Elementary Linear Algebra

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Instructor’s Solutions Manual for

SIXTH EDITION

Larson

Houghton Mifflin Harcourt Publishing Company

Boston New York

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Publisher: Richard Stratton

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Houghton Mifflin Harcourt Publishing Company hereby grants you permission to reproduce the Houghton Mifflin Harcourt Publishing material contained in this work in classroom quantities, solely for use with the accompanying Houghton Mifflin Harcourt Publishing textbook. All reproductions must include the Houghton Mifflin Harcourt Publishing copyright notice, and no fee may be collected except to cover the cost of duplication. If you wish to make any other use of this material, including reproducing or transmitting the material or portions thereof in any form or by any electronic or mechanical means including any information storage or retrieval system, you must obtain prior written permission from Houghton Mifflin Harcourt Publishing Company, unless such use is expressly permitted by federal copyright law. If you wish to reproduce material acknowledging a rights holder other than Houghton Mifflin Harcourt Publishing Company, you must obtain permission from the rights holder. Address inquiries to College Permissions, Houghton Mifflin Harcourt Publishing Company, 222 Berkeley Street, Boston, MA 02116-3764.

ISBN 13: 978-0-547-19855-2 ISBN 10: 0-547-19855-8

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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.

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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

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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

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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 a x1 + a y2 = b,it is linear in the variables x and y.

3. Because the equation cannot be written in the form

1 2 ,

a x+ a y = b it is not linear in the variables x and y. 5. Because the equation cannot be written in the form

1 2 ,

a x+ a y = b it is not linear in the variables x and y. 7. Choosing y as the free variable, let y = and obtain t

2 4 0 2 4 2 . x t x t x t − = = =

So, you can describe the solution set as x = 2tand ,

y = t where t is any real number.

9. Choosing y and z as the free variables, let y = and s ,

z = t and obtain x + + = or 1s t 1 x = − − So, s t. you can describe the solution set as x = − − 1 s t,

y = and ,s 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= 2or 1 5.

x = So, the system has exactly one solution:

1 5

x = and x2 = 3.

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

3 2 2 0 3 . y y + = = Finally, by substituting 3 2

y = and 0z = into Equation 1, you obtain 3 2 3 2 0 0 . x x − + − = =

So, the system has exactly one solution: 3 3 2, 2,

x = y = and 0.z =

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

1 2 1 2 3 2 0 5 2 0 x x x x x + = + + =

The first equation is multiplied by 1 2. 1 2 1 2 3 1 2 0 5 2 0 x x x x x + = + + =

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

1 2 3 2 1 2 1 2 0 0 x x x x + = = − +

The second equation is multiplied by 2.−

1 2 2 3 1 2 0 2 0 x x x x + = − =

To represent the solutions, choose x to be the free 3 variable and represent it by the parameter t. Because

2 2 3

x = x and x1 = −12x2,you can describe the solution set as

1 , 2 2 ,

x = −t x = t x3 = t,where t is any real number. 17. 2 4 2 x y x y + = − =

Adding the first equation to the second equation produces a new second equation, 3x = 6,or 2.x = So,

0,

y = and the solution is x = 2, y = 0.This is the point where the two lines intersect.

x 1 2 3 3 4 −2 −2 −4 −4 4 x − y = 2 2x + y = 4 y

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Section 1.1 Introduction to Systems of Linear Equations 3 19. 1 2 2 5 x y x y − = − + =

Adding 2 times the first equation to the second equation produces a new second equation.

1

0 7

x− y = =

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. 21. 3 5 7 2 9 x y x y − = + =

Adding the first equation to 5 times the second equation produces a new second equation, 13x = 52,or 4.x = So, 2 4

( )

+ y = 9,or 1,y = and the solution is: x = 4,

1.

y = This is the point where the two lines intersect. 23. 2 5 5 11 x y x y − = − =

Subtracting the first equation from the second equation produces a new second equation, 3x = 6,or 2.x = So,

( )

2 2 − y = 5,or 1,y = − and the solution is: x = 2, 1.

y = − This is the point where the two lines intersect.

25. 3 1 1 4 3 2 12 x y x y + ₊ − ₌ − =

Multiplying the first equation by 12 produces a new first equation. 3 4 7 2 12 x y x y + = − =

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

5.

x = So, 2 5

( )

− y = 12,or 2,y = − and the solution is: 5,x = y = − This is the point where the two lines 2. intersect. 27. 0.05 0.03 0.07 0.07 0.02 0.16 x y x y − = + =

Multiplying the first equation by 200 and the second equation by 300 produces new equations.

