Student Solutions Manual for ELEMENTARY LINEAR ALGEBRA SIXTH EDITION PDF

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Student Solutions Manual for ELEMENTARY LINEAR ALGEBRA SIXTH EDITION Larson Houghton Mifflin Harcourt Publishing Company Boston New York Publisher: Richard Stratton Senior Sponsoring Editor: Cathy Cantin Marketing Manager: Jennifer Jones Development Editor: Janine Tangney Associate Editor: Jeannine ...

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Syst ems of Linear Equat ions and Mat rices Sect ion 1.1 Exercise Set 1.1 gdfgfd dfgdfg 02Sol Linear Algebra Anson Li

Student Solutions Manual for

ELEMENTARY LINEAR ALGEBRA SIXTH EDITION Larson

Houghton Mifflin Harcourt Publishing Company Boston

New York

Publisher: Richard Stratton Senior Sponsoring Editor: Cathy Cantin Marketing Manager: Jennifer Jones Development Editor: Janine Tangney Associate Editor: Jeannine Lawless Project Editor: Kerry Falvey New Title Project Manager: Pat O’Neill

Copyright © 2009 by Houghton Mifflin Harcourt Publishing Company. All rights reserved. No part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying and recording, or any information storage or retrieval system without the prior written permission of Houghton Mifflin Harcourt Publishing Company unless such copying is expressly permitted by federal copyright law. Address inquiries to College Permissions, Houghton Mifflin Harcourt Publishing Company, 222 Berkeley Street, Boston, MA 02116-3764. Printed in the United States of America. ISBN 13: 978-0-618-78377-9 ISBN 10: 0-618-78377-6 123456789-???-12 11 10 09

PREFACE This Student Solutions Manual is designed as a supplement to Elementary Linear Algebra, Sixth Edition, by Ron Larson and David C. Falvo. All references to chapters, theorems, and exercises relate to the main text. Solutions to every odd-numbered exercise in the text are given with all essential algebraic steps included. Although this supplement is not a substitute for good study habits, it can be valuable when incorporated into a well-planned course of study. We have made every effort to see that the solutions are correct. However, we would appreciate hearing about any errors or other suggestions for improvement. Good luck with your study of elementary linear algebra. Ron Larson Larson Texts, Inc.

iii

CONTENTS

iv

Chapter 1

Systems of Linear Equations..................................................................1

Chapter 2

Matrices.................................................................................................26

Chapter 3

Determinants.........................................................................................63

Chapter 4

Vector Spaces .......................................................................................96

Chapter 5

Inner Product Spaces ..........................................................................137

Chapter 6

Linear Transformations ......................................................................185

Chapter 7

Eigenvalues and Eigenvectors ...........................................................219

C H A P T E R 1 Systems of Linear Equations Section 1.1

Introduction to Systems of Linear Equations........................................2

Section 1.2

Gaussian Elimination and Gauss-Jordan Elimination ..........................9

Section 1.3

Applications of Systems of Linear Equations .....................................15

Review Exercises ..........................................................................................................21

Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.

C H A P T E R 1 Systems of Linear Equations Section 1.1 Introduction to Systems of Linear Equations 1. Because the equation is in the form a1 x + a2 y = b, it is linear in the variables x and y. 3. Because the equation cannot be written in the form a1 x + a2 y = b, it is not linear in the variables x and y. 5. Because the equation cannot be written in the form a1 x + a2 y = b, it is not linear in the variables x and y.

15. Begin by rewriting the system in row-echelon form. The equations are interchanged.

2 x1 +

5 x1 + 2 x2 + x3 = 0 The first equation is multiplied by 12 . x1 +

Adding −5 times the first equation to the second equation produces a new second equation.

2 x − 4t = 0 2 x = 4t

x1 +

x = 2t.

The second equation is multiplied by −2.

9. Choosing y and z as the free variables, let y = s and z = t , and obtain x + s + t = 1 or x = 1 − s − t. So, you can describe the solution set as x = 1 − s − t , y = s and z = t , where s and t are any real numbers. 11. From Equation 2 you have x2 = 3. Substituting this value into Equation 1 produces x1 − 3 = 2 or x1 = 5. So, the system has exactly one solution: x1 = 5 and x2 = 3.

1x 2 2

x1 +

= 0

x2 − 2 x3 = 0 To represent the solutions, choose x3 to be the free variable and represent it by the parameter t. Because x2 = 2 x3 and x1 = − 12 x2 , you can describe the solution set as x1 = −t , x2 = 2t , x3 = t , where t is any real number.

17. 2 x + y = 4

13. From Equation 3 you can conclude that z = 0. Substituting this value into Equation 2 produces

x − y = 2 y

2y + 0 = 3 4 3 2 1

y = 32 . 3 2

and z = 0 into Equation

1, you obtain 3 2

−4

−0 = 0

and z = 0.

2

−2

3, 2

y =

3, 2

x−y=2 x 3 4

−2 −4

x = 32 .

So, the system has exactly one solution: x =

= 0

1x 2 2

− 12 x2 + x3 = 0

So, you can describe the solution set as x = 2t and y = t , where t is any real number.

−x +

= 0

1x 2 2

5 x1 + 2 x2 + x3 = 0

7. Choosing y as the free variable, let y = t and obtain

Finally, by substituting y =

= 0

x2

2x + y = 4

Adding the first equation to the second equation produces a new second equation, 3x = 6, or x = 2. So, y = 0, and the solution is x = 2, y = 0. This is the point where the two lines intersect.

Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.

Section 1.1 x− y =1

19.

25.

−2 x + 2 y = 5 y

Introduction to Systems of Linear Equations

−2x + 2y = 5

x +3 y −1 + =1 4 3 2 x − y = 12 y

4 3

−4

x

−2

1 2 3 4 −3 −4

Adding 2 times the first equation to the second equation produces a new second equation. x − y =1

Because the second equation is a false statement, you can conclude that the original system of equations has no solution. Geometrically, the two lines are parallel.

y = 9

6 7

x+3 y−1 + =1 4 3

Multiplying the first equation by 12 produces a new first equation. 2x −

y = 12

Adding the first equation to 4 times the second equation produces a new second equation, 11x = 55, or

27. 0.05 x − 0.03 y = 0.07

10

0.07 x + 0.02 y = 0.16

2x + y = 9

y

4 2 −2

1 2 3

is: x = 5, y = −2. This is the point where the two lines intersect.

y

6

x −2 −4 −6 −8 −10 −12

x = 5. So, 2(5) − y = 12, or y = −2, and the solution

21. 3x − 5 y = 7

8

2x − y = 12

3x + 4 y = 7

0 = 7

2x +

6 4

x−y=1

1

3

3x − 5y = 7 x

2

4

6

8 10

Adding the first equation to 5 times the second equation produces a new second equation, 13x = 52, or x = 4. So, 2( 4) + y = 9, or y = 1, and the solution is: x = 4,

y = 1. This is the point where the two lines intersect. 23. 2 x − y = 5

1 2

4 5 6 7 8

0.07x + 0.02y = 0.16

Multiplying the first equation by 200 and the second equation by 300 produces new equations. 21x + 6 y = 48

y

−4

x −2 −4 −6

10 x − 6 y = 14

5 x − y = 11

−6 −4 −2 −2

0.05x − 0.03y = 0.07 8 6 4 2

x 4

6

5x − y = 11

2x − y = 5

Adding the first equation to the second equation produces a new second equation, 31x = 62, or x = 2. So, 10( 2) − 6 y = 14, or y = 1, and the solution is: x = 2, y = 1. This is the point where the two lines intersect.

−12

Subtracting the first equation from the second equation produces a new second equation, 3x = 6, or x = 2. So, 2( 2) − y = 5, or y = −1, and the solution is: x = 2, y = −1. This is the point where the two lines intersect.

Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.

Chapter 1

4

29.

Systems of Linear Equations 35. (a)

x y + =1 4 6 x − y = 3

3

−4

y 8

x y + =1 4 6

6

−3

2 x 1

3

4

x−y=3

Adding 6 times the first equation to the second equation produces a new second equation, 52 x = 9, or x = 18 . 5 18 − y = 3, or y = 3 , and the solution is: x = 18 , 5 5 5 = 53. This is the point where the two lines intersect.

So, y

31. (a) −3x − y = 3

3

−4

4

−3

33. (a)

2x − 8y = 3

4

1 2x

41. Dividing the first equation by 9 produces a new first equation.

+y=0

x− 1x 5

(b) The system is consistent (c) The solution is approximately x =

1, 2

y = − 14 .

(d) Adding − 14 times the first equation to the second equation produces the new second equation 3 y = − 34 , or y = − 14 . So, x = 12 , and the solution 1, 2

37. Adding −3 times the first equation to the second equation produces a new second equation. x1 − x2 = 0 x2 = −1 Now, using back-substitution you can conclude that the system has exactly one solution: x1 = −1 and x2 = −1.

Adding −2 times the first equation to the second equation produces a new second equation. u + 2v = 120 −3v = −120 Solving the second equation you have v = 40. Substituting this value into the first equation gives u + 80 = 120 or u = 40. So, the system has exactly one solution: u = 40 and v = 40.

3

−3

(e) The solutions in (c) and (d) are consistent.

2u + v = 120.

6x + 2y = 1

−4

(b) The system is consistent. (c) There are infinite solutions. (d) The second equation is the result of multiplying both sides of the first equation by 0.2. A parametric representation of the solution set is given by x = 94 + 2t , y = t , where t is any real number.

39. Interchanging the two equations produces the system u + 2v = 120

(b) The system is inconsistent.

is x =

4

0.8x − 1.6y = 1.8

4

−1 −2

4x − 8y = 9

y = − 14 .

(e) The solutions in (c) and (d) are the same.

+

1y 3 2y 5

= − 19 = − 13

Adding − 15 times the first equation to the second equation produces a new second equation. x −

1y 3 7 y 15

= − 19 = − 14 45

Multiplying the second equation by

15 7

produces a new

second equation. x −

1y 3

= − 19

y = − 23

Now, using back-substitution, you can substitute y = − 23 into the first equation to obtain x +

2 9

= − 19 or x = − 13 .

So, you can conclude that the system has exactly one solution: x = − 13 and y = − 23 . Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.

Section 1.1 43. To begin, change the form of the first equation. 1x 2

+

1y 3

=

23 6

x − 2y = 5

2y 3

=

23 3

x − 2y = 5

47. Adding −2 times the first equation to the second equation yields a new second equation. x + y + z = 6

− z = 0

3x

Adding −3 times the first equation to the third equation yields a new third equation. x + y + z = 6 −3 y −

Subtracting the first equation from the second equation yields a new second equation. x +

2y 3

=

23 3

−3 y − 4 z = −18

Dividing the second equation by − 83 yields a new second

y +

equation. 2y 3

=

Now, using back-substitution you can conclude that the system has exactly one solution: x = 7 and y = 1. 45. Multiplying the first equation by 50 and the second equation by 100 produces a new system.

x1 − 2.5 x2 = −9.5

4 x2 =

1z 3

3

=

−3 y − 4 z = −18

23 3

y = 1

3x1 +

z = −9

Dividing the second equation by −3 yields a new second equation. x + y + z = 6

− 83 y = − 83

x +

5

−3 y − z = −9

Multiplying the first equation by 2 yields a new first equation. x +

Introduction to Systems of Linear Equations

52

Adding −3 times the first equation to the second equation produces a new second equation. x1 − 2.5 x2 = −9.5

11.5 x2 = 80.5 Now, using back-substitution, you can conclude that the system has exactly one solution: x1 = 8 and x2 = 7.

Adding 3 times the second equation to the third equation yields a new third equation. x+ y + z = 6 y + 13 z = 3 −3 z = −9 Dividing the third equation by −3 yields a new third equation. x+ y + z = 6 y + 13 z = 3 z = 3 Now, using back-substitution you can conclude that the system has exactly one solution: x = 1, y = 2, and z = 3.

49. Dividing the first equation by 3 yields a new first equation.

x1 − x1 +

2x 3 2

+

4x 3 3

=

1 3

x2 − 2 x3 = 3

2 x1 − 3 x2 + 6 x3 = 8 Subtracting the first equation from the second equation yields a new second equation. x1 −

2x 3 2

+

4x 3 3

=

1 3

5x 3 2

−

10 x 3 3

=

8 3

2 x1 − 3 x2 + 6 x3 = 8 Adding −2 times the first equation to the third equation yields a new third equation. x1 −

2x 3 2

+

4x 3 3

=

1 3

5x 3 2

−

10 x 3 3

=

8 3

− 53 x2 +

10 x 3 3

=

22 3

At this point you should recognize that Equations 2 and 3 cannot both be satisfied. So, the original system of equations has no solution.

Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.

6

Chapter 1

Systems of Linear Equations

51. Dividing the first equation by 2 yields a new first equation.

x1 +

1x 2 2

3x 2 3

−

= 2

+ 2 x3 = 10

4 x1

55. Adding −2 times the first equation to the second, 3 times the first equation to the third, and −1 times the first equation to the fourth, produces

x +

−

= 2

3x 2 3

−2 x2 + 8 x3 = 2 −2 x1 +

3x2 − 13 x3 = −8

y − 2z

1x 2 2

−

= 2

3x 2 3

−2 x2 + 8 x3 = 2 4 x2 − 16 x3 = −4 Dividing the second equation by −2 yields a new second equation. x1 +

1x 2 2

3x 2 3

−

= 2

x2 − 4 x3 = −1 4 x2 − 16 x3 = −4 Adding −4 times the second equation to the third equation produces a new third equation. x1 +

1x 2 2

3x 2 3

−

= 2

x + y +

0 = 0 Adding − 12 times the second equation to the first equation produces a new first equation.

+

1x 2 3

=

5 2

w =

6

18 z + 26w = 106 3w =

6.

Using back-substitution, you find the original system has exactly one solution: x = 1, y = 0, z = 3, and w = 2. Answers may vary slightly for Exercises 57–63. 57. Using a computer software program or graphing utility, you obtain x1 = −15, x2 = 40, x3 = 45, x4 = −75

59. Using a computer software program or graphing utility, you obtain x = −1.2, y = −0.6, z = 2.4.

x1 =

1, 5

x2 = − 54 , x3 =

1. 2

63. Using a computer software program or graphing utility, you obtain x = 6.8813, y = −163.3111, z = −210.2915,

x2 − 4 x3 = −1 Choosing x3 = t as the free variable, you can describe the solution as x1 =

z +

y − 2 z − 3w = −12

61. Using a computer software program or graphing utility, you obtain

x2 − 4 x3 = −1

x1

= −6.

Adding −7 times the second equation to the third, and −1 times the second equation to the fourth, produces

Adding 2 times the first equation to the third equation produces a new third equation. x1 +

6

7 y + 4 z + 5w = 22

Adding −4 times the first equation to the second equation produces a new second equation. 1x 2 2

z + w =

y − 2 z − 3w = −12

−2 x1 + 3 x2 − 13 x3 = −8

x1 +

y +

5 2

w = −59.2913.

− 12 t , x2 = 4t − 1, and

x3 = t , where t is any real number. 53. Adding −5 times the first equation to the second equation yields a new second equation.

x − 3y + 2z =

18

0 = −72 Because the second equation is a false statement, you can conclude that the original system of equations has no solution.

Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.

Section 1.1 65. x = y = z = 0 is clearly a solution.

Dividing the first equation by 4 yields a new first equation. x+

3y 4

+

17 z 4

= 0

Introduction to Systems of Linear Equations 67. x = y = z = 0 is clearly a solution.

Dividing the first equation by 5 yields a new first equation. x +

y −

1z 5

= 0

5 x + 4 y + 22 z = 0

10 x + 5 y + 2 z = 0

4 x + 2 y + 19 z = 0

5 x + 15 y − 9 z = 0

Adding −5 times the first equation to the second equation yields a new second equation.

x +

7

Adding −10 times the first equation to the second equation yields a new second equation. x +

y −

3y 4

+

17 z 4

= 0

1y 4

+

3z 4

= 0

−5 y + 4 z = 0

4 x + 2 y + 19 z = 0