the system is consistent for all h.
Ç −4
h È
~
−3ÙÚ È 0
É
and only if h = 6.
Ç −4
22. È
É 2
12
−6
12
0
h
h
h ÙÙ The system is consistent if and only if −3 + = 0, that is, if
2
−3 +
2Ú
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description applies to a single solution. The solution set consists of all possible
solutions. Only in special cases does the solution set consist of exactly one solution. Mark a
statement True only if the statement is always true.
d. True. See the box before Example 2.
24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
c. False. The definition of equivalent systems is in the second paragraph after equation (2).
d. True. By definition, a consistent system has at least one solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
6
CHAPTER 1
• Linear Equations in Linear Algebra
7 g
7
g
7
g
Ç 1 −4
Ç1 −4
Ç1 −4
È
Ù
È
Ù
È
Ù
h Ù ~ È0
h
3 −5 h Ù ~ È 0
3 −5
3 −5
25. È 0
Ù
ÈÉ −2
5 −9 k ÙÚ ÈÉ0 −3
5 k + 2 g ÙÚ ÈÉ 0
0
0 k + 2 g + h ÙÚ
Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix
above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution
if and only if k + 2g + h = 0.
26. Row reduce the augmented matrix for the given system:
Ç2
Èc
É
4
d
Ç1
~È
Ù
g Ú Éc
f
2
Ç1
~È
Ù
g Ú É0
f /2
d
2
f /2
d − 2c
g − c ( f / 2) ÙÚ
This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f
and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus
d 2c.
27. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible
since a is nonzero. Then replace R2 by R2 + (–c)R1.
Ça
È
Éc
b
d
Ç1
~È
Ù
g Ú Éc
f
Ç1
~È
Ù
g Ú É0
b/a
f /a
d
b/a
f /a
g − c( f / a ) ÚÙ
d − c(b / a )
The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity
g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be
written as ad – bc ≠ 0, or ad ≠ bc.
28. A basic principle of this section is that row operations do not affect the solution set of a linear
system. Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and
then perform any elementary row operations to produce other augmented matrices. Here are three
examples. The fact that they are all row equivalent proves that they all have the solution set (3, –2, –
1).
Ç1
È0
È
ÉÈ0
0
1
0
0
0
1
3 Ç1
−2ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 0
0
1
0
0
0
1
3 Ç1
4ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 2
0
1
0
0
0
1
3
4ÙÙ
5ÚÙ
29. Swap R1 and R3; swap R1 and R3.
30. Multiply R3 by –1/5; multiply R3 by –5.
31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.
32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2.
33. The first equation was given. The others are:
T2 = (T1 + 20 + 40 + T3 )/4,
or
4T2 − T1 − T3 = 60
T3 = (T4 + T2 + 40 + 30)/4,
or
4T3 − T4 − T2 = 70
T4 = (10 + T1 + T3 + 30)/4,
or
4T4 − T1 − T3 = 40
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
7
Rearranging,
4T1
−
T2
−T1
+
4T2
−T2
−
−T1
T4
= 30
− T3
+ 4T3
−
T4
= 60
= 70
−
+ 4T4
= 40
T3
34. Begin by interchanging R1 and R4, then create zeros in the first column:
Ç 4
È −1
È
È 0
È
É −1
−1
0
−1
4
−1
−1
4
0
−1
0
−1
4
Ç −1
Ù
60 È −1
Ù~È
70 Ù È 0
Ù È
40 Ú É 4
0
−1
4
4
−1
−1
4
0
−1
−1
0
−1
30
Ç −1
Ù
60 È 0
Ù~È
70 Ù È 0
Ù È
30 Ú É 0
40
0
−1
4
4
−1
0
4
−4
−1
−1
−4
15
40
20 Ù
Ù
70 Ù
Ù
190 Ú
Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
−1
0
4
−1
−1
−1
−4
15
Ç1
Ù
È
5
0
Ù~È
70 Ù È 0
Ù È
190 Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
−4
14
Ç1
Ù
È
5
0
Ù~È
75Ù È 0
Ù È
195Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
0
12
−40
5Ù
Ù
75Ù
Ù
270 Ú
Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
0
0
4
−1
−2
0
0
1
Ç1
Ù
5 È0
Ù~È
75Ù È0
Ù È
22.5Ú É0
−40
0
1
0
1
0
0
4
0
0
0
0
1
Ç1
Ù
27.5 È 0
Ù~È
120 Ù È0
Ù È
22.5Ú É0
50
0
1
0
1
0
0
1
0
0
0
0
1
50
27.5Ù
Ù
30 Ù
Ù
22.5Ú
The last step is to replace R1 by R1 + (–1)R3:
Ç1
È0
~È
È0
È
É0
0
0
0
1
0
0
1
0
0
0
0
1
20.0
27.5Ù
Ù . The solution is (20, 27.5, 30, 22.5).
30.0 Ù
Ù
22.5Ú
Notes: The Study Guide includes a “Mathematical Note” about statements, “If … , then … .”
This early in the course, students typically use single row operations to reduce a matrix. As a result,
even the small grid for Exercise 34 leads to about 80 multiplications or additions (not counting operations
with zero). This exercise should give students an appreciation for matrix programs such as MATLAB.
Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not
already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in
connection with an LU factorization.
For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB
notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/84+/89
calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section
1.1 describes how to access the data that is available for all numerical exercises in the text. This feature
has the ability to save students time if they regularly have their matrix program at hand when studying
linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale.
These commands are included in the text data sets, available from the text web site,
www.pearsonhighered.com/lay.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
8
CHAPTER 1
1.2
• Linear Equations in Linear Algebra
SOLUTIONS
Notes: The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises
7–14, in preparation for Section 1.5.)
1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c.
2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.
Ç1
3. ÈÈ 2
ÈÉ 3
2
4
6
4
6
9
Ç1
~ ÈÈ0
ÈÉ0
Ç1
È
4. È 2
ÉÈ 4
2
4
5
8 Ç1
8ÙÙ ~ ÈÈ0
12 ÙÚ ÈÉ0
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
4 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
2
0
0
5 Ç1
4 ÙÙ ~ ÈÈ0
2 ÚÙ ÉÈ0
2
0
−3
4
−3
−12
2
1
0
4
4
1
5 Ç1
6 ÙÙ ~ ÈÈ 0
2 ÙÚ ÈÉ0
̈
0 ÙÚ
Ç̈
5. È
É0
* Ç̈
,
̈ÙÚ ÈÉ 0
* Ç0
,
0 ÙÚ ÈÉ 0
Ç1
7. È
É3
3
9
7 Ç1
~
6 ÙÚ ÈÉ 0
4
7
4
−2
−3
4
1
0
2
0
0
4
5
4
2
0
0
3
0
2
1
0
8 Ç1
−8ÙÙ ~ ÈÈ0
−12 ÙÚ ÈÉ0
0
1
0
2
0
0
4
1
−3
−8
4ÙÙ . Pivot cols 1 and 3.
0ÙÚ
5
Ç1
Ù
È
−6 Ù ~ È0
−18ÚÙ ÉÈ0
4
0
1
2
−3
0
5 Ç1
−2 ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
7 Ç1
~
−15ÙÚ ÈÉ0
Corresponding system of equations:
x1
3
0
0
0
1
+ 3 x2
x3
2
4
6
4
6
9
1
Pivot cols
−2 ÙÙ .
1, 2, and 3
2 ÙÚ
* Ç̈
̈ÙÙ , ÈÈ 0
0 ÚÙ ÉÈ 0
4
1
Ç1
È2
È
ÈÉ 3
5
Ç1
Ù
È
− 18Ù ~ È 0
−6 ÚÙ ÉÈ 0
4
− 12
−3
0
1
0
Ç̈
6. ÈÈ 0
ÉÈ 0
4
−5
8
4ÙÙ
−12ÙÚ
* Ç0
0 ÙÙ , ÈÈ0
0ÚÙ ÉÈ0
7 Ç1
~
3ÙÚ ÈÉ 0
8
8ÙÙ
12 ÙÚ
2
1
0
4
4
−3
Ç1
È2
È
ÈÉ 4
2
4
5
5
6ÙÙ
−6ÚÙ
4
5
4
5
4 ÙÙ
2 ÙÚ
̈
0 ÙÙ
0 ÙÚ
3
0
0
1
−5
3ÙÚ
= −5
=
3
The basic variables (corresponding to the pivot positions) are x1 and x3. The remaining variable x2 is
free. Solve for the basic variables in terms of the free variable. The general solution is
Ê x1 = −5 − 3 x2
Í
Ë x2 is free
Íx = 3
Ì 3
Note: Exercise 7 is paired with Exercise 10.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
8.
Ç 1
È −3
É
Ç 1 −3 0 −5 Ç 1
~È
~
Ù
−2 0 −6 Ú É 0
7 0
1 0
3ÙÚ ÈÉ 0
x
= 4
Corresponding system of equations: 1
x2
= 3
−3
0
Ç1
~È
Ù
9Ú É0
−5
−3
0
−5
0
0
1
0
• Solutions
9
4
3ÙÚ
The basic variables (corresponding to the pivot positions) are x1 and x2. The remaining variable x3 is
free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic
variables do not depend on the value of the free variable.
Ê x1 = 4
Í
General solution: Ë x2 = 3
Í x is free
Ì 3
Note: A common error in Exercise 8 is to assume that x3 is zero. To avoid this, identify the basic
variables first. Any remaining variables are free. (This type of computation will arise in Chapter 5.)
Ç0
9. È
É1
1
−2
−3
4
Ç1
~È
Ù
−6 Ú É 0
3
Corresponding system:
−3
4
1
−2
x1
x2
Ç1
~È
Ù
3Ú É 0
−6
− 2 x3
= 3
− 2 x3
= 3
0
−2
1
−2
3
3ÙÚ
Ê x1 = 3 + 2 x3
Í
Basic variables: x1, x2; free variable: x3. General solution: Ë x2 = 3 + 2 x3
Í x is free
Ì 3
Ç 1
10. È
É −2
−2
−1
4
−5
Ç1
~È
Ù
6 Ú É0
4
Corresponding system:
x1
−2
−1
0
−7
Ç1
~È
Ù
14 Ú É 0
− 2 x2
= 2
4
x3
−2
0
0
1
2
−2 ÙÚ
= −2
Ê x1 = 2 + 2 x2
Í
Basic variables: x1, x3; free variable: x2. General solution: Ë x2 is free
Í x = −2
Ì 3
Ç3
È
11. È9
ÈÉ6
−2
−6
−4
4
12
8
0 Ç3
0 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
−2
0
0
x1
Corresponding system:
−
4
0
0
0 Ç1
0ÙÙ ~ ÈÈ 0
0ÙÚ ÈÉ 0
2
x2
3
+
4
x3
3
0
0
−2 3
0
0
43
0
0
0
0ÙÙ
0ÙÚ
= 0
= 0
= 0
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
10
CHAPTER 1
• Linear Equations in Linear Algebra
2
4
Ê
Í x1 = 3 x2 − 3 x3
Í
Basic variable: x1; free variables x2, x3. General solution: Ë x2 is free
Í x is free
Í 3
Ì
12. Since the bottom row of the matrix is equivalent to the equation 0 = 1, the system has no solutions.
Ç1
È0
13. È
È0
È
É0
−3
0
−1
0
1
0
0
0
0
1
−4
9
0
0
0
0
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
−2
−3
0
0
9
1
0
0
0
0
1
−4
9
0
0
0
0
x1
x2
Corresponding system:
x4
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
2
− 3 x5
= 5
− 4 x5
=
0
0
0
−3
1
0
0
0
0
1
−4
9
0
0
0
0
5
1Ù
Ù
4Ù
Ù
0Ú
1
+ 9 x5 = 4
0 = 0
Ê x1 = 5 + 3x5
Í x = 1 + 4x
5
ÍÍ 2
Basic variables: x1, x2, x4; free variables: x3, x5. General solution: Ë x3 is free
Íx = 4 − 9x
5
Í 4
ÍÌ x5 is free
Note: The Study Guide discusses the common mistake x3 = 0.
Ç1
È0
14. È
È0
È
É0
0
1
−5
4
0
−1
−8
0
0
0
0
0
0
0
1
0
3 Ç1
6 ÙÙ ÈÈ0
~
0 Ù È0
Ù È
0 Ú É0
−5
4
0
−1
0
0
0
0
0
0
0
0
1
0
− 5 x3
x1
Corresponding system:
0
1
x2
+ 4 x3
3
6 ÙÙ
0Ù
Ù
0Ú
= 3
−
x4
= 6
x5 = 0
0 = 0
Ê x1 = 3 + 5 x3
Íx = 6 − 4x + x
3
4
ÍÍ 2
Basic variables: x1, x2, x5; free variables: x3, x4. General solution: Ë x3 is free
Í x is free
Í 4
ÍÌ x5 = 0
15. a. The system is consistent. There are many solutions because x3 is a free variable.
b. The system is consistent. There are many solutions because x1 is a free variable.
16. a. The system is inconsistent. (The rightmost column of the augmented matrix is a pivot column).
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
• Solutions
11
b. The system is consistent. There are many solutions because x2 is a free variable.
4
Ç 1 −1 4 Ç 1 −1
17. È
The system has a solution for all values of h since the augmented
~È
Ù
3 h Ú É0 1 h + 8ÙÚ
É −2
column cannot be a pivot column.
1 Ç1
1
−3
Ç 1 −3
18. È
If 3h + 6 is zero, that is, if h = –2, then the system has a
~È
Ù
6 −2 Ú É 0 3h + 6 − h − 2 ÙÚ
Éh
solution, because 0 equals 0. When h ≠ −2, the system has a solution since the augmented column
cannot be a pivot column. Thus the system has a solution for all values of h.
2
h
Ç 1 h 2 Ç1
19. È
~È
Ù
Ù
É 4 8 k Ú É 0 8 − 4 h k − 8Ú
a. When h = 2 and k ≠ 8, the augmented column is a pivot column, and the system is inconsistent.
b. When h ≠ 2, the system is consistent and has a unique solution. There are no free variables.
c. When h = 2 and k = 8, the system is consistent and has many solutions.
−3
1
Ç 1 −3 1 Ç 1
20. È
~È
Ù
h k Ú É 0 h + 6 k − 2 ÙÚ
É2
a. When h = –6 and k ≠ 2, the system is inconsistent, because the augmented column is a pivot
column.
b. When h ≠ −6, the system is consistent and has a unique solution. There are no free variables.
c. When h = –6 and k = 2, the system is consistent and has many solutions.
21. a.
b.
c.
d.
e.
False. See Theorem 1.
False. See the second paragraph of the section.
True. Basic variables are defined after equation (4).
True. This statement is at the beginning of Parametric Descriptions of Solution Sets.
False. The row shown corresponds to the equation 5x4 = 0, which does not by itself lead to a
contradiction. So the system might be consistent or it might be inconsistent.
22. a. True. See Theorem 1.
b. False. See Theorem 2.
c. False. See the beginning of the subsection Pivot Positions. The pivot positions in a matrix are
determined completely by the positions of the leading entries in the nonzero rows of any echelon
form obtained from the matrix.
d. True. See the paragraph just before Example 4.
e. False. The existence of at least one solution is not related to the presence or absence of free
variables. If the system is inconsistent, the solution set is empty. See the solution of Practice
Problem 2.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
12
CHAPTER 1
• Linear Equations in Linear Algebra
23. Since there are four pivots (one in each row), the augmented matrix must reduce to the form
Ç1
È0
È
È0
È
É0
0
1
0
0
0
0
0
0
1
0
0
1
x1
a
Ù
bÙ
and so
cÙ
Ù
dÚ
x2
x3
x4
=
=
=
a
b
c
=
d
No matter what the values of a, b, c, and d, the solution exists and is unique.
24. The system is consistent because there is not a pivot in column 5, which means that there is not a row
of the form [0 0 0 0 1]. Since the matrix is the augmented matrix for a system, Theorem 2 shows
that the system has a solution.
25. If the coefficient matrix has a pivot position in every row, then there is a pivot position in the bottom
row, and there is no room for a pivot in the augmented column. So, the system is consistent, by
Theorem 2.
26. Since the coefficient matrix has three pivot columns, there is a pivot in each row of the coefficient
matrix. Thus the augmented matrix will not have a row of the form [0 0 0 0 0 1], and the
system is consistent.
27. “If a linear system is consistent, then the solution is unique if and only if every column in the
coefficient matrix is a pivot column; otherwise there are infinitely many solutions. ”
This statement is true because the free variables correspond to nonpivot columns of the coefficient
matrix. The columns are all pivot columns if and only if there are no free variables. And there are no
free variables if and only if the solution is unique, by Theorem 2.
28. Every column in the augmented matrix except the rightmost column is a pivot column, and the
rightmost column is not a pivot column.
29. An underdetermined system always has more variables than equations. There cannot be more basic
variables than there are equations, so there must be at least one free variable. Such a variable may be
assigned infinitely many different values. If the system is consistent, each different value of a free
variable will produce a different solution, and the system will not have a unique solution. If the
system is inconsistent, it will not have any solution.
30. Example:
x1
2 x1
+
x2
+ 2 x2
+
x3
+ 2 x3
= 4
= 5
31. Yes, a system of linear equations with more equations than unknowns can be consistent.
x1 +
x2 = 2
Example (in which x1 = x2 = 1): x1 −
x2 = 0
3 x1 + 2 x2 = 5
32. According to the numerical note in Section 1.2, when n = 20 the reduction to echelon form takes
about 2(20)3/3 5,333 flops, while further reduction to reduced echelon form needs at most (20)2 =
400 flops. Of the total flops, the “backward phase” is about 400/5733 = .07 or about 7%. When n =
200, the estimates are 2(200)3/3 5,333,333 flops for the reduction to echelon form and (200)2 =
40,000 flops for the backward phase. The fraction associated with the backward phase is about
(4×104) /(5.3×106) = .007, or about .7%.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
3 −1 Ç 2
3 −1 Ç 2
3 −1
Ç2
È6
Ù
È
Ù
È
5
0 Ù ~ È 0 −4
3Ù ~ È 0 −4
3ÙÙ
È
ÈÉ 2 −5
7 ÙÚ ÈÉ 0 −8
8ÙÚ ÈÉ 0
0
2ÙÚ
The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in
common.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
Ç2
È0
È
ÈÉ 2
4
1
3
4
−2
0
4 Ç2
−2 ÙÙ ~ ÈÈ 0
0 ÙÚ ÈÉ 0
4
1
−1
4
−2
−4
4 Ç2
−2 ÙÙ ~ ÈÈ 0
−4ÙÚ ÈÉ 0
4
1
0
4
−2
−6
4
−2ÙÙ
−6ÙÚ
The system is consistent, and using the argument from Example 2, there is only one solution. So the
three planes have only one point in common.
4
h
Ç 1 h 4 Ç1
19. È
~È
Ù
Ù Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no
É3 6 8Ú É 0 6 − 3h −4 Ú
solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
h −5 Ç 1
h
−5
Ç1
Write c for −8 − 2h. If c = 0, that is, if h = –4, then the system
20. È
~È
Ù
6 Ú É 0 −8 − 2h 16 ÚÙ
É 2 −8
has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.
−2
4
Ç 1 4 −2 Ç 1
21. È
~È
Ù
Ù Write c for h − 12 . Then the second equation cx2 = 0 has a solution
É3 h −6 Ú É 0 h − 12 0 Ú
for every value of c. So the system is consistent for all h.
Ç −4
h È
~
−3ÙÚ È 0
É
and only if h = 6.
Ç −4
22. È
É 2
12
−6
12
0
h
h
h ÙÙ The system is consistent if and only if −3 + = 0, that is, if
2
−3 +
2Ú
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description applies to a single solution. The solution set consists of all possible
solutions. Only in special cases does the solution set consist of exactly one solution. Mark a
statement True only if the statement is always true.
d. True. See the box before Example 2.
24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
c. False. The definition of equivalent systems is in the second paragraph after equation (2).
d. True. By definition, a consistent system has at least one solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
6
CHAPTER 1
• Linear Equations in Linear Algebra
7 g
7
g
7
g
Ç 1 −4
Ç1 −4
Ç1 −4
È
Ù
È
Ù
È
Ù
h Ù ~ È0
h
3 −5 h Ù ~ È 0
3 −5
3 −5
25. È 0
Ù
ÈÉ −2
5 −9 k ÙÚ ÈÉ0 −3
5 k + 2 g ÙÚ ÈÉ 0
0
0 k + 2 g + h ÙÚ
Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix
above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution
if and only if k + 2g + h = 0.
26. Row reduce the augmented matrix for the given system:
Ç2
Èc
É
4
d
Ç1
~È
Ù
g Ú Éc
f
2
Ç1
~È
Ù
g Ú É0
f /2
d
2
f /2
d − 2c
g − c ( f / 2) ÙÚ
This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f
and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus
d 2c.
27. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible
since a is nonzero. Then replace R2 by R2 + (–c)R1.
Ça
È
Éc
b
d
Ç1
~È
Ù
g Ú Éc
f
Ç1
~È
Ù
g Ú É0
b/a
f /a
d
b/a
f /a
g − c( f / a ) ÚÙ
d − c(b / a )
The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity
g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be
written as ad – bc ≠ 0, or ad ≠ bc.
28. A basic principle of this section is that row operations do not affect the solution set of a linear
system. Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and
then perform any elementary row operations to produce other augmented matrices. Here are three
examples. The fact that they are all row equivalent proves that they all have the solution set (3, –2, –
1).
Ç1
È0
È
ÉÈ0
0
1
0
0
0
1
3 Ç1
−2ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 0
0
1
0
0
0
1
3 Ç1
4ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 2
0
1
0
0
0
1
3
4ÙÙ
5ÚÙ
29. Swap R1 and R3; swap R1 and R3.
30. Multiply R3 by –1/5; multiply R3 by –5.
31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.
32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2.
33. The first equation was given. The others are:
T2 = (T1 + 20 + 40 + T3 )/4,
or
4T2 − T1 − T3 = 60
T3 = (T4 + T2 + 40 + 30)/4,
or
4T3 − T4 − T2 = 70
T4 = (10 + T1 + T3 + 30)/4,
or
4T4 − T1 − T3 = 40
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
7
Rearranging,
4T1
−
T2
−T1
+
4T2
−T2
−
−T1
T4
= 30
− T3
+ 4T3
−
T4
= 60
= 70
−
+ 4T4
= 40
T3
34. Begin by interchanging R1 and R4, then create zeros in the first column:
Ç 4
È −1
È
È 0
È
É −1
−1
0
−1
4
−1
−1
4
0
−1
0
−1
4
Ç −1
Ù
60 È −1
Ù~È
70 Ù È 0
Ù È
40 Ú É 4
0
−1
4
4
−1
−1
4
0
−1
−1
0
−1
30
Ç −1
Ù
60 È 0
Ù~È
70 Ù È 0
Ù È
30 Ú É 0
40
0
−1
4
4
−1
0
4
−4
−1
−1
−4
15
40
20 Ù
Ù
70 Ù
Ù
190 Ú
Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
−1
0
4
−1
−1
−1
−4
15
Ç1
Ù
È
5
0
Ù~È
70 Ù È 0
Ù È
190 Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
−4
14
Ç1
Ù
È
5
0
Ù~È
75Ù È 0
Ù È
195Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
0
12
−40
5Ù
Ù
75Ù
Ù
270 Ú
Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
0
0
4
−1
−2
0
0
1
Ç1
Ù
5 È0
Ù~È
75Ù È0
Ù È
22.5Ú É0
−40
0
1
0
1
0
0
4
0
0
0
0
1
Ç1
Ù
27.5 È 0
Ù~È
120 Ù È0
Ù È
22.5Ú É0
50
0
1
0
1
0
0
1
0
0
0
0
1
50
27.5Ù
Ù
30 Ù
Ù
22.5Ú
The last step is to replace R1 by R1 + (–1)R3:
Ç1
È0
~È
È0
È
É0
0
0
0
1
0
0
1
0
0
0
0
1
20.0
27.5Ù
Ù . The solution is (20, 27.5, 30, 22.5).
30.0 Ù
Ù
22.5Ú
Notes: The Study Guide includes a “Mathematical Note” about statements, “If … , then … .”
This early in the course, students typically use single row operations to reduce a matrix. As a result,
even the small grid for Exercise 34 leads to about 80 multiplications or additions (not counting operations
with zero). This exercise should give students an appreciation for matrix programs such as MATLAB.
Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not
already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in
connection with an LU factorization.
For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB
notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/84+/89
calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section
1.1 describes how to access the data that is available for all numerical exercises in the text. This feature
has the ability to save students time if they regularly have their matrix program at hand when studying
linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale.
These commands are included in the text data sets, available from the text web site,
www.pearsonhighered.com/lay.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
8
CHAPTER 1
1.2
• Linear Equations in Linear Algebra
SOLUTIONS
Notes: The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises
7–14, in preparation for Section 1.5.)
1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c.
2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.
Ç1
3. ÈÈ 2
ÈÉ 3
2
4
6
4
6
9
Ç1
~ ÈÈ0
ÈÉ0
Ç1
È
4. È 2
ÉÈ 4
2
4
5
8 Ç1
8ÙÙ ~ ÈÈ0
12 ÙÚ ÈÉ0
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
4 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
2
0
0
5 Ç1
4 ÙÙ ~ ÈÈ0
2 ÚÙ ÉÈ0
2
0
−3
4
−3
−12
2
1
0
4
4
1
5 Ç1
6 ÙÙ ~ ÈÈ 0
2 ÙÚ ÈÉ0
̈
0 ÙÚ
Ç̈
5. È
É0
* Ç̈
,
̈ÙÚ ÈÉ 0
* Ç0
,
0 ÙÚ ÈÉ 0
Ç1
7. È
É3
3
9
7 Ç1
~
6 ÙÚ ÈÉ 0
4
7
4
−2
−3
4
1
0
2
0
0
4
5
4
2
0
0
3
0
2
1
0
8 Ç1
−8ÙÙ ~ ÈÈ0
−12 ÙÚ ÈÉ0
0
1
0
2
0
0
4
1
−3
−8
4ÙÙ . Pivot cols 1 and 3.
0ÙÚ
5
Ç1
Ù
È
−6 Ù ~ È0
−18ÚÙ ÉÈ0
4
0
1
2
−3
0
5 Ç1
−2 ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
7 Ç1
~
−15ÙÚ ÈÉ0
Corresponding system of equations:
x1
3
0
0
0
1
+ 3 x2
x3
2
4
6
4
6
9
1
Pivot cols
−2 ÙÙ .
1, 2, and 3
2 ÙÚ
* Ç̈
̈ÙÙ , ÈÈ 0
0 ÚÙ ÉÈ 0
4
1
Ç1
È2
È
ÈÉ 3
5
Ç1
Ù
È
− 18Ù ~ È 0
−6 ÚÙ ÉÈ 0
4
− 12
−3
0
1
0
Ç̈
6. ÈÈ 0
ÉÈ 0
4
−5
8
4ÙÙ
−12ÙÚ
* Ç0
0 ÙÙ , ÈÈ0
0ÚÙ ÉÈ0
7 Ç1
~
3ÙÚ ÈÉ 0
8
8ÙÙ
12 ÙÚ
2
1
0
4
4
−3
Ç1
È2
È
ÈÉ 4
2
4
5
5
6ÙÙ
−6ÚÙ
4
5
4
5
4 ÙÙ
2 ÙÚ
̈
0 ÙÙ
0 ÙÚ
3
0
0
1
−5
3ÙÚ
= −5
=
3
The basic variables (corresponding to the pivot positions) are x1 and x3. The remaining variable x2 is
free. Solve for the basic variables in terms of the free variable. The general solution is
Ê x1 = −5 − 3 x2
Í
Ë x2 is free
Íx = 3
Ì 3
Note: Exercise 7 is paired with Exercise 10.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
8.
Ç 1
È −3
É
Ç 1 −3 0 −5 Ç 1
~È
~
Ù
−2 0 −6 Ú É 0
7 0
1 0
3ÙÚ ÈÉ 0
x
= 4
Corresponding system of equations: 1
x2
= 3
−3
0
Ç1
~È
Ù
9Ú É0
−5
−3
0
−5
0
0
1
0
• Solutions
9
4
3ÙÚ
The basic variables (corresponding to the pivot positions) are x1 and x2. The remaining variable x3 is
free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic
variables do not depend on the value of the free variable.
Ê x1 = 4
Í
General solution: Ë x2 = 3
Í x is free
Ì 3
Note: A common error in Exercise 8 is to assume that x3 is zero. To avoid this, identify the basic
variables first. Any remaining variables are free. (This type of computation will arise in Chapter 5.)
Ç0
9. È
É1
1
−2
−3
4
Ç1
~È
Ù
−6 Ú É 0
3
Corresponding system:
−3
4
1
−2
x1
x2
Ç1
~È
Ù
3Ú É 0
−6
− 2 x3
= 3
− 2 x3
= 3
0
−2
1
−2
3
3ÙÚ
Ê x1 = 3 + 2 x3
Í
Basic variables: x1, x2; free variable: x3. General solution: Ë x2 = 3 + 2 x3
Í x is free
Ì 3
Ç 1
10. È
É −2
−2
−1
4
−5
Ç1
~È
Ù
6 Ú É0
4
Corresponding system:
x1
−2
−1
0
−7
Ç1
~È
Ù
14 Ú É 0
− 2 x2
= 2
4
x3
−2
0
0
1
2
−2 ÙÚ
= −2
Ê x1 = 2 + 2 x2
Í
Basic variables: x1, x3; free variable: x2. General solution: Ë x2 is free
Í x = −2
Ì 3
Ç3
È
11. È9
ÈÉ6
−2
−6
−4
4
12
8
0 Ç3
0 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
−2
0
0
x1
Corresponding system:
−
4
0
0
0 Ç1
0ÙÙ ~ ÈÈ 0
0ÙÚ ÈÉ 0
2
x2
3
+
4
x3
3
0
0
−2 3
0
0
43
0
0
0
0ÙÙ
0ÙÚ
= 0
= 0
= 0
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
10
CHAPTER 1
• Linear Equations in Linear Algebra
2
4
Ê
Í x1 = 3 x2 − 3 x3
Í
Basic variable: x1; free variables x2, x3. General solution: Ë x2 is free
Í x is free
Í 3
Ì
12. Since the bottom row of the matrix is equivalent to the equation 0 = 1, the system has no solutions.
Ç1
È0
13. È
È0
È
É0
−3
0
−1
0
1
0
0
0
0
1
−4
9
0
0
0
0
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
−2
−3
0
0
9
1
0
0
0
0
1
−4
9
0
0
0
0
x1
x2
Corresponding system:
x4
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
2
− 3 x5
= 5
− 4 x5
=
0
0
0
−3
1
0
0
0
0
1
−4
9
0
0
0
0
5
1Ù
Ù
4Ù
Ù
0Ú
1
+ 9 x5 = 4
0 = 0
Ê x1 = 5 + 3x5
Í x = 1 + 4x
5
ÍÍ 2
Basic variables: x1, x2, x4; free variables: x3, x5. General solution: Ë x3 is free
Íx = 4 − 9x
5
Í 4
ÍÌ x5 is free
Note: The Study Guide discusses the common mistake x3 = 0.
Ç1
È0
14. È
È0
È
É0
0
1
−5
4
0
−1
−8
0
0
0
0
0
0
0
1
0
3 Ç1
6 ÙÙ ÈÈ0
~
0 Ù È0
Ù È
0 Ú É0
−5
4
0
−1
0
0
0
0
0
0
0
0
1
0
− 5 x3
x1
Corresponding system:
0
1
x2
+ 4 x3
3
6 ÙÙ
0Ù
Ù
0Ú
= 3
−
x4
= 6
x5 = 0
0 = 0
Ê x1 = 3 + 5 x3
Íx = 6 − 4x + x
3
4
ÍÍ 2
Basic variables: x1, x2, x5; free variables: x3, x4. General solution: Ë x3 is free
Í x is free
Í 4
ÍÌ x5 = 0
15. a. The system is consistent. There are many solutions because x3 is a free variable.
b. The system is consistent. There are many solutions because x1 is a free variable.
16. a. The system is inconsistent. (The rightmost column of the augmented matrix is a pivot column).
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
• Solutions
11
b. The system is consistent. There are many solutions because x2 is a free variable.
4
Ç 1 −1 4 Ç 1 −1
17. È
The system has a solution for all values of h since the augmented
~È
Ù
3 h Ú É0 1 h + 8ÙÚ
É −2
column cannot be a pivot column.
1 Ç1
1
−3
Ç 1 −3
18. È
If 3h + 6 is zero, that is, if h = –2, then the system has a
~È
Ù
6 −2 Ú É 0 3h + 6 − h − 2 ÙÚ
Éh
solution, because 0 equals 0. When h ≠ −2, the system has a solution since the augmented column
cannot be a pivot column. Thus the system has a solution for all values of h.
2
h
Ç 1 h 2 Ç1
19. È
~È
Ù
Ù
É 4 8 k Ú É 0 8 − 4 h k − 8Ú
a. When h = 2 and k ≠ 8, the augmented column is a pivot column, and the system is inconsistent.
b. When h ≠ 2, the system is consistent and has a unique solution. There are no free variables.
c. When h = 2 and k = 8, the system is consistent and has many solutions.
−3
1
Ç 1 −3 1 Ç 1
20. È
~È
Ù
h k Ú É 0 h + 6 k − 2 ÙÚ
É2
a. When h = –6 and k ≠ 2, the system is inconsistent, because the augmented column is a pivot
column.
b. When h ≠ −6, the system is consistent and has a unique solution. There are no free variables.
c. When h = –6 and k = 2, the system is consistent and has many solutions.
21. a.
b.
c.
d.
e.
False. See Theorem 1.
False. See the second paragraph of the section.
True. Basic variables are defined after equation (4).
True. This statement is at the beginning of Parametric Descriptions of Solution Sets.
False. The row shown corresponds to the equation 5x4 = 0, which does not by itself lead to a
contradiction. So the system might be consistent or it might be inconsistent.
22. a. True. See Theorem 1.
b. False. See Theorem 2.
c. False. See the beginning of the subsection Pivot Positions. The pivot positions in a matrix are
determined completely by the positions of the leading entries in the nonzero rows of any echelon
form obtained from the matrix.
d. True. See the paragraph just before Example 4.
e. False. The existence of at least one solution is not related to the presence or absence of free
variables. If the system is inconsistent, the solution set is empty. See the solution of Practice
Problem 2.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
12
CHAPTER 1
• Linear Equations in Linear Algebra
23. Since there are four pivots (one in each row), the augmented matrix must reduce to the form
Ç1
È0
È
È0
È
É0
0
1
0
0
0
0
0
0
1
0
0
1
x1
a
Ù
bÙ
and so
cÙ
Ù
dÚ
x2
x3
x4
=
=
=
a
b
c
=
d
No matter what the values of a, b, c, and d, the solution exists and is unique.
24. The system is consistent because there is not a pivot in column 5, which means that there is not a row
of the form [0 0 0 0 1]. Since the matrix is the augmented matrix for a system, Theorem 2 shows
that the system has a solution.
25. If the coefficient matrix has a pivot position in every row, then there is a pivot position in the bottom
row, and there is no room for a pivot in the augmented column. So, the system is consistent, by
Theorem 2.
26. Since the coefficient matrix has three pivot columns, there is a pivot in each row of the coefficient
matrix. Thus the augmented matrix will not have a row of the form [0 0 0 0 0 1], and the
system is consistent.
27. “If a linear system is consistent, then the solution is unique if and only if every column in the
coefficient matrix is a pivot column; otherwise there are infinitely many solutions. ”
This statement is true because the free variables correspond to nonpivot columns of the coefficient
matrix. The columns are all pivot columns if and only if there are no free variables. And there are no
free variables if and only if the solution is unique, by Theorem 2.
28. Every column in the augmented matrix except the rightmost column is a pivot column, and the
rightmost column is not a pivot column.
29. An underdetermined system always has more variables than equations. There cannot be more basic
variables than there are equations, so there must be at least one free variable. Such a variable may be
assigned infinitely many different values. If the system is consistent, each different value of a free
variable will produce a different solution, and the system will not have a unique solution. If the
system is inconsistent, it will not have any solution.
30. Example:
x1
2 x1
+
x2
+ 2 x2
+
x3
+ 2 x3
= 4
= 5
31. Yes, a system of linear equations with more equations than unknowns can be consistent.
x1 +
x2 = 2
Example (in which x1 = x2 = 1): x1 −
x2 = 0
3 x1 + 2 x2 = 5
32. According to the numerical note in Section 1.2, when n = 20 the reduction to echelon form takes
about 2(20)3/3 5,333 flops, while further reduction to reduced echelon form needs at most (20)2 =
400 flops. Of the total flops, the “backward phase” is about 400/5733 = .07 or about 7%. When n =
200, the estimates are 2(200)3/3 5,333,333 flops for the reduction to echelon form and (200)2 =
40,000 flops for the backward phase. The fraction associated with the backward phase is about
(4×104) /(5.3×106) = .007, or about .7%.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
• Solutions
13
33. For a quadratic polynomial p(t) = a0 + a1t + a2t2 to exactly fit the data (1, 6), (2, 15), and (3, 28), the
coefficients a0, a1, a2 must satisfy the systems of equations given in the text. Row reduce the
augmented matrix:
Ç1
È1
È
ÈÉ1
1
2
3
Ç1
~ ÈÈ0
ÉÈ0
6 Ç1
15ÙÙ ~ ÈÈ0
28ÚÙ ÉÈ0
1
4
9
1
1
0
0
0
1
1
1
2
4 Ç1
3ÙÙ ~ ÈÈ0
2 ÚÙ ÉÈ0
6 Ç1
9ÙÙ ~ ÈÈ0
22ÚÙ ÉÈ0
1
3
8
0
1
0
0
0
1
1
1
0
1
3
2
6 Ç1
9ÙÙ ~ ÈÈ 0
4ÚÙ ÉÈ 0
1
1
0
1
3
1
6
9ÙÙ
2ÚÙ
1
3ÙÙ
2 ÙÚ
The polynomial is p(t) = 1 + 3t + 2t2.
34. [M] The system of equations to be solved is:
a0
+
a1 ⋅ 0
+
a2 ⋅ 02
+
a3 ⋅ 03
+
a4 ⋅ 0 4
+
a5 ⋅ 05
=
a0
+
a1 ⋅ 2
+
a2 ⋅ 22
+
a3 ⋅ 23
+
a4 ⋅ 2 4
+
a5 ⋅ 25
= 2.90
2
+
3
a3 ⋅ 4
+
a4 ⋅ 4
4
+
a5 ⋅ 4
5
= 14.8
+
a3 ⋅ 63
+
a4 ⋅ 6 4
+
a5 ⋅ 65
= 39.6
+
3
+
4
+
5
= 74.3
a0
+
a1 ⋅ 4
+
a2 ⋅ 4
a0
+
a1 ⋅ 6
+
a2 ⋅ 62
+
2
a0
a0
a1 ⋅ 8
+
a2 ⋅ 8
+ a1 ⋅ 10 + a2 ⋅ 10
2
a3 ⋅ 8
+ a3 ⋅ 10
3
a4 ⋅ 8
+ a4 ⋅ 10
4
a5 ⋅ 8
5
+ a5 ⋅ 10
=
0
119
The unknowns are a0, a1, …, a5. Use technology to compute the reduced echelon of the augmented
matrix:
Ç1 0
È1 2
È
È1 4
È
È1 6
È1 8
È
ÉÈ1 10
Ç1
È0
È
È0
~È
È0
È0
È
ÈÉ 0
0
4
0
8
0
16
0
32
16
64
256
1024
36
216
1296
7776
64
512
4096
32768
2
10
3
4
105
0
0
0
0
0
2
4
8
16
32
0
0
8
0
48
48
224
576
960
4800
0
0
192
2688
26880
0
0
480
7680
90240
10
10
0 Ç1
2.9 Ù È 0
Ù È
14.8Ù È 0
Ù~È
39.6 Ù È 0
74.3Ù È 0
Ù È
119 ÚÙ ÈÉ 0
Ç1
Ù
2.9 È 0
Ù È
9Ù È0
Ù~È
3.9 Ù È 0
8.7 Ù È 0
Ù È
14.5ÙÚ ÈÉ 0
0
0
2
0
0
4
8
0
8
48
0
16
224
0
32
960
0
0
0
24
48
80
192
480
960
1248
4032
9920
7680
32640
99840
0
0
0
0
0
2
4
8
16
32
0
0
8
0
48
48
224
576
960
4800
0
0
0
384
7680
0
0
0
1920
42240
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
0
2.9 Ù
Ù
9Ù
Ù
30.9 Ù
62.7 Ù
Ù
104.5ÙÚ
0
2.9 Ù
Ù
9Ù
Ù
3.9 Ù
−6.9 Ù
Ù
−24.5ÚÙ
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
3 −1 Ç 2
3 −1 Ç 2
3 −1
Ç2
È6
Ù
È
Ù
È
5
0 Ù ~ È 0 −4
3Ù ~ È 0 −4
3ÙÙ
È
ÈÉ 2 −5
7 ÙÚ ÈÉ 0 −8
8ÙÚ ÈÉ 0
0
2ÙÚ
The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in
common.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
Ç2
È0
È
ÈÉ 2
4
1
3
4
−2
0
4 Ç2
−2 ÙÙ ~ ÈÈ 0
0 ÙÚ ÈÉ 0
4
1
−1
4
−2
−4
4 Ç2
−2 ÙÙ ~ ÈÈ 0
−4ÙÚ ÈÉ 0
4
1
0
4
−2
−6
4
−2ÙÙ
−6ÙÚ
The system is consistent, and using the argument from Example 2, there is only one solution. So the
three planes have only one point in common.
4
h
Ç 1 h 4 Ç1
19. È
~È
Ù
Ù Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no
É3 6 8Ú É 0 6 − 3h −4 Ú
solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
h −5 Ç 1
h
−5
Ç1
Write c for −8 − 2h. If c = 0, that is, if h = –4, then the system
20. È
~È
Ù
6 Ú É 0 −8 − 2h 16 ÚÙ
É 2 −8
has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.
−2
4
Ç 1 4 −2 Ç 1
21. È
~È
Ù
Ù Write c for h − 12 . Then the second equation cx2 = 0 has a solution
É3 h −6 Ú É 0 h − 12 0 Ú
for every value of c. So the system is consistent for all h.
Ç −4
h È
~
−3ÙÚ È 0
É
and only if h = 6.
Ç −4
22. È
É 2
12
−6
12
0
h
h
h ÙÙ The system is consistent if and only if −3 + = 0, that is, if
2
−3 +
2Ú
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description applies to a single solution. The solution set consists of all possible
solutions. Only in special cases does the solution set consist of exactly one solution. Mark a
statement True only if the statement is always true.
d. True. See the box before Example 2.
24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
c. False. The definition of equivalent systems is in the second paragraph after equation (2).
d. True. By definition, a consistent system has at least one solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
6
CHAPTER 1
• Linear Equations in Linear Algebra
7 g
7
g
7
g
Ç 1 −4
Ç1 −4
Ç1 −4
È
Ù
È
Ù
È
Ù
h Ù ~ È0
h
3 −5 h Ù ~ È 0
3 −5
3 −5
25. È 0
Ù
ÈÉ −2
5 −9 k ÙÚ ÈÉ0 −3
5 k + 2 g ÙÚ ÈÉ 0
0
0 k + 2 g + h ÙÚ
Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix
above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution
if and only if k + 2g + h = 0.
26. Row reduce the augmented matrix for the given system:
Ç2
Èc
É
4
d
Ç1
~È
Ù
g Ú Éc
f
2
Ç1
~È
Ù
g Ú É0
f /2
d
2
f /2
d − 2c
g − c ( f / 2) ÙÚ
This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f
and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus
d 2c.
27. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible
since a is nonzero. Then replace R2 by R2 + (–c)R1.
Ça
È
Éc
b
d
Ç1
~È
Ù
g Ú Éc
f
Ç1
~È
Ù
g Ú É0
b/a
f /a
d
b/a
f /a
g − c( f / a ) ÚÙ
d − c(b / a )
The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity
g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be
written as ad – bc ≠ 0, or ad ≠ bc.
28. A basic principle of this section is that row operations do not affect the solution set of a linear
system. Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and
then perform any elementary row operations to produce other augmented matrices. Here are three
examples. The fact that they are all row equivalent proves that they all have the solution set (3, –2, –
1).
Ç1
È0
È
ÉÈ0
0
1
0
0
0
1
3 Ç1
−2ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 0
0
1
0
0
0
1
3 Ç1
4ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 2
0
1
0
0
0
1
3
4ÙÙ
5ÚÙ
29. Swap R1 and R3; swap R1 and R3.
30. Multiply R3 by –1/5; multiply R3 by –5.
31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.
32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2.
33. The first equation was given. The others are:
T2 = (T1 + 20 + 40 + T3 )/4,
or
4T2 − T1 − T3 = 60
T3 = (T4 + T2 + 40 + 30)/4,
or
4T3 − T4 − T2 = 70
T4 = (10 + T1 + T3 + 30)/4,
or
4T4 − T1 − T3 = 40
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
7
Rearranging,
4T1
−
T2
−T1
+
4T2
−T2
−
−T1
T4
= 30
− T3
+ 4T3
−
T4
= 60
= 70
−
+ 4T4
= 40
T3
34. Begin by interchanging R1 and R4, then create zeros in the first column:
Ç 4
È −1
È
È 0
È
É −1
−1
0
−1
4
−1
−1
4
0
−1
0
−1
4
Ç −1
Ù
60 È −1
Ù~È
70 Ù È 0
Ù È
40 Ú É 4
0
−1
4
4
−1
−1
4
0
−1
−1
0
−1
30
Ç −1
Ù
60 È 0
Ù~È
70 Ù È 0
Ù È
30 Ú É 0
40
0
−1
4
4
−1
0
4
−4
−1
−1
−4
15
40
20 Ù
Ù
70 Ù
Ù
190 Ú
Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
−1
0
4
−1
−1
−1
−4
15
Ç1
Ù
È
5
0
Ù~È
70 Ù È 0
Ù È
190 Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
−4
14
Ç1
Ù
È
5
0
Ù~È
75Ù È 0
Ù È
195Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
0
12
−40
5Ù
Ù
75Ù
Ù
270 Ú
Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
0
0
4
−1
−2
0
0
1
Ç1
Ù
5 È0
Ù~È
75Ù È0
Ù È
22.5Ú É0
−40
0
1
0
1
0
0
4
0
0
0
0
1
Ç1
Ù
27.5 È 0
Ù~È
120 Ù È0
Ù È
22.5Ú É0
50
0
1
0
1
0
0
1
0
0
0
0
1
50
27.5Ù
Ù
30 Ù
Ù
22.5Ú
The last step is to replace R1 by R1 + (–1)R3:
Ç1
È0
~È
È0
È
É0
0
0
0
1
0
0
1
0
0
0
0
1
20.0
27.5Ù
Ù . The solution is (20, 27.5, 30, 22.5).
30.0 Ù
Ù
22.5Ú
Notes: The Study Guide includes a “Mathematical Note” about statements, “If … , then … .”
This early in the course, students typically use single row operations to reduce a matrix. As a result,
even the small grid for Exercise 34 leads to about 80 multiplications or additions (not counting operations
with zero). This exercise should give students an appreciation for matrix programs such as MATLAB.
Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not
already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in
connection with an LU factorization.
For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB
notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/84+/89
calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section
1.1 describes how to access the data that is available for all numerical exercises in the text. This feature
has the ability to save students time if they regularly have their matrix program at hand when studying
linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale.
These commands are included in the text data sets, available from the text web site,
www.pearsonhighered.com/lay.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
8
CHAPTER 1
1.2
• Linear Equations in Linear Algebra
SOLUTIONS
Notes: The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises
7–14, in preparation for Section 1.5.)
1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c.
2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.
Ç1
3. ÈÈ 2
ÈÉ 3
2
4
6
4
6
9
Ç1
~ ÈÈ0
ÈÉ0
Ç1
È
4. È 2
ÉÈ 4
2
4
5
8 Ç1
8ÙÙ ~ ÈÈ0
12 ÙÚ ÈÉ0
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
4 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
2
0
0
5 Ç1
4 ÙÙ ~ ÈÈ0
2 ÚÙ ÉÈ0
2
0
−3
4
−3
−12
2
1
0
4
4
1
5 Ç1
6 ÙÙ ~ ÈÈ 0
2 ÙÚ ÈÉ0
̈
0 ÙÚ
Ç̈
5. È
É0
* Ç̈
,
̈ÙÚ ÈÉ 0
* Ç0
,
0 ÙÚ ÈÉ 0
Ç1
7. È
É3
3
9
7 Ç1
~
6 ÙÚ ÈÉ 0
4
7
4
−2
−3
4
1
0
2
0
0
4
5
4
2
0
0
3
0
2
1
0
8 Ç1
−8ÙÙ ~ ÈÈ0
−12 ÙÚ ÈÉ0
0
1
0
2
0
0
4
1
−3
−8
4ÙÙ . Pivot cols 1 and 3.
0ÙÚ
5
Ç1
Ù
È
−6 Ù ~ È0
−18ÚÙ ÉÈ0
4
0
1
2
−3
0
5 Ç1
−2 ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
7 Ç1
~
−15ÙÚ ÈÉ0
Corresponding system of equations:
x1
3
0
0
0
1
+ 3 x2
x3
2
4
6
4
6
9
1
Pivot cols
−2 ÙÙ .
1, 2, and 3
2 ÙÚ
* Ç̈
̈ÙÙ , ÈÈ 0
0 ÚÙ ÉÈ 0
4
1
Ç1
È2
È
ÈÉ 3
5
Ç1
Ù
È
− 18Ù ~ È 0
−6 ÚÙ ÉÈ 0
4
− 12
−3
0
1
0
Ç̈
6. ÈÈ 0
ÉÈ 0
4
−5
8
4ÙÙ
−12ÙÚ
* Ç0
0 ÙÙ , ÈÈ0
0ÚÙ ÉÈ0
7 Ç1
~
3ÙÚ ÈÉ 0
8
8ÙÙ
12 ÙÚ
2
1
0
4
4
−3
Ç1
È2
È
ÈÉ 4
2
4
5
5
6ÙÙ
−6ÚÙ
4
5
4
5
4 ÙÙ
2 ÙÚ
̈
0 ÙÙ
0 ÙÚ
3
0
0
1
−5
3ÙÚ
= −5
=
3
The basic variables (corresponding to the pivot positions) are x1 and x3. The remaining variable x2 is
free. Solve for the basic variables in terms of the free variable. The general solution is
Ê x1 = −5 − 3 x2
Í
Ë x2 is free
Íx = 3
Ì 3
Note: Exercise 7 is paired with Exercise 10.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
8.
Ç 1
È −3
É
Ç 1 −3 0 −5 Ç 1
~È
~
Ù
−2 0 −6 Ú É 0
7 0
1 0
3ÙÚ ÈÉ 0
x
= 4
Corresponding system of equations: 1
x2
= 3
−3
0
Ç1
~È
Ù
9Ú É0
−5
−3
0
−5
0
0
1
0
• Solutions
9
4
3ÙÚ
The basic variables (corresponding to the pivot positions) are x1 and x2. The remaining variable x3 is
free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic
variables do not depend on the value of the free variable.
Ê x1 = 4
Í
General solution: Ë x2 = 3
Í x is free
Ì 3
Note: A common error in Exercise 8 is to assume that x3 is zero. To avoid this, identify the basic
variables first. Any remaining variables are free. (This type of computation will arise in Chapter 5.)
Ç0
9. È
É1
1
−2
−3
4
Ç1
~È
Ù
−6 Ú É 0
3
Corresponding system:
−3
4
1
−2
x1
x2
Ç1
~È
Ù
3Ú É 0
−6
− 2 x3
= 3
− 2 x3
= 3
0
−2
1
−2
3
3ÙÚ
Ê x1 = 3 + 2 x3
Í
Basic variables: x1, x2; free variable: x3. General solution: Ë x2 = 3 + 2 x3
Í x is free
Ì 3
Ç 1
10. È
É −2
−2
−1
4
−5
Ç1
~È
Ù
6 Ú É0
4
Corresponding system:
x1
−2
−1
0
−7
Ç1
~È
Ù
14 Ú É 0
− 2 x2
= 2
4
x3
−2
0
0
1
2
−2 ÙÚ
= −2
Ê x1 = 2 + 2 x2
Í
Basic variables: x1, x3; free variable: x2. General solution: Ë x2 is free
Í x = −2
Ì 3
Ç3
È
11. È9
ÈÉ6
−2
−6
−4
4
12
8
0 Ç3
0 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
−2
0
0
x1
Corresponding system:
−
4
0
0
0 Ç1
0ÙÙ ~ ÈÈ 0
0ÙÚ ÈÉ 0
2
x2
3
+
4
x3
3
0
0
−2 3
0
0
43
0
0
0
0ÙÙ
0ÙÚ
= 0
= 0
= 0
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
10
CHAPTER 1
• Linear Equations in Linear Algebra
2
4
Ê
Í x1 = 3 x2 − 3 x3
Í
Basic variable: x1; free variables x2, x3. General solution: Ë x2 is free
Í x is free
Í 3
Ì
12. Since the bottom row of the matrix is equivalent to the equation 0 = 1, the system has no solutions.
Ç1
È0
13. È
È0
È
É0
−3
0
−1
0
1
0
0
0
0
1
−4
9
0
0
0
0
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
−2
−3
0
0
9
1
0
0
0
0
1
−4
9
0
0
0
0
x1
x2
Corresponding system:
x4
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
2
− 3 x5
= 5
− 4 x5
=
0
0
0
−3
1
0
0
0
0
1
−4
9
0
0
0
0
5
1Ù
Ù
4Ù
Ù
0Ú
1
+ 9 x5 = 4
0 = 0
Ê x1 = 5 + 3x5
Í x = 1 + 4x
5
ÍÍ 2
Basic variables: x1, x2, x4; free variables: x3, x5. General solution: Ë x3 is free
Íx = 4 − 9x
5
Í 4
ÍÌ x5 is free
Note: The Study Guide discusses the common mistake x3 = 0.
Ç1
È0
14. È
È0
È
É0
0
1
−5
4
0
−1
−8
0
0
0
0
0
0
0
1
0
3 Ç1
6 ÙÙ ÈÈ0
~
0 Ù È0
Ù È
0 Ú É0
−5
4
0
−1
0
0
0
0
0
0
0
0
1
0
− 5 x3
x1
Corresponding system:
0
1
x2
+ 4 x3
3
6 ÙÙ
0Ù
Ù
0Ú
= 3
−
x4
= 6
x5 = 0
0 = 0
Ê x1 = 3 + 5 x3
Íx = 6 − 4x + x
3
4
ÍÍ 2
Basic variables: x1, x2, x5; free variables: x3, x4. General solution: Ë x3 is free
Í x is free
Í 4
ÍÌ x5 = 0
15. a. The system is consistent. There are many solutions because x3 is a free variable.
b. The system is consistent. There are many solutions because x1 is a free variable.
16. a. The system is inconsistent. (The rightmost column of the augmented matrix is a pivot column).
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
• Solutions
11
b. The system is consistent. There are many solutions because x2 is a free variable.
4
Ç 1 −1 4 Ç 1 −1
17. È
The system has a solution for all values of h since the augmented
~È
Ù
3 h Ú É0 1 h + 8ÙÚ
É −2
column cannot be a pivot column.
1 Ç1
1
−3
Ç 1 −3
18. È
If 3h + 6 is zero, that is, if h = –2, then the system has a
~È
Ù
6 −2 Ú É 0 3h + 6 − h − 2 ÙÚ
Éh
solution, because 0 equals 0. When h ≠ −2, the system has a solution since the augmented column
cannot be a pivot column. Thus the system has a solution for all values of h.
2
h
Ç 1 h 2 Ç1
19. È
~È
Ù
Ù
É 4 8 k Ú É 0 8 − 4 h k − 8Ú
a. When h = 2 and k ≠ 8, the augmented column is a pivot column, and the system is inconsistent.
b. When h ≠ 2, the system is consistent and has a unique solution. There are no free variables.
c. When h = 2 and k = 8, the system is consistent and has many solutions.
−3
1
Ç 1 −3 1 Ç 1
20. È
~È
Ù
h k Ú É 0 h + 6 k − 2 ÙÚ
É2
a. When h = –6 and k ≠ 2, the system is inconsistent, because the augmented column is a pivot
column.
b. When h ≠ −6, the system is consistent and has a unique solution. There are no free variables.
c. When h = –6 and k = 2, the system is consistent and has many solutions.
21. a.
b.
c.
d.
e.
False. See Theorem 1.
False. See the second paragraph of the section.
True. Basic variables are defined after equation (4).
True. This statement is at the beginning of Parametric Descriptions of Solution Sets.
False. The row shown corresponds to the equation 5x4 = 0, which does not by itself lead to a
contradiction. So the system might be consistent or it might be inconsistent.
22. a. True. See Theorem 1.
b. False. See Theorem 2.
c. False. See the beginning of the subsection Pivot Positions. The pivot positions in a matrix are
determined completely by the positions of the leading entries in the nonzero rows of any echelon
form obtained from the matrix.
d. True. See the paragraph just before Example 4.
e. False. The existence of at least one solution is not related to the presence or absence of free
variables. If the system is inconsistent, the solution set is empty. See the solution of Practice
Problem 2.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
12
CHAPTER 1
• Linear Equations in Linear Algebra
23. Since there are four pivots (one in each row), the augmented matrix must reduce to the form
Ç1
È0
È
È0
È
É0
0
1
0
0
0
0
0
0
1
0
0
1
x1
a
Ù
bÙ
and so
cÙ
Ù
dÚ
x2
x3
x4
=
=
=
a
b
c
=
d
No matter what the values of a, b, c, and d, the solution exists and is unique.
24. The system is consistent because there is not a pivot in column 5, which means that there is not a row
of the form [0 0 0 0 1]. Since the matrix is the augmented matrix for a system, Theorem 2 shows
that the system has a solution.
25. If the coefficient matrix has a pivot position in every row, then there is a pivot position in the bottom
row, and there is no room for a pivot in the augmented column. So, the system is consistent, by
Theorem 2.
26. Since the coefficient matrix has three pivot columns, there is a pivot in each row of the coefficient
matrix. Thus the augmented matrix will not have a row of the form [0 0 0 0 0 1], and the
system is consistent.
27. “If a linear system is consistent, then the solution is unique if and only if every column in the
coefficient matrix is a pivot column; otherwise there are infinitely many solutions. ”
This statement is true because the free variables correspond to nonpivot columns of the coefficient
matrix. The columns are all pivot columns if and only if there are no free variables. And there are no
free variables if and only if the solution is unique, by Theorem 2.
28. Every column in the augmented matrix except the rightmost column is a pivot column, and the
rightmost column is not a pivot column.
29. An underdetermined system always has more variables than equations. There cannot be more basic
variables than there are equations, so there must be at least one free variable. Such a variable may be
assigned infinitely many different values. If the system is consistent, each different value of a free
variable will produce a different solution, and the system will not have a unique solution. If the
system is inconsistent, it will not have any solution.
30. Example:
x1
2 x1
+
x2
+ 2 x2
+
x3
+ 2 x3
= 4
= 5
31. Yes, a system of linear equations with more equations than unknowns can be consistent.
x1 +
x2 = 2
Example (in which x1 = x2 = 1): x1 −
x2 = 0
3 x1 + 2 x2 = 5
32. According to the numerical note in Section 1.2, when n = 20 the reduction to echelon form takes
about 2(20)3/3 5,333 flops, while further reduction to reduced echelon form needs at most (20)2 =
400 flops. Of the total flops, the “backward phase” is about 400/5733 = .07 or about 7%. When n =
200, the estimates are 2(200)3/3 5,333,333 flops for the reduction to echelon form and (200)2 =
40,000 flops for the backward phase. The fraction associated with the backward phase is about
(4×104) /(5.3×106) = .007, or about .7%.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
• Solutions
13
33. For a quadratic polynomial p(t) = a0 + a1t + a2t2 to exactly fit the data (1, 6), (2, 15), and (3, 28), the
coefficients a0, a1, a2 must satisfy the systems of equations given in the text. Row reduce the
augmented matrix:
Ç1
È1
È
ÈÉ1
1
2
3
Ç1
~ ÈÈ0
ÉÈ0
6 Ç1
15ÙÙ ~ ÈÈ0
28ÚÙ ÉÈ0
1
4
9
1
1
0
0
0
1
1
1
2
4 Ç1
3ÙÙ ~ ÈÈ0
2 ÚÙ ÉÈ0
6 Ç1
9ÙÙ ~ ÈÈ0
22ÚÙ ÉÈ0
1
3
8
0
1
0
0
0
1
1
1
0
1
3
2
6 Ç1
9ÙÙ ~ ÈÈ 0
4ÚÙ ÉÈ 0
1
1
0
1
3
1
6
9ÙÙ
2ÚÙ
1
3ÙÙ
2 ÙÚ
The polynomial is p(t) = 1 + 3t + 2t2.
34. [M] The system of equations to be solved is:
a0
+
a1 ⋅ 0
+
a2 ⋅ 02
+
a3 ⋅ 03
+
a4 ⋅ 0 4
+
a5 ⋅ 05
=
a0
+
a1 ⋅ 2
+
a2 ⋅ 22
+
a3 ⋅ 23
+
a4 ⋅ 2 4
+
a5 ⋅ 25
= 2.90
2
+
3
a3 ⋅ 4
+
a4 ⋅ 4
4
+
a5 ⋅ 4
5
= 14.8
+
a3 ⋅ 63
+
a4 ⋅ 6 4
+
a5 ⋅ 65
= 39.6
+
3
+
4
+
5
= 74.3
a0
+
a1 ⋅ 4
+
a2 ⋅ 4
a0
+
a1 ⋅ 6
+
a2 ⋅ 62
+
2
a0
a0
a1 ⋅ 8
+
a2 ⋅ 8
+ a1 ⋅ 10 + a2 ⋅ 10
2
a3 ⋅ 8
+ a3 ⋅ 10
3
a4 ⋅ 8
+ a4 ⋅ 10
4
a5 ⋅ 8
5
+ a5 ⋅ 10
=
0
119
The unknowns are a0, a1, …, a5. Use technology to compute the reduced echelon of the augmented
matrix:
Ç1 0
È1 2
È
È1 4
È
È1 6
È1 8
È
ÉÈ1 10
Ç1
È0
È
È0
~È
È0
È0
È
ÈÉ 0
0
4
0
8
0
16
0
32
16
64
256
1024
36
216
1296
7776
64
512
4096
32768
2
10
3
4
105
0
0
0
0
0
2
4
8
16
32
0
0
8
0
48
48
224
576
960
4800
0
0
192
2688
26880
0
0
480
7680
90240
10
10
0 Ç1
2.9 Ù È 0
Ù È
14.8Ù È 0
Ù~È
39.6 Ù È 0
74.3Ù È 0
Ù È
119 ÚÙ ÈÉ 0
Ç1
Ù
2.9 È 0
Ù È
9Ù È0
Ù~È
3.9 Ù È 0
8.7 Ù È 0
Ù È
14.5ÙÚ ÈÉ 0
0
0
2
0
0
4
8
0
8
48
0
16
224
0
32
960
0
0
0
24
48
80
192
480
960
1248
4032
9920
7680
32640
99840
0
0
0
0
0
2
4
8
16
32
0
0
8
0
48
48
224
576
960
4800
0
0
0
384
7680
0
0
0
1920
42240
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
0
2.9 Ù
Ù
9Ù
Ù
30.9 Ù
62.7 Ù
Ù
104.5ÙÚ
0
2.9 Ù
Ù
9Ù
Ù
3.9 Ù
−6.9 Ù
Ù
−24.5ÚÙ
14
CHAPTER 1
Ç1
È0
È
È0
~È
È0
È0
È
ÉÈ 0
• Linear Equations in Linear Algebra
0
0
0
0
0
2
4
8
16
32
0
0
8
0
48
48
224
576
960
4800
0
0
0
384
7680
0
0
0
0
3840
Ç1
Ù
2.9 È 0
Ù È
9 Ù È0
Ù~È
3.9 Ù È 0
−6.9 Ù È 0
Ù È
10 ÚÙ ÉÈ 0
0
0
0
0
0
0
2
4
8
16
32
0
0
8
0
48
48
224
576
960
4800
0
0
0
384
7680
0
0
0
0
1
0
2.9 Ù
Ù
9Ù
Ù
3.9 Ù
−6.9 Ù
Ù
.0026 ÚÙ
0
0 0
0
0
Ç1 0 0
Ç1 0 0 0 0 0
È0 2 4
Ù
È
8
16 0
2.8167 Ù
1.7125ÙÙ
È
È0 1 0 0 0 0
È 0 0 8 48 224 0
È0 0 1 0 0 0 −1.1948Ù
6.5000 Ù
~È
Ù ~A~ È
Ù
.6615Ù
È 0 0 0 48 576 0 −8.6000 Ù
È0 0 0 1 0 0
È0 0 0
È0 0 0 0 1 0
0 384 0 −26.900 Ù
−.0701Ù
È
Ù
È
Ù
0
0 1 .002604 ÙÚ
.0026 ÙÚ
ÈÉ 0 0 0
ÈÉ0 0 0 0 0 1
Thus p(t) = 1.7125t – 1.1948t2 + .6615t3 – .0701t4 + .0026t5, and p(7.5) = 64.6 hundred lb.
Notes: In Exercise 34, if the coefficients are retained to higher accuracy than shown here, then p(7.5) =
64.8. If a polynomial of lower degree is used, the resulting system of equations is overdetermined. The
augmented matrix for such a system is the same as the one used to find p, except that at least column 6 is
missing. When the augmented matrix is row reduced, the sixth row of the augmented matrix will be
entirely zero except for a nonzero entry in the augmented column, indicating that no solution exists.
Exercise 34 requires 25 row operations. It should give students an appreciation for higher-level
commands such as gauss and bgauss, discussed in Section 1.4 of the Study Guide. The command
ref (reduced echelon form) is available, but I recommend postponing that command until Chapter 2.
The Study Guide includes a “Mathematical Note” about the phrase, “If and only if,” used in Theorem
2.
1.3
SOLUTIONS
Notes: The key exercises are 11–16, 19–22, 25, and 26. A discussion of Exercise 25 will help students
understand the notation [a1 a2 a3], {a1, a2, a3}, and Span{a1, a2, a3}.
Ç −1 Ç −3 Ç−1 + ( −3) Ç −4
1. u + v = È Ù + È Ù = È
Ù=È Ù.
É 2 Ú É −1Ú É 2 + ( −1) Ú É 1Ú
Using the definitions carefully,
Ç −1
Ç −3 Ç −1 Ç ( −2)(−3) Ç −1 + 6 Ç 5
u − 2 v = È Ù + ( −2) È Ù = È Ù + È
Ù=È
Ù = È Ù , or, more quickly,
É 2Ú
É −1Ú É 2 Ú É ( −2)( −1) Ú É 2 + 2 Ú É 4 Ú
Ç −1
Ç −3 Ç −1 + 6 Ç 5
u − 2v = È Ù − 2 È Ù = È
Ù = È Ù . The intermediate step is often not written.
É 2Ú
É −1Ú É 2 + 2 Ú É 4 Ú
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
INSTRUCTORÕS
SOLUTIONS MANUAL
THOMAS POLASKI
Winthrop University
JUDITH MCDONALD
Washington State University
L INEAR ALGEBRA
AND I TS
A PPLICATIONS
F OURTH E DITION
David C. Lay
University of Maryland
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2012, 2006, 1997 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-38888-9
ISBN-10: 0-321-38888-7
1 2 3 4 5 6 BB 15 14 13 12 11
_____________________________________________________
Contents
CHAPTER 1
Linear Equations in Linear Algebra 1
CHAPTER 2
Matrix Algebra
87
CHAPTER 3
Determinants
167
CHAPTER 4
Vector Spaces
197
CHAPTER 5
Eigenvalues and Eigenvectors
273
CHAPTER 6
Orthogonality and Least Squares
357
CHAPTER 7
Symmetric Matrices and Quadratic Forms
CHAPTER 8
The Geometry of Vector Spaces
405
453
iii
1.1
SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7
−2 x1 − 7 x2 = −5
Ç 1
È −2
É
5
−7
7
−5ÙÚ
Replace R2 by R2 + (2)R1 and obtain:
x1 + 5 x2 = 7
3x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/3:
x2 = 3
x1
Replace R1 by R1 + (–5)R2:
= −8
x2 = 3
Ç1
È0
É
5
3
9 ÙÚ
Ç1
È0
É
5
7
7
1
3 ÙÚ
Ç1
È0
É
0
−8
Ç1
È5
É
2
1
3ÙÚ
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
3x1 + 6 x2 = −3
5 x1 + 7 x2 = 10
Ç3
È5
É
6
7
−3
10 ÙÚ
Scale R1 by 1/3 and obtain:
Replace R2 by R2 + (–5)R1:
Scale R2 by –1/3:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = −1
5 x1 + 7 x2 = 10
x1 + 2 x2 = −1
−3x2 = 15
x1 + 2 x2 = −1
x2 = −5
x1
= 9
x2 = −5
Ç1
È0
É
−1
10 ÙÚ
7
2
−3
−1
15ÙÚ
Ç1
È0
É
2
1
−1
−5ÙÚ
Ç1
È0
É
0
9
1
−5 ÙÚ
The solution is (x1, x2) = (9, –5), or simply (9, –5).
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1
2
CHAPTER 1
• Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
Ç1
È1
É
x1 + 2 x2 = 4
x1 − x2 = 1
2
−1
4
1ÙÚ
x1 + 2 x2 = 4
Replace R2 by R2 + (–1)R1 and obtain:
−3 x2 = −3
x1 + 2 x2 = 4
Scale R2 by –1/3:
x2 = 1
= 2
x1
Replace R1 by R1 + (–2)R2:
x2 = 1
Ç1
È0
É
Ç1
È0
É
Ç1
È0
É
2
4
−3ÙÚ
−3
2
4
1
1 ÙÚ
0
2
1ÙÚ
1
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
x1 + 2 x2 = −13
3 x1 − 2 x2 =
1
Ç1
È3
É
2
−13
−2
1ÙÚ
Replace R2 by R2 + (–3)R1 and obtain:
Scale R2 by –1/8:
Replace R1 by R1 + (–2)R2:
x1 + 2 x2 = − 13
− 8 x2 =
40
x1 + 2 x2 = − 13
x1
x2 =
−5
=
−3
x2 =
−5
Ç1
È0
É
2
−8
Ç1
È0
É
2
Ç1
È0
É
0
1
1
−13
40 ÙÚ
−13
−5ÙÚ
−3
−5 ÙÚ
The point of intersection is (x1, x2) = (–3, –5).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do
not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to
eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by
its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
4
0 −1
Ç 1 −6
È0
2 −7
0
4 ÙÙ
È
. After that, the next step is to scale the fourth row by –1/7.
produces
È0
0
1
2 −3Ù
È
Ù
0
0 −7 14 Ú
É0
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third
column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0
x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row
operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
8. The standard row operations are:
Ç1
È0
È
È0
È
É0
−5
4
0
1
0
0
0
3
0
1
0
2
Ç1
È0
~È
È0
È
É0
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
0
−5
1
0
0
0
1
0
0
0
0
0
1
Ç1
Ù
È
0Ù È0
~
0Ù È0
Ù È
0Ú É0
−5
4
0
1
0
0
0
3
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0 Ç1
0ÙÙ ÈÈ0
~
0Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
3
0
0
0
1
Ç1
Ù
È
0 Ù È0
~
0 Ù È0
Ù È
0Ú É0
0
−5
4
0
1
0
0
0
1
0
0
0
1
0
0 ÙÙ
0Ù
Ù
0Ú
0
0ÙÙ
0Ù
Ù
0Ú
The solution set contains one solution: (0, 0, 0, 0).
9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:
Ç1
È0
È
È0
È
É0
−1
1
0
−2
0
0
0
0
1
0
−3
1
−5 Ç 1
−7 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
4Ú É0
−1
1
0
−2
0
0
0
0
1
0
0
1
−5
−7 ÙÙ
14 Ù
Ù
4Ú
Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:
Ç1
È0
~È
È0
È
É0
−1
1
0
0
0
0
0
0
1
0
0
1
−5 Ç 1
21ÙÙ ÈÈ 0
~
14 Ù È 0
Ù È
4 Ú É0
0
1
0
0
0
0
1
0
0
0
16
21ÙÙ
0 14 Ù
Ù
1 4Ú
The solution set contains one solution: (16, 21, 14, 4).
10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3
and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the
final step, replace R1 by R1 + (-3)R2.
Ç1
È0
È
È0
È
É0
3
1
0
0
−2
3
0
0
1
0
0
1
−7 Ç 1
6 ÙÙ ÈÈ 0
~
2Ù È0
Ù È
−2 Ú É 0
3
1
0
0
0
0
0
0
1
0
0
1
−11 Ç 1
12 ÙÙ ÈÈ0
~
2Ù È0
Ù È
−2 Ú É0
0
1
0
0
0
0
0
0
1
0
0
1
−47
12ÙÙ
2Ù
Ù
−2Ú
The solution set contains one solution: (–47, 12, 2, –2).
11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.
Ç0
È1
È
ÉÈ 2
1
4
7
5
3
1
−4 Ç 1
−2 ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ 2
4
1
7
3
5
1
−2 Ç 1
−4ÙÙ ~ ÈÈ 0
−2 ÚÙ ÉÈ0
4
1
−1
3
5
−5
−2 Ç 1
−4 ÙÙ ~ ÈÈ 0
2 ÚÙ ÉÈ 0
4
1
0
3
5
0
−2
−4ÙÙ
−2ÚÙ
The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.
The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
CHAPTER 1
• Linear Equations in Linear Algebra
12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
Ç 1
È 2
È
ÈÉ −2
−5
−7
1
4
3
7
−3 Ç 1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÈÉ0
−5
3
−9
4
−5
15
−3 Ç 1
4ÙÙ ~ ÈÈ0
−7 ÙÚ ÈÉ0
−5
3
0
4
−5
0
−3
4ÙÙ
5ÙÚ
The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.
The solution set is empty.
Ç1
13. ÈÈ 2
ÈÉ 0
Ç1
~ È0
È
ÈÉ0
Ç2
È
14. È 0
ÉÈ 3
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
7 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ0
−3
9
5
0
2
1
0
1
0
−3
5
1
0
1
0
8 Ç1
−2 ÙÙ ~ ÈÈ0
−1ÙÚ ÉÈ0
−8 Ç 1
3ÙÙ ~ ÈÈ0
−4 ÚÙ ÉÈ 3
−6
2
−2
0
1
6
−3
2
1
−4 Ç 1
3ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
15
5
0
2
1
0
1
0
0
0
1
−3
2
−2
0
1
6
0
1
0
−3
0
1
8 Ç1
−9 ÙÙ ~ ÈÈ0
−2 ÙÚ ÈÉ 0
−3
5
15
0
1
2
8 Ç1
−2 ÙÙ ~ ÈÈ0
−9ÙÚ ÈÉ 0
0
1
0
−3
5
5
8
−2 ÙÙ
−5ÙÚ
0
1
0
−3
2
−5
−4
3ÙÙ
−10ÚÙ
5
3Ù . The solution is (5, 3, –1).
Ù
−1ÙÚ
−4 Ç 1
3ÙÙ ~ ÈÈ0
−4ÚÙ ÉÈ0
−4 Ç 1
−1ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
−3
2
7
0
1
6
0
1
0
0
0
1
−4 Ç 1
3ÙÙ ~ ÈÈ 0
8ÚÙ ÉÈ 0
2
−1ÙÙ . The solution is (2, –1, 2).
2 ÙÚ
15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–
1)R3.
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0 5 Ç 1 −6
0 0
5
Ç 1 −6
È 0
Ù
È
Ù
È
Ù
È
1 −4 1 0 Ù È 0
1 −4 1 0 Ù È0
1 −4 1 0 Ù È 0
1 −4 1
0 ÙÙ
È
~
~
~
È −1
6
1 5 3Ù È 0
0
1 5 8Ù È0
0
1 5 8Ù È 0
0
1 5
8Ù
È
Ù È
Ù È
Ù È
Ù
5 4 0 Ú É0 −1
5 4 0 Ú É0
0
1 5 0Ú É0
0
0 0 −8Ú
É 0 −1
The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.
16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2
before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace
R4 by R4 + (–1)R3.
Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10 Ç 2 0 0 −4 −10
È 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0 ÙÙ ÈÈ 0 3 3
0
0ÙÙ ÈÈ 0 3 3
0
0 ÙÙ
È
~
~
~
È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù È 0 0 1
−1Ù
4
4
4
4
È
Ù È
Ù È
Ù È
Ù
−9 Ú
1
5Ú É 0 2 3 −5 −10Ú É 0 0 1 −5 −10Ú É 0 0 0 −9
É −3 2 3
The system is now in triangular form and has a solution. In fact, using the argument from Example 2,
one can see that the solution is unique.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations:
3 −1 Ç 2
3 −1 Ç 2
3 −1
Ç2
È6
Ù
È
Ù
È
5
0 Ù ~ È 0 −4
3Ù ~ È 0 −4
3ÙÙ
È
ÈÉ 2 −5
7 ÙÚ ÈÉ 0 −8
8ÙÚ ÈÉ 0
0
2ÙÚ
The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in
common.
18. Row reduce the augmented matrix corresponding to the given system of three equations:
Ç2
È0
È
ÈÉ 2
4
1
3
4
−2
0
4 Ç2
−2 ÙÙ ~ ÈÈ 0
0 ÙÚ ÈÉ 0
4
1
−1
4
−2
−4
4 Ç2
−2 ÙÙ ~ ÈÈ 0
−4ÙÚ ÈÉ 0
4
1
0
4
−2
−6
4
−2ÙÙ
−6ÙÚ
The system is consistent, and using the argument from Example 2, there is only one solution. So the
three planes have only one point in common.
4
h
Ç 1 h 4 Ç1
19. È
~È
Ù
Ù Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no
É3 6 8Ú É 0 6 − 3h −4 Ú
solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
h −5 Ç 1
h
−5
Ç1
Write c for −8 − 2h. If c = 0, that is, if h = –4, then the system
20. È
~È
Ù
6 Ú É 0 −8 − 2h 16 ÚÙ
É 2 −8
has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.
−2
4
Ç 1 4 −2 Ç 1
21. È
~È
Ù
Ù Write c for h − 12 . Then the second equation cx2 = 0 has a solution
É3 h −6 Ú É 0 h − 12 0 Ú
for every value of c. So the system is consistent for all h.
Ç −4
h È
~
−3ÙÚ È 0
É
and only if h = 6.
Ç −4
22. È
É 2
12
−6
12
0
h
h
h ÙÙ The system is consistent if and only if −3 + = 0, that is, if
2
−3 +
2Ú
23. a. True. See the remarks following the box titled Elementary Row Operations.
b. False. A 5 × 6 matrix has five rows.
c. False. The description applies to a single solution. The solution set consists of all possible
solutions. Only in special cases does the solution set consist of exactly one solution. Mark a
statement True only if the statement is always true.
d. True. See the box before Example 2.
24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that
transforms one matrix into the other.
b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.
c. False. The definition of equivalent systems is in the second paragraph after equation (2).
d. True. By definition, a consistent system has at least one solution.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
6
CHAPTER 1
• Linear Equations in Linear Algebra
7 g
7
g
7
g
Ç 1 −4
Ç1 −4
Ç1 −4
È
Ù
È
Ù
È
Ù
h Ù ~ È0
h
3 −5 h Ù ~ È 0
3 −5
3 −5
25. È 0
Ù
ÈÉ −2
5 −9 k ÙÚ ÈÉ0 −3
5 k + 2 g ÙÚ ÈÉ 0
0
0 k + 2 g + h ÙÚ
Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix
above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution
if and only if k + 2g + h = 0.
26. Row reduce the augmented matrix for the given system:
Ç2
Èc
É
4
d
Ç1
~È
Ù
g Ú Éc
f
2
Ç1
~È
Ù
g Ú É0
f /2
d
2
f /2
d − 2c
g − c ( f / 2) ÙÚ
This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f
and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus
d 2c.
27. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible
since a is nonzero. Then replace R2 by R2 + (–c)R1.
Ça
È
Éc
b
d
Ç1
~È
Ù
g Ú Éc
f
Ç1
~È
Ù
g Ú É0
b/a
f /a
d
b/a
f /a
g − c( f / a ) ÚÙ
d − c(b / a )
The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity
g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be
written as ad – bc ≠ 0, or ad ≠ bc.
28. A basic principle of this section is that row operations do not affect the solution set of a linear
system. Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and
then perform any elementary row operations to produce other augmented matrices. Here are three
examples. The fact that they are all row equivalent proves that they all have the solution set (3, –2, –
1).
Ç1
È0
È
ÉÈ0
0
1
0
0
0
1
3 Ç1
−2ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 0
0
1
0
0
0
1
3 Ç1
4ÙÙ ~ ÈÈ 2
−1ÚÙ ÉÈ 2
0
1
0
0
0
1
3
4ÙÙ
5ÚÙ
29. Swap R1 and R3; swap R1 and R3.
30. Multiply R3 by –1/5; multiply R3 by –5.
31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.
32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2.
33. The first equation was given. The others are:
T2 = (T1 + 20 + 40 + T3 )/4,
or
4T2 − T1 − T3 = 60
T3 = (T4 + T2 + 40 + 30)/4,
or
4T3 − T4 − T2 = 70
T4 = (10 + T1 + T3 + 30)/4,
or
4T4 − T1 − T3 = 40
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.1
• Solutions
7
Rearranging,
4T1
−
T2
−T1
+
4T2
−T2
−
−T1
T4
= 30
− T3
+ 4T3
−
T4
= 60
= 70
−
+ 4T4
= 40
T3
34. Begin by interchanging R1 and R4, then create zeros in the first column:
Ç 4
È −1
È
È 0
È
É −1
−1
0
−1
4
−1
−1
4
0
−1
0
−1
4
Ç −1
Ù
60 È −1
Ù~È
70 Ù È 0
Ù È
40 Ú É 4
0
−1
4
4
−1
−1
4
0
−1
−1
0
−1
30
Ç −1
Ù
60 È 0
Ù~È
70 Ù È 0
Ù È
30 Ú É 0
40
0
−1
4
4
−1
0
4
−4
−1
−1
−4
15
40
20 Ù
Ù
70 Ù
Ù
190 Ú
Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
−1
0
4
−1
−1
−1
−4
15
Ç1
Ù
È
5
0
Ù~È
70 Ù È 0
Ù È
190 Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
−4
14
Ç1
Ù
È
5
0
Ù~È
75Ù È 0
Ù È
195Ú É 0
−40
0
1
−4
1
0
0
4
−1
−2
0
0
12
−40
5Ù
Ù
75Ù
Ù
270 Ú
Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4:
Ç1
È0
~È
È0
È
É0
0
1
−4
1
0
0
4
−1
−2
0
0
1
Ç1
Ù
5 È0
Ù~È
75Ù È0
Ù È
22.5Ú É0
−40
0
1
0
1
0
0
4
0
0
0
0
1
Ç1
Ù
27.5 È 0
Ù~È
120 Ù È0
Ù È
22.5Ú É0
50
0
1
0
1
0
0
1
0
0
0
0
1
50
27.5Ù
Ù
30 Ù
Ù
22.5Ú
The last step is to replace R1 by R1 + (–1)R3:
Ç1
È0
~È
È0
È
É0
0
0
0
1
0
0
1
0
0
0
0
1
20.0
27.5Ù
Ù . The solution is (20, 27.5, 30, 22.5).
30.0 Ù
Ù
22.5Ú
Notes: The Study Guide includes a “Mathematical Note” about statements, “If … , then … .”
This early in the course, students typically use single row operations to reduce a matrix. As a result,
even the small grid for Exercise 34 leads to about 80 multiplications or additions (not counting operations
with zero). This exercise should give students an appreciation for matrix programs such as MATLAB.
Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not
already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in
connection with an LU factorization.
For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB
notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/84+/89
calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section
1.1 describes how to access the data that is available for all numerical exercises in the text. This feature
has the ability to save students time if they regularly have their matrix program at hand when studying
linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale.
These commands are included in the text data sets, available from the text web site,
www.pearsonhighered.com/lay.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
8
CHAPTER 1
1.2
• Linear Equations in Linear Algebra
SOLUTIONS
Notes: The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises
7–14, in preparation for Section 1.5.)
1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c.
2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.
Ç1
3. ÈÈ 2
ÈÉ 3
2
4
6
4
6
9
Ç1
~ ÈÈ0
ÈÉ0
Ç1
È
4. È 2
ÉÈ 4
2
4
5
8 Ç1
8ÙÙ ~ ÈÈ0
12 ÙÚ ÈÉ0
Ç1
~ ÈÈ0
ÈÉ0
8 Ç1
4 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
2
0
0
5 Ç1
4 ÙÙ ~ ÈÈ0
2 ÚÙ ÉÈ0
2
0
−3
4
−3
−12
2
1
0
4
4
1
5 Ç1
6 ÙÙ ~ ÈÈ 0
2 ÙÚ ÈÉ0
̈
0 ÙÚ
Ç̈
5. È
É0
* Ç̈
,
̈ÙÚ ÈÉ 0
* Ç0
,
0 ÙÚ ÈÉ 0
Ç1
7. È
É3
3
9
7 Ç1
~
6 ÙÚ ÈÉ 0
4
7
4
−2
−3
4
1
0
2
0
0
4
5
4
2
0
0
3
0
2
1
0
8 Ç1
−8ÙÙ ~ ÈÈ0
−12 ÙÚ ÈÉ0
0
1
0
2
0
0
4
1
−3
−8
4ÙÙ . Pivot cols 1 and 3.
0ÙÚ
5
Ç1
Ù
È
−6 Ù ~ È0
−18ÚÙ ÉÈ0
4
0
1
2
−3
0
5 Ç1
−2 ÙÙ ~ ÈÈ0
2 ÙÚ ÈÉ0
7 Ç1
~
−15ÙÚ ÈÉ0
Corresponding system of equations:
x1
3
0
0
0
1
+ 3 x2
x3
2
4
6
4
6
9
1
Pivot cols
−2 ÙÙ .
1, 2, and 3
2 ÙÚ
* Ç̈
̈ÙÙ , ÈÈ 0
0 ÚÙ ÉÈ 0
4
1
Ç1
È2
È
ÈÉ 3
5
Ç1
Ù
È
− 18Ù ~ È 0
−6 ÚÙ ÉÈ 0
4
− 12
−3
0
1
0
Ç̈
6. ÈÈ 0
ÉÈ 0
4
−5
8
4ÙÙ
−12ÙÚ
* Ç0
0 ÙÙ , ÈÈ0
0ÚÙ ÉÈ0
7 Ç1
~
3ÙÚ ÈÉ 0
8
8ÙÙ
12 ÙÚ
2
1
0
4
4
−3
Ç1
È2
È
ÈÉ 4
2
4
5
5
6ÙÙ
−6ÚÙ
4
5
4
5
4 ÙÙ
2 ÙÚ
̈
0 ÙÙ
0 ÙÚ
3
0
0
1
−5
3ÙÚ
= −5
=
3
The basic variables (corresponding to the pivot positions) are x1 and x3. The remaining variable x2 is
free. Solve for the basic variables in terms of the free variable. The general solution is
Ê x1 = −5 − 3 x2
Í
Ë x2 is free
Íx = 3
Ì 3
Note: Exercise 7 is paired with Exercise 10.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
8.
Ç 1
È −3
É
Ç 1 −3 0 −5 Ç 1
~È
~
Ù
−2 0 −6 Ú É 0
7 0
1 0
3ÙÚ ÈÉ 0
x
= 4
Corresponding system of equations: 1
x2
= 3
−3
0
Ç1
~È
Ù
9Ú É0
−5
−3
0
−5
0
0
1
0
• Solutions
9
4
3ÙÚ
The basic variables (corresponding to the pivot positions) are x1 and x2. The remaining variable x3 is
free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic
variables do not depend on the value of the free variable.
Ê x1 = 4
Í
General solution: Ë x2 = 3
Í x is free
Ì 3
Note: A common error in Exercise 8 is to assume that x3 is zero. To avoid this, identify the basic
variables first. Any remaining variables are free. (This type of computation will arise in Chapter 5.)
Ç0
9. È
É1
1
−2
−3
4
Ç1
~È
Ù
−6 Ú É 0
3
Corresponding system:
−3
4
1
−2
x1
x2
Ç1
~È
Ù
3Ú É 0
−6
− 2 x3
= 3
− 2 x3
= 3
0
−2
1
−2
3
3ÙÚ
Ê x1 = 3 + 2 x3
Í
Basic variables: x1, x2; free variable: x3. General solution: Ë x2 = 3 + 2 x3
Í x is free
Ì 3
Ç 1
10. È
É −2
−2
−1
4
−5
Ç1
~È
Ù
6 Ú É0
4
Corresponding system:
x1
−2
−1
0
−7
Ç1
~È
Ù
14 Ú É 0
− 2 x2
= 2
4
x3
−2
0
0
1
2
−2 ÙÚ
= −2
Ê x1 = 2 + 2 x2
Í
Basic variables: x1, x3; free variable: x2. General solution: Ë x2 is free
Í x = −2
Ì 3
Ç3
È
11. È9
ÈÉ6
−2
−6
−4
4
12
8
0 Ç3
0 ÙÙ ~ ÈÈ0
0 ÙÚ ÈÉ0
−2
0
0
x1
Corresponding system:
−
4
0
0
0 Ç1
0ÙÙ ~ ÈÈ 0
0ÙÚ ÈÉ 0
2
x2
3
+
4
x3
3
0
0
−2 3
0
0
43
0
0
0
0ÙÙ
0ÙÚ
= 0
= 0
= 0
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
10
CHAPTER 1
• Linear Equations in Linear Algebra
2
4
Ê
Í x1 = 3 x2 − 3 x3
Í
Basic variable: x1; free variables x2, x3. General solution: Ë x2 is free
Í x is free
Í 3
Ì
12. Since the bottom row of the matrix is equivalent to the equation 0 = 1, the system has no solutions.
Ç1
È0
13. È
È0
È
É0
−3
0
−1
0
1
0
0
0
0
1
−4
9
0
0
0
0
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
−2
−3
0
0
9
1
0
0
0
0
1
−4
9
0
0
0
0
x1
x2
Corresponding system:
x4
Ç1
Ù
È
1
0
Ù~È
4Ù È0
Ù È
0Ú É0
2
− 3 x5
= 5
− 4 x5
=
0
0
0
−3
1
0
0
0
0
1
−4
9
0
0
0
0
5
1Ù
Ù
4Ù
Ù
0Ú
1
+ 9 x5 = 4
0 = 0
Ê x1 = 5 + 3x5
Í x = 1 + 4x
5
ÍÍ 2
Basic variables: x1, x2, x4; free variables: x3, x5. General solution: Ë x3 is free
Íx = 4 − 9x
5
Í 4
ÍÌ x5 is free
Note: The Study Guide discusses the common mistake x3 = 0.
Ç1
È0
14. È
È0
È
É0
0
1
−5
4
0
−1
−8
0
0
0
0
0
0
0
1
0
3 Ç1
6 ÙÙ ÈÈ0
~
0 Ù È0
Ù È
0 Ú É0
−5
4
0
−1
0
0
0
0
0
0
0
0
1
0
− 5 x3
x1
Corresponding system:
0
1
x2
+ 4 x3
3
6 ÙÙ
0Ù
Ù
0Ú
= 3
−
x4
= 6
x5 = 0
0 = 0
Ê x1 = 3 + 5 x3
Íx = 6 − 4x + x
3
4
ÍÍ 2
Basic variables: x1, x2, x5; free variables: x3, x4. General solution: Ë x3 is free
Í x is free
Í 4
ÍÌ x5 = 0
15. a. The system is consistent. There are many solutions because x3 is a free variable.
b. The system is consistent. There are many solutions because x1 is a free variable.
16. a. The system is inconsistent. (The rightmost column of the augmented matrix is a pivot column).
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
• Solutions
11
b. The system is consistent. There are many solutions because x2 is a free variable.
4
Ç 1 −1 4 Ç 1 −1
17. È
The system has a solution for all values of h since the augmented
~È
Ù
3 h Ú É0 1 h + 8ÙÚ
É −2
column cannot be a pivot column.
1 Ç1
1
−3
Ç 1 −3
18. È
If 3h + 6 is zero, that is, if h = –2, then the system has a
~È
Ù
6 −2 Ú É 0 3h + 6 − h − 2 ÙÚ
Éh
solution, because 0 equals 0. When h ≠ −2, the system has a solution since the augmented column
cannot be a pivot column. Thus the system has a solution for all values of h.
2
h
Ç 1 h 2 Ç1
19. È
~È
Ù
Ù
É 4 8 k Ú É 0 8 − 4 h k − 8Ú
a. When h = 2 and k ≠ 8, the augmented column is a pivot column, and the system is inconsistent.
b. When h ≠ 2, the system is consistent and has a unique solution. There are no free variables.
c. When h = 2 and k = 8, the system is consistent and has many solutions.
−3
1
Ç 1 −3 1 Ç 1
20. È
~È
Ù
h k Ú É 0 h + 6 k − 2 ÙÚ
É2
a. When h = –6 and k ≠ 2, the system is inconsistent, because the augmented column is a pivot
column.
b. When h ≠ −6, the system is consistent and has a unique solution. There are no free variables.
c. When h = –6 and k = 2, the system is consistent and has many solutions.
21. a.
b.
c.
d.
e.
False. See Theorem 1.
False. See the second paragraph of the section.
True. Basic variables are defined after equation (4).
True. This statement is at the beginning of Parametric Descriptions of Solution Sets.
False. The row shown corresponds to the equation 5x4 = 0, which does not by itself lead to a
contradiction. So the system might be consistent or it might be inconsistent.
22. a. True. See Theorem 1.
b. False. See Theorem 2.
c. False. See the beginning of the subsection Pivot Positions. The pivot positions in a matrix are
determined completely by the positions of the leading entries in the nonzero rows of any echelon
form obtained from the matrix.
d. True. See the paragraph just before Example 4.
e. False. The existence of at least one solution is not related to the presence or absence of free
variables. If the system is inconsistent, the solution set is empty. See the solution of Practice
Problem 2.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
12
CHAPTER 1
• Linear Equations in Linear Algebra
23. Since there are four pivots (one in each row), the augmented matrix must reduce to the form
Ç1
È0
È
È0
È
É0
0
1
0
0
0
0
0
0
1
0
0
1
x1
a
Ù
bÙ
and so
cÙ
Ù
dÚ
x2
x3
x4
=
=
=
a
b
c
=
d
No matter what the values of a, b, c, and d, the solution exists and is unique.
24. The system is consistent because there is not a pivot in column 5, which means that there is not a row
of the form [0 0 0 0 1]. Since the matrix is the augmented matrix for a system, Theorem 2 shows
that the system has a solution.
25. If the coefficient matrix has a pivot position in every row, then there is a pivot position in the bottom
row, and there is no room for a pivot in the augmented column. So, the system is consistent, by
Theorem 2.
26. Since the coefficient matrix has three pivot columns, there is a pivot in each row of the coefficient
matrix. Thus the augmented matrix will not have a row of the form [0 0 0 0 0 1], and the
system is consistent.
27. “If a linear system is consistent, then the solution is unique if and only if every column in the
coefficient matrix is a pivot column; otherwise there are infinitely many solutions. ”
This statement is true because the free variables correspond to nonpivot columns of the coefficient
matrix. The columns are all pivot columns if and only if there are no free variables. And there are no
free variables if and only if the solution is unique, by Theorem 2.
28. Every column in the augmented matrix except the rightmost column is a pivot column, and the
rightmost column is not a pivot column.
29. An underdetermined system always has more variables than equations. There cannot be more basic
variables than there are equations, so there must be at least one free variable. Such a variable may be
assigned infinitely many different values. If the system is consistent, each different value of a free
variable will produce a different solution, and the system will not have a unique solution. If the
system is inconsistent, it will not have any solution.
30. Example:
x1
2 x1
+
x2
+ 2 x2
+
x3
+ 2 x3
= 4
= 5
31. Yes, a system of linear equations with more equations than unknowns can be consistent.
x1 +
x2 = 2
Example (in which x1 = x2 = 1): x1 −
x2 = 0
3 x1 + 2 x2 = 5
32. According to the numerical note in Section 1.2, when n = 20 the reduction to echelon form takes
about 2(20)3/3 5,333 flops, while further reduction to reduced echelon form needs at most (20)2 =
400 flops. Of the total flops, the “backward phase” is about 400/5733 = .07 or about 7%. When n =
200, the estimates are 2(200)3/3 5,333,333 flops for the reduction to echelon form and (200)2 =
40,000 flops for the backward phase. The fraction associated with the backward phase is about
(4×104) /(5.3×106) = .007, or about .7%.
Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
1.2
• Solutions
13
33. For a quadratic polynomial p(t) = a0 + a1t + a2t2 to exactly fit the data (1, 6), (2, 15), and (3, 28), the
coefficients a0, a1, a2 must satisfy the systems of equations given in the text. Row reduce the
augmented matrix:
Ç1
È1
È
ÈÉ1
1
2
3
Ç1
~ ÈÈ0
ÉÈ0
6 Ç1
15ÙÙ ~ ÈÈ0
28ÚÙ ÉÈ0
1
4
9
1
1
0
0
0
1
1
1
2
4 Ç1
3ÙÙ ~ ÈÈ0
2 ÚÙ ÉÈ0
6 Ç1
9ÙÙ ~ ÈÈ0
22ÚÙ ÉÈ0
1
3
8
0
1
0
0
0
1
1
1
0
1
3
2
6 Ç1
9ÙÙ ~ ÈÈ 0
4ÚÙ ÉÈ 0
1
1
0
1
3
1
6
9ÙÙ
2ÚÙ
1
3ÙÙ
2 ÙÚ
The polynomial is p(t) = 1 + 3t + 2t2.
34. [M] The system of equations to be solved is:
a0
+
a1 ⋅ 0
+
a2 ⋅ 02
+
a3 ⋅ 03
+
a4 ⋅ 0 4
+
a5 ⋅ 05
=
a0
+
a1 ⋅ 2
+
a2 ⋅ 22
+
a3 ⋅ 23
+
a4 ⋅ 2 4
+
a5 ⋅ 25
= 2.90
2
+
3
a3 ⋅ 4
+
a4 ⋅ 4
4
+
a5 ⋅ 4
5
= 14.8
+
a3 ⋅ 63
+
a4 ⋅ 6 4
+
a5 ⋅ 65
= 39.6
+
3
+
4
+
5
= 74.3
a0
+
a1 ⋅ 4
+
a2 ⋅ 4
a0
+
a1 ⋅ 6
+
a2 ⋅ 62
+
2
a0
a0
a1 ⋅ 8
+
a2 ⋅ 8
+ a1 ⋅ 10 + a2 ⋅ 10
2
a3 ⋅ 8
+ a3 ⋅ 10
3
a4 ⋅ 8
+ a4 ⋅ 10
4
a5 ⋅ 8
5
+ a5 ⋅ 10
=
0
119
The unknowns are a0, a1, …, a5. Use technology to compute the reduced echelon of the augmented
matrix:
Ç1 0
È1 2
È
È1 4
È
È1 6
È1 8
È
ÉÈ1 10
Ç1
È0
È
È0
~È
È0
È0
È
ÈÉ 0
0
4
0
8
0
16
0
32
16
64
256
1024
36
216
1296
7776
64
512
4096
32768
2
10
3
4
105
0
0
0
0
0
2
4
8
16
32
0
0
8
0
48
48
224
576
960
4800
0
0
192
2688
26880
0
0
480
7680
90240
10
10
0 Ç1
2.9 Ù È 0
Ù È
14.8Ù È 0
Ù~È
39.6 Ù È 0
74.3Ù È 0
Ù È
119 ÚÙ ÈÉ 0
Ç1
Ù
2.9 È 0
Ù È
9Ù È0
Ù~È
3.9 Ù È 0
8.7 Ù È 0
Ù È
14.5ÙÚ ÈÉ 0
0
0
2
0
0
4
8
0
8
48
0
16
224
0
32
960
0
0
0
24
48
80
192
480
960
1248
4032
9920
7680
32640
99840
0
0
0
0
0
2
4
8
16
32
0
0
8
0
48
48
224
576
960
4800
0
0
0
384
7680
0
0
0
1920
42240
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
0
2.9 Ù
Ù
9Ù
Ù
30.9 Ù
62.7 Ù
Ù
104.5ÙÚ
0
2.9 Ù
Ù
9Ù
Ù
3.9 Ù
−6.9 Ù
Ù
−24.5ÚÙ
14
CHAPTER 1
Ç1
È0
È
È0
~È
È0
È0
È
ÉÈ 0
• Linear Equations in Linear Algebra
0
0
0
0
0
2
4
8
16
32
0
0
8
0
48
48
224
576
960
4800
0
0
0
384
7680
0
0
0
0
3840
Ç1
Ù
2.9 È 0
Ù È
9 Ù È0
Ù~È
3.9 Ù È 0
−6.9 Ù È 0
Ù È
10 ÚÙ ÉÈ 0
0
0
0
0
0
0
2
4
8
16
32
0
0
8
0
48
48
224
576
960
4800
0
0
0
384
7680
0
0
0
0
1
0
2.9 Ù
Ù
9Ù
Ù
3.9 Ù
−6.9 Ù
Ù
.0026 ÚÙ
0
0 0
0
0
Ç1 0 0
Ç1 0 0 0 0 0
È0 2 4
Ù
È
8
16 0
2.8167 Ù
1.7125ÙÙ
È
È0 1 0 0 0 0
È 0 0 8 48 224 0
È0 0 1 0 0 0 −1.1948Ù
6.5000 Ù
~È
Ù ~A~ È
Ù
.6615Ù
È 0 0 0 48 576 0 −8.6000 Ù
È0 0 0 1 0 0
È0 0 0
È0 0 0 0 1 0
0 384 0 −26.900 Ù
−.0701Ù
È
Ù
È
Ù
0
0 1 .002604 ÙÚ
.0026 ÙÚ
ÈÉ 0 0 0
ÈÉ0 0 0 0 0 1
Thus p(t) = 1.7125t – 1.1948t2 + .6615t3 – .0701t4 + .0026t5, and p(7.5) = 64.6 hundred lb.
Notes: In Exercise 34, if the coefficients are retained to higher accuracy than shown here, then p(7.5) =
64.8. If a polynomial of lower degree is used, the resulting system of equations is overdetermined. The
augmented matrix for such a system is the same as the one used to find p, except that at least column 6 is
missing. When the augmented matrix is row reduced, the sixth row of the augmented matrix will be
entirely zero except for a nonzero entry in the augmented column, indicating that no solution exists.
Exercise 34 requires 25 row operations. It should give students an appreciation for higher-level
commands such as gauss and bgauss, discussed in Section 1.4 of the Study Guide. The command
ref (reduced echelon form) is available, but I recommend postponing that command until Chapter 2.
The Study Guide includes a “Mathematical Note” about the phrase, “If and only if,” used in Theorem
2.
1.3
SOLUTIONS
Notes: The key exercises are 11–16, 19–22, 25, and 26. A discussion of Exercise 25 will help students
understand the notation [a1 a2 a3], {a1, a2, a3}, and Span{a1, a2, a3}.
Ç −1 Ç −3 Ç−1 + ( −3) Ç −4
1. u + v = È Ù + È Ù = È
Ù=È Ù.
É 2 Ú É −1Ú É 2 + ( −1) Ú É 1Ú
Using the definitions carefully,
Ç −1
Ç −3 Ç −1 Ç ( −2)(−3) Ç −1 + 6 Ç 5
u − 2 v = È Ù + ( −2) È Ù = È Ù + È
Ù=È
Ù = È Ù , or, more quickly,
É 2Ú
É −1Ú É 2 Ú É ( −2)( −1) Ú É 2 + 2 Ú É 4 Ú
Ç −1
Ç −3 Ç −1 + 6 Ç 5
u − 2v = È Ù − 2 È Ù = È
Ù = È Ù . The intermediate step is often not written.
É 2Ú
É −1Ú É 2 + 2 Ú É 4 Ú
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1.3
• Solutions
15
Ç 3 Ç 2 Ç3 + 2 Ç5
2. u + v = È Ù + È Ù = È
Ù =È Ù.
É 2 Ú É −1Ú É 2 − 1 Ú É 1Ú
Using the definitions carefully,
Ç3
Ç 2 Ç 3 Ç ( −2)(2)
Ç3 − 4
Ç −1
=È
= È Ù , or, more quickly,
u − 2 v = È Ù + ( −2) È Ù = È Ù + È
Ù
Ù
É2Ú
É −1Ú É 2 Ú É (−2)(−1) Ú É 2 + 2 Ú É 4 Ú
Ç3
Ç 2 Ç3 − 4
Ç −1
= È Ù . The intermediate step is often not written.
u − 2v = È Ù − 2 È Ù = È
Ù
É2Ú
É −1Ú É 2 + 2 Ú É 4 Ú
3.
4.
Ç 3 x1 Ç 5 x2
Ç 2
Ç 3
Ç 5 Ç 2
È
Ù
È
Ù
È
Ù
È
Ù
È
Ù
5. x1 È −2 Ù + x2 È 0Ù = È −3Ù , È −2 x1 Ù + È 0 Ù = ÈÈ −3ÙÙ ,
ÈÉ 8 x1 ÙÚ ÈÉ −9 x2 ÙÚ ÈÉ 8ÙÚ
ÈÉ 8ÙÚ
ÈÉ −9ÙÚ ÈÉ 8ÙÚ
3 x1 + 5 x2 = 2
−2 x1
= −3
8 x1 − 9 x2 = 8
Ç3x1 + 5 x2
Ç 2
È −2 x Ù = È −3Ù
1 Ù
È
È Ù
ÈÉ8 x1 − 9 x2 ÙÚ ÈÉ 8ÙÚ
Usually the intermediate steps are not displayed.
Ç 3
Ç7
Ç −2 Ç 0
6. x1 È Ù + x2 È Ù + x3 È Ù = È Ù ,
É −2 Ú
É 3Ú
É 1Ú É 0 Ú
3 x2 + 7 x2 − 2 x3 =
−2 x1 + 3 x2 +
x3 =
Ç 3x1 Ç7 x2
Ç −2 x3
Ç0
È −2 x Ù + È 3x Ù + È x Ù = È0Ù ,
É 1Ú É 2 Ú É 3 Ú É Ú
0
0
Ç3x1 + 7 x2 − 2 x3
Ç0
È −2 x + 3x + x Ù = È0Ù
É Ú
2
3Ú
É 1
Usually the intermediate steps are not displayed.
7. See the figure below. Since the grid can be extended in every direction, the figure suggests that every
vector in R2 can be written as a linear combination of u and v.
To write a vector a as a linear combination of u and v, imagine walking from the origin to a along
the grid "streets" and keep track of how many "blocks" you travel in the u-direction and how many in
the v-direction.
a. To reach a from the origin, you might travel 1 unit in the u-direction and –2 units in the vdirection (that is, 2 units in the negative v-direction). Hence a = u – 2v.
b. To reach b from the origin, travel 2 units in the u-direction and –2 units in the v-direction. So
b = 2u – 2v. Or, use the fact that b is 1 unit in the u-direction from a, so that
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.

GOOG

good one

jul

grazie

nooo