10 6 14 21 6 48 x y x y − = + =

Adding the first equation to the second equation produces a new second equation, 31x = 62,or 2.x = So, 10 2

( )

− 6y = 14,or 1,y = and the solution is:

2,

x = y = 1.This is the point where the two lines intersect. −2 2 4 6 8 10 x 2 4 6 3x − 5y = 7 2x + y = 9 8 10 y 1 2 4 5 6 7 8 −2 −4 −6 2 4 6 8 0.05x − 0.03y = 0.07 0.07x + 0.02y = 0.16 x y x y −2 −4 −6 4 6 2x − y = 5 5x − y = 11 −2 −4 −12 1 2 3 6 7 −2 −4 −6 −8 −10 −12 4 6 x + 3 4 y − 1 3 = 1 + 2x − y = 12 x y x 1 3 4 −2 −4 −3 −4 2 −2x + 2y = 5 x − y = 1 y 1 3 4

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4 Chapter 1 Systems of Linear Equations 29. 1 4 6 3 x y x y + = − =

Adding 6 times the first equation to the second equation produces a new second equation, 5

2x = 9,or x = 185.

So, 18

5 − y = 3,or y = 35,and the solution is: x = 185, 3

5.

y = This is the point where the two lines intersect. 31. (a)

(b) The system is inconsistent. 33. (a)

(b) The system is consistent

(c) The solution is approximately 1 1 2, 4.

x = y = − (d) Adding 1

4

− times the first equation to the second equation produces the new second equation

3 4 3y = − or , 1 4. y = − So, 1 2,

x = and the solution is 1

2,

x = 1

4.

y = −

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

35. (a)

(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

9

4 2 , ,

x = + t y = t where t is any real number. (e) The solutions in (c) and (d) are consistent. 37. Adding 3− times the first equation to the second

equation produces a new second equation.

1 2 2 0 1 x x x − = = −

Now, using back-substitution you can conclude that the system has exactly one solution: x1 = − and 1 x2 = − 1. 39. Interchanging the two equations produces the system

2 120 . 2 120 u v u v + = + =

2 120 3 120 u v v + = − = −

Solving the second equation you have v = 40. Substituting this value into the first equation gives

80 120

u+ = or 40.u = So, the system has exactly one solution: u = 40and 40.v =

41. Dividing the first equation by 9 produces a new first equation. 1 1 3 9 1 2 1 5 5 3 x y x y − = − + = − Adding 1 5

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

1 1 3 9 7 14 15 45 x y y − = − = −

Multiplying the second equation by 15

7 produces a new second equation. 1 1 3 9 2 3 x y y − = − = −

Now, using back-substitution, you can substitute 2 3

y = − into the first equation to obtain 2 1

9 9

x+ = − or 1 3.

x = − So, you can conclude that the system has exactly one

−1 1 3 4 −2 2 4 6 8 x − y = 3 x 4 + = 1 y 6 x y −4 −3 4 3 −3x − y = 3 6 + 2 = 1x y −4 −3 4 3 2x − 8y = 3 1 2x + y = 0 −4 −3 4 3 4x − 8y = 9 0.8x − 1.6y = 1.8

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Section 1.1 Introduction to Systems of Linear Equations 5

43. To begin, change the form of the first equation.

23 1 1 2 3 6 2 5 x y x y + = − =

Multiplying the first equation by 2 yields a new first equation. 23 2 3 3 2 5 x y x y + = − =

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

23 2 3 3 8 8 3 3 x y y + = − = −

Dividing the second equation by 8 3

− yields a new second equation. 23 2 3 3 1 x y y + = =

Now, using back-substitution you can conclude that the system has exactly one solution:

7

x = and 1.y =

45. Multiplying the first equation by 50 and the second equation by 100 produces a new system.

1 2 1 2 2.5 9.5 3 4 52 x x x x − = − + =

1 2 2 2.5 9.5 11.5 80.5 x x x − = − =

Now, using back-substitution, you can conclude that the system has exactly one solution:

1 8

x = and x2 = 7.

47. Adding 2− times the first equation to the second equation yields a new second equation.

6 3 9 3 0 x y z y z x z + + = − − = − − =

Adding − times the first equation to the third equation 3 yields a new third equation.

6 3 9 3 4 18 x y z y z y z + + = − − = − − − = −

Dividing the second equation by 3− yields a new second equation. 1 3 6 3 3 4 18 x y z y z y z + + = + ₌ − − = −

Adding 3 times the second equation to the third equation yields a new third equation.

1 3 6 3 3 9 x y z y z z + + = + = − = −

Dividing the third equation by 3− yields a new third equation. 1 3 6 3 3 x y z y z z + + = + = =

Now, using back-substitution you can conclude that the system has exactly one solution:

1,x = y = 2,and 3.z =

49. Dividing the first equation by 3 yields a new first equation. 1 2 3 1 2 3 1 2 3 2 4 1 3 3 3 2 3 2 3 6 8 x x x x x x x x x − + = + − = − + =

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

1 2 3 2 3 1 2 3 2 4 1 3 3 3 5 10 8 3 3 3 2 3 6 8 x x x x x x x x − + = − = − + =

1 2 3 2 3 2 3 2 4 1 3 3 3 5 10 8 3 3 3 5 10 22 3 3 3 x x x x x x x − + = − = − + =

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

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6 Chapter 1 Systems of Linear Equations

51. Dividing the first equation by 2 yields a new first equation. 1 2 3 1 3 1 2 3 1 3 2 2 2 4 2 10 2 3 13 8 x x x x x x x x + ₋ ₌ + = − + − = −

1 ₂ ₃ 2 3 1 2 3 1 3 2 2 2 2 8 2 2 3 13 8 x _x _x x x x x x + − = − + = − + − = −

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

1 2 3 2 3 2 3 1 3 2 2 2 2 8 2 4 16 4 x x x x x x x + − = − + = − = −

Dividing the second equation by 2− yields a new second equation. 1 2 3 2 3 2 3 1 3 2 2 2 4 1 4 16 4 x _x _x x x x x + − = − = − − = −

Adding − times the second equation to the third 4 equation produces a new third equation.

1 2 3 2 3 1 3 2 2 2 4 1 0 0 x x x x x + − = − = − = Adding 1 2

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

1 3 2 3 1 5 2 2 4 1 x x x x + = − = −

Choosing x₃ = as the free variable, you can describe t the solution as x1 = ₂5 − ₂1t, x2 = 4t − and 1,

3 ,

x = t where t is any real number.

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

3 2 18

0 72

x− y + z = = −

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

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

6 2 3 12 7 4 5 22 . 2 6 x y z w y z w y z w y z + + + = − − = − + + = − = −

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

− times the second equation to the fourth, produces 6 2 3 12 18 26 106 3 6. x y z w y z w z w w + + + = − − = − + = =

Using back-substitution, you find the original system has exactly one solution: x =1, y = 0, z = 3,and 2.w = Answers may vary slightly for Exercises 57–63.

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

1 15, 2 40, 3 45, 4 75

x = − x = x = x = −

1.2, 0.6, 2.4. x = − y = − z =

1 1₅, 2 4₅, 3 1₂. x = x = − x =

63. Using a computer software program or graphing utility, you obtain 6.8813, 163.3111, x = y = − z = −210.2915, 59.2913. w = −

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Section 1.1 Introduction to Systems of Linear Equations 7

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

Dividing the first equation by 4 yields a new first equation. 3 17 4 4 0 5 4 22 0 4 2 19 0 x y z x y z x y z + ₊ ₌ + + = + + =

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

3 17 4 4 1 3 4 4 0 0 4 2 19 0 x y z y z x y z + ₊ ₌ + = + + =

3 17 4 4 1 3 4 4 0 0 2 0 x _y _z y z y z + + = + = − + =

Multiplying the second equation by 4 yields a new second equation. 3 17 4 4 0 3 0 2 0 x y z y z y z + ₊ ₌ + = − + =

Adding the second equation to the third equation yields a new third equation.

3 17 4 4 0 3 0 5 0 x _y _z y z z + + = + = =

Dividing the third equation by 5 yields a new third equation. 3 17 4 4 0 3 0 0 x y z y z z + ₊ ₌ + = =

Now, using back-substitution, you can conclude that the system has exactly one solution: x = 0, y = 0,and

0. z =

67. x = y = z = is clearly a solution. 0

Dividing the first equation by 5 yields a new first equation. 1 5 0 10 5 2 0 5 15 9 0 x y z x y z x y z + − = + + = + − =

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

1 5 0 5 4 0 5 15 9 0 x y z y z x y z + − ₌ − + = + − =

1 5 0 5 4 0 10 8 0 x y z y z y z + − = − + = − =

Dividing the second equation by 5− yields a new second equation. 1 5 4 5 0 0 10 8 0 x y z y z y z + − = − = − =

Adding −10times the second equation to the third equation yields a new third equation.

1 5 4 5 0 0 0 0 x y z y z + − = − = =

Adding − times the second equation to the first 1 equation yields a new first equation.

3 5 4 5 0 0 x z y z + = − =

Choosing z = as the free variable you find the solution t to be 3

5 ,

x = − t 4

5,

y = t and ,z = t where t is any real number.

69. (a) True. You can describe the entire solution set using parametric representation.

ax+ by = c

Choosing y = as the free variable, the solution is t ,

c b

x t

a a

= − y = t,where t is any real number. (b) False. For example, consider the system

1 2 3 1 2 3 1 2 x x x x x x + + = + + =

which is an inconsistent system.

(c) False. A consistent system may have only one solution.

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71. Because x1 = and t x2 = 3t− 4 = 3x1 − 4,one answer is the system

1 2 1 2 3 4 . 3 4 x x x x − = − + = −

Letting x2 = t,you get 1

4 4

.

3 3 3

t t

x = + = +

73. Substituting A =1xand B =1 yinto the original system yields 12 12 7 . 3 4 0 A B A B − = + =

Reduce this system to row-echelon form. Dividing the first equation by 12 yields a new first equation.

7 12 3 4 0 A B A B − = + =

Adding − times the first equation to the second 3 equation yields a new second equation.

7 12 7 4 7 A B B − = = −

Dividing the second equation by 7 yields a new second equation. 7 12 1 4 A B B − = = −

So, A = −1 4and B =1 3.Because A = 1xand 1 ,

B = y the solution of the original system of equations is: 3x = and y = − 4.

75. Substituting A =1 ,x B = 1 ,y and C = 1 ,z into the original system yields

2 3 4 4 2 10 2 3 13 8. A B C A C A B C + − = + = − + − = −

Reduce this system to row-echelon form.

1 3 2 2 2 2 8 2 4 16 4 A B C B C B C + − = − + = − = − 3 1 2 2 2 4 1 A B C B C + − = − = −

Letting t = Cbe the free variable, you have ,

C = t B = 4t − and 1

(

5

)

. 2 t

A = − + So, the solution to the original problem is

2 , 5 x t = − 1 , 4 1 y t = − 1 , z t = where 5, , 0.1 4 t ≠

77. Reduce the system to row-echelon form. Dividing the first equation by cosθyields a new second equation.

(

)

(

)

sin 1 cos cos sin cos 0 x y x y θ θ θ θ θ ⎛ ⎞ + _⎜ _⎟ = ⎝ ⎠ − + =

Multiplying the first equation by sinθand adding to the second equation yields a new second equation.

sin 1 cos cos 1 sin cos cos x y y θ θ θ θ θ θ ⎛ ⎞ + _⎜ _⎟ = ⎝ ⎠ ⎛ ⎞ ₌ ⎜ ⎟ ⎝ ⎠ 2 2 2

sin sin cos 1

Because cos

cos cos cos

θ _θ θ θ

θ θ θ

⎛ ₊ ₌ + ₌ ⎞

⎜ ⎟

⎝ ⎠

Multiplying the second equation by cosθyields a new second equation. sin 1 cos cos sin x y y θ θ θ θ ⎛ ⎞ + _⎜ _⎟ = ⎝ ⎠ =

Substituting siny = θinto the first equation yields

2 2 sin _sin 1 cos cos 1 sin cos cos . cos cos x x θ _θ θ θ θ θ _θ θ θ ⎛ ⎞ + ⎜ ⎟ = ⎝ ⎠ − = = =

So, the solution of the original system of equations is: cos

x = θ and y = sin .θ

79. For this system to have an infinite number of solutions both equations need to be equivalent.

Multiply the second equation by 2.−

4 6 2 2 6 x ky kx y + = − − =

So, when k = − the system will have an infinite 2 number of solutions.

81. Reduce the system to row-echelon form.

(

2

)

0 1 0 x ky k y + = − = 2 0 , 1 0 0 x ky k y + = − ≠ = 2 0 1 0 0, x k y = − ≠ =

If ₁₋ _k2 _≠ _0,_{that is if}_k _{≠ ± the system will have}_1, exactly one solution.

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Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination 9

83. To begin, reduce the system to row-echelon form.

(

)

2 6 8 3 14 x y kz k z + + = − = −

This system will have no solution if 8− 3k = 0,that is,

8 3.

k =

85. Reducing the system to row-echelon form, you have

(

)

(

)

(

)

(

2

)

3 1 1 1 1 1 1 3 x y kz k y k z k y k z k + + = − + − = − − + − = −

(

)

(

)

(

2

)

3 1 1 1 . 2 3 x y kz k y k z k k z k + + = − + − = − − − + = −

If ₋_k2 ₋_k ₊ ₂ ₌ _0,_{then there is no solution. So, if} 1

k = or 2,k = − there is not a unique solution. 87. (a) All three of the lines will intersect in exactly one

point (corresponding to the solution point). (b) All three of the lines will coincide (every point on

these lines is a solution point). (c) The three lines have no common point.

89. Answers vary.

(

Hint Choose three different values for x : and solve the resulting system of linear equations in the variables a, b, and c .

)

91. 4 3 5 6 13 x y x y − = − − = 4 3 14 28 x y y − = − = 4 3 2 x y y − = − = 5 2 x y = =

At each step, the lines always intersect at

( )

5, 2 , which is the solution to the system of equations.

93. Solve each equation for y.

1 100 1 99 2 2 y x y x = + = −

The graphs are misleading because, while they appear parallel, when the equations are solved for y they have slightly different slopes.

Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination

1. Because the matrix has 3 rows and 3 columns, it has size 3×3.

3. Because the matrix has 2 rows and 4 columns, it has size 2× 4.

5. Because the matrix has 1 row and 5 columns, it has size 1 5.×

7. Because the matrix has 4 rows and 5 columns, it has size 4× 5.

9. The matrix satisfies all three conditions in the definition of row-echelon form. Moreover, because each column that has a leading 1 (columns one and two) has zeros elsewhere, the matrix is in reduced row-echelon form. 11. Because the matrix has two non-zero rows without

leading 1’s, it is not in row-echelon form.

13. Because the matrix has a non-zero row without a leading 1, it is not in row-echelon form.

15. Because the matrix is in reduced row-echelon form, convert back to a system of linear equations

1 2 0 2 x x = =

So, the solution is: x₁ = and 0 x2 = 2.

x x − 4y = −3 5 −2 −4 34 −4 −3 −2 −5 3 4 5 14y = 28 y x x − 4y = −3 5 −2 −4 34 −4 −3 −2 −5 3 4 5 y = 2 y x −2 −4 123 −4 −3 −2 −5 3 1 4 5 y = 2 x = 5 y 4 x x − 4y = −3 5x − 6y = 13 5 −2 −4 3 4 2 −4 −5 3 4 5 y

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17. Because the matrix is in row-echelon form, convert back to a system of linear equations.

1 2 2 3 3 3 2 1 1 x x x x x − = − = = −

Solve this system by back-substitution.

( )

2 2 1 2 1 1 x x − − = = −

Substituting x2 = − into equation 1, 1

( )

1 1 3 1 . 2 x x − − = =

So, the solution is: x1 = 2, x2 = − and 1, x3 = − 1. 19. Interchange the first and second rows.

1 1 1 0 2 1 1 3 0 1 2 1 − ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

Interchange the second and third rows.

1 1 1 0 0 1 2 1 2 1 1 3 − ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎣ ⎦

Add − times the first row to the third row to produce a 2 new third row.

1 1 1 0 0 1 2 1 0 3 3 3 − ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎣ ⎦

Add − times the second row to the third row to produce 3 a new third row.

1 1 1 0 0 1 2 1 0 0 9 0 − ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎣ ⎦

Divide the third row by 9− to produce a new third row. 1 1 1 0 0 1 2 1 0 0 1 0 − ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

Convert back to a system of linear equations.

1 2 3 2 3 3 0 2 1 0 x x x x x x − + = + = =

( )

2 3 1 2 3 1 2 1 2 0 1 1 0 1 x x x x x = − = − = = − = − =

So, the solution is: x1 = 1, x2 = 1,and x3 = 0.

21. Because the matrix is in row-echelon form, convert back to a system of linear equations.

1 2 4 2 3 4 3 4 4 2 4 2 3 2 1 4 x x x x x x x x x + + = + + = + = =

( )

3 4 2 3 4 1 2 4 1 2 1 2 4 7 3 2 3 2 7 4 13 4 2 4 2 13 4 26 x x x x x x x x = − = − = − = − − = − − − = = − − = − − = −

So, the solution is: x1 = −26, x2 = 13, x3 = − and 7, 4 4.

x =

23. The augmented matrix for this system is 1 2 7 . 2 1 8 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦

Adding − times the first row to the second row yields a 2 new second row.

1 2 7

0 3 6

⎡ ⎤

⎢ ₋ ₋ ⎥

⎣ ⎦

Dividing the second row by 3− yields a new second row. 1 2 7

0 1 2

⎡ ⎤

⎢ ⎥

⎣ ⎦

Converting back to a system of linear equations produces

2 7 2. x y y + = =

Finally, using back-substitution you find that x = and 3 2.

y =

25. The augmented matrix for this system is 1 2 1.5 . 2 4 3 − ⎡ ⎤ ⎢ ₋ ⎥ ⎣ ⎦

Gaussian elimination produces the following. 1 2 1.5 2 4 3 − ⎡ ⎤ _⇒ ⎢ ₋ ⎥ ⎣ ⎦ 3 2 1 2 2 4 3 − ⎡ − ⎤ ⇒ ⎢ ⎥ − ⎢ ⎥ ⎣ ⎦ 3 2 1 2 0 0 6 − ⎡ − ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

Because the second row of this matrix corresponds to the equation 0 = 6,you can conclude that the original system has no solution.

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27. The augmented matrix for this system is

3 5 22 3 4 4 . 4 8 32 − − ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎣ ⎦

Dividing the first row by 3− yields a new first row.

5 22 3 3 1 3 4 4 4 8 32 ⎡ − ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎣ ⎦

Adding − times the first row to the second row 3 yields a new second row.

5 22 3 3 1 0 9 18 4 8 32 ⎡ − ⎤ ⎢ ⎥ − ⎢ ⎥ ⎢ ₋ ⎥ ⎣ ⎦

Adding − times the first row to the third row 4 yields a new third row.

5 22 3 3 8 4 3 3 1 0 9 18 0 ⎡ − ⎤ ⎢ ⎥ − ⎢ ⎥ ⎢ ₋ ⎥ ⎣ ⎦

Dividing the second row by 9 yields a new second row. 5 22 3 3 8 4 3 3 1 0 1 2 0 ⎡ − ⎤ ⎢ ⎥ − ⎢ ⎥ ⎢ ₋ ⎥ ⎣ ⎦ Adding 4

3times the second row to the third row

yields a new third row.

5 22 3 3 1 0 1 2 0 0 0 ⎡ − ⎤ ⎢ ⎥ − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

Converting back to a system of linear equations produces 5 22 3 3 2. x y y − = = −

Finally, using back-substitution you find that the solution is: 4x = and 2.y = −

29. The augmented matrix for this system is

1 0 3 2 3 1 2 5 . 2 2 1 4 − − ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

Gaussian elimination produces the following.

1 0 3 2 1 0 3 2 1 0 3 2 3 1 2 5 0 1 7 11 0 1 7 11 2 2 1 4 0 2 7 8 0 0 7 14 − − − − − − ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ₋ ⎥_⇒ ⎢ ⎥_⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ − − ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Back substitution now yields

( )

3 2 3 1 3 2 11 7 11 7 2 3 2 3 2 3 2 4. x x x x x = = − = − = − = − + = − + =

So, the solution is: x1 = 4, x2 = −3, and x3 = 2. 31. The augmented matrix for this system is

1 1 5 3 1 0 2 1 . 2 1 1 0 − ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ − − ⎥ ⎣ ⎦

Subtracting the first row from the second row yields a new second row.

1 1 5 3 0 1 3 2 2 1 1 0 − ⎡ ⎤ ⎢ ₋ ₋ ⎥ ⎢ ⎥ ⎢ − − ⎥ ⎣ ⎦

Adding − times the first row to the third row 2 yields a new third row.

1 1 5 3 0 1 3 2 0 3 9 6 − ⎡ ⎤ ⎢ ₋ ₋ ⎥ ⎢ ⎥ ⎢ − − ⎥ ⎣ ⎦

Multiplying the second row by 1− yields a new second row. 1 1 5 3 0 1 3 2 0 3 9 6 − ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ − − ⎥ ⎣ ⎦

Adding 3 times the second row to the third row yields a new third row.

1 1 5 3 0 1 3 2 0 0 0 0 − ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

Adding − times the second row to the first row 1 yields a new first row.

1 0 2 1 0 1 3 2 0 0 0 0 − ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

1 3 2 3 2 1 3 2. x x x x − = − =

Finally, choosing x3 = as the free variable you can t describe the solution as x1 = +1 2 ,t x2 = 2+ 3 ,t and

3 ,

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33. The augmented matrix for this system is 4 12 7 20 22 . 3 9 5 28 30 − − ⎡ ⎤ ⎢ ₋ ₋ ⎥ ⎣ ⎦

Dividing the first row by 4 yields a new first row.

7 11 2 4 1 3 5 3 9 5 28 30 − ⎡ − ⎤ ⎢ ⎥ − − ⎢ ⎥ ⎣ ⎦

Adding − times the first row to the second row yields a 3 new second row.

7 11 2 4 1 27 4 2 1 3 5 0 0 13 − ⎡ − ⎤ ⎢ ⎥ − ⎢ ⎥ ⎣ ⎦

Multiplying the second row by 4 yields a new second row. 7 11 2 4 1 3 5 0 0 1 52 54 − ⎡ − ⎤ ⎢ ⎥ − ⎢ ⎥ ⎣ ⎦

7 11 4 2 3 5 52 54. x y z w z w + − − = − =

Choosing y = and ws = as the free variables you t can describe the solution as x =100− 3s + 96 ,t

, 54 52 ,

y = s z = + t and ,w = t where s and t are any real numbers.

35. The augmented matrix for this system is 3 3 12 6 1 1 4 2 . 2 5 20 10 1 2 8 4 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢− ⎥ ⎣ ⎦

Gaussian elimination produces the following.

3 3 12 6 1 1 4 2 1 1 4 2 1 1 4 2 2 5 20 10 2 5 20 10 1 2 8 4 1 2 8 4 1 1 4 2 1 1 4 2 0 0 0 0 0 3 12 6 0 3 12 6 0 0 0 0 0 3 12 6 0 0 0 0 1 1 4 2 1 0 0 0 0 1 4 2 0 1 4 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ _⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢− ⎥ ⎢− ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⇒ _⎢ _⎥ ⇒ _⎢ _⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⇒ _⎢ _⎥ ⇒ _⎢ _⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

Letting z = be the free variable, the solution is: t

0, 2 4 , ,

x = y = − t z = t where t is any real number.

37. Using a computer software program or graphing utility, the augmented matrix reduces to

1 0 0 0.5278 23.5361 0 1 0 4.1111 18.5444 . 0 0 1 2.1389 7.4306 − ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎣ ⎦

Letting x4 = be the free variable, you obtain t 1 2 3 4 23.5361 0.5278 18.5444 4.1111 7.43062 2.1389

, where is any real number.

x t x t x t x t t = + = + = + =

1 0 0 0 0 2 0 1 0 0 0 2 . 0 0 1 0 0 3 0 0 0 1 0 5 0 0 0 0 1 1 ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ − ⎥ ⎢ ⎥ ⎣ ⎦

So, the solution is: x1 = 2, x2 = −2,x3 = 3, x4 = − 5, and x5 =1.

1 0 0 0 0 0 1 0 1 0 0 0 0 2 0 0 1 0 0 0 6 . 0 0 0 1 0 0 3 0 0 0 0 1 0 4 0 0 0 0 0 1 3 ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ − ⎥ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

So, the solution is: x1 = 1,x2 = −2, x3 = 6,

4 3, 5 4,

x = − x = − and x6 = 3.

43. The corresponding system of equations is 1 2 3 0 0 0 0 x x x = + = =

Letting x3 = be the free variable, the solution is: t

1 0, 2 , 3 ,

x = x = −t x = t where t is any real number. 45. The corresponding system of equations is

1 4 3 0 0 0 0 x x x + = = =

Letting x4 = and t x3 = be the free variables, the s solution is: x1 = −t, x2 = s, x3 = 0, x4 = t.

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47. (a) Because there are two rows and three columns, there are two equations and two variables. (b) Gaussian elimination produces the following.

1 2 1 2 1 2 7 3 4 1 0 4 3 7 0 1 4 3 k k k k k ⎡ ⎤ ⎡ ⎤ _⇒ ⎡ ⎤ _⇒ _⎢ _⎥ ⎢₋ ⎥ ⎢ ₊ ⎥ _⎢ _⎥ ⎣ ⎦ ⎣ ⎦ _⎢_⎣ ₊ _⎥_⎦

The system is consistent if 4+ 3k ≠ 0,or if 4 3.

k ≠ −

(c) Because there are two rows and three columns, there are two equations and three variables. (d) Gaussian elimination produces the following.

1 2 0 1 2 0 1 2 0 7 3 4 1 0 0 4 3 7 0 0 1 0 4 3 k k k k k ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ _⎢ _⎥ ⇒ ⇒ ⎢₋ ⎥ ⎢ ₊ ⎥ _⎢ _⎥ ⎣ ⎦ ⎣ ⎦ _⎢_⎣ ₊ _⎥_⎦ Notice that 4+3k ≠ 0,or 4 3. k ≠ − But if 4 3

− is substituted for k in the original matrix. Guassian elimination produces the following:

4 4 3 3 1 2 0 1 2 0 3 4 1 0 0 0 7 0 ⎡ − ⎤ ⎡ − ⎤ ⇒ ⎢ ⎥ ⎢ ⎥ − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

The system is consistent. So, the original system is consistent for all real k. 49. Begin by forming the augmented matrix for the system

1 1 0 2 0 1 1 2 . 1 0 1 2 0 a b c ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

Then use Gauss-Jordan elimination as follows.

1 1 0 2 1 1 0 2 0 1 1 2 0 1 1 2 0 1 1 0 0 1 1 0 0 0 2 1 1 0 2 1 1 0 2 0 1 1 2 0 1 1 2 0 0 2 2 0 0 2 2 0 2 0 0 2 1 1 0 2 1 1 0 2 0 1 1 2 0 1 1 2 0 0 1 1 0 0 1 1 0 0 2 0 0 0 a b c b a c a b a c a a b c b a b c b a b c ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ _⇒ ⎢ ⎥ ⎢ − ⎥ ⎢ − ⎥ ⎢ ⎥ ⎢ ₋ ₋ ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⇒ _⎢ _⎥ ⇒ _⎢ _⎥ ⎢ ₋ ₋ ⎥ ⎢ ₋ ₊ ₋ ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⇒ _⎢ _⎥ ⇒ _⎢ _⎥ ⎢ ₋ ₊ ₋ ⎥ ⎢ ₊ ₊ ⎥ ⎣ ⎦ ⎣ ⎦ ⇒ 1 1 0 2 1 0 0 1 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 1 0 0 0 a b c 0 0 0 a b c ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ _⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ₊ ₊ ⎥ ⎢ ₊ ₊ ⎥ ⎣ ⎦ ⎣ ⎦

Converting back to a system of linear equations 1 1 1 0 . x y z a b c = = = = + + The system

(a) will have a unique solution if a+ b+c = 0, (b) will have no solution if a +b+ c ≠ 0,and (c) cannot have an infinite number of solutions.

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51. Solve each pair of equations by Gaussian elimination as follows. (a) Equations 1 and 2:

5 8 6 3 5 8 6 3 1 0 4 2 5 16 0 1 1 1 0 0 ⎡ ⎤ − ⎡ ⎤ ⇒ ⎢ ⎥ ⇒ ⎢ ⎥ ₋ ₋ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 8 5 3 6 8 5 3 6 , , x t y t z t = − = − + = (b) Equations 1 and 3: 36 11 14 14 13 40 14 14 1 0 4 2 5 16 0 1 1 3 2 6 ⎡ ⎤ − ⎡ ⎤ ⇒ ⎢ ⎥ ⇒ ⎢₋ ₋ ⎥ ₋ ₋ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 18 11 7 14 20 13 7 14 , , x t y t z t = − = − + = (c) Equations 2 and 3: 1 1 0 0 1 0 1 3 1 3 2 6 0 1 1 3 ⎡ ⎤ ⎡ ⎤ ⇒ ⇒ ⎢₋ ₋ ⎥ ⎢ ₋ ₋ ⎥ ⎣ ⎦ ⎣ ⎦ x_y 3₃ t,_t_, z t = − = − + =

(d) Each of these systems has an infinite number of solutions. 53. Use Gauss-Jordan elimination as follows.

1 2 1 2 1 2 1 0

1 2 0 4 0 1 0 1

⎡ ⎤ _⇒ ⎡ ⎤ _⇒ ⎡ ⎤ _⇒ ⎡ ⎤

⎢₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

55. Begin by finding all possible first rows.

[

0 0 , 0 1 , 1 0 , 1

] [

k

]

.

For each of these, examine the possible second rows.

0 0 0 1 1 0 1 , , , 0 0 0 0 0 1 0 0 k ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

These represent all possible 2× reduced row-echelon 2 matrices.

57. (a) True. In the notation m× n m, is the number of rows of the matrix. So, a 6× matrix has six rows. 3 (b) True. On page 19, after Example 4, the sentence

reads, “It can be shown that every matrix is row-equivalent to a matrix in row-echelon form.” (c) False. Consider the row-echelon form

1 0 0 0 0 0 1 0 0 1 0 0 1 0 2 0 0 0 1 3 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

which gives the solution x1 = 0,x2 = 1, x3 = 2, 4

and x = 3.

(d) True. Theorem 1.1 states that if a homogeneous system has fewer equations than variables, then it must have an infinite number of solutions.

59. First, a and c cannot both be zero. So, assume 0,

a ≠ and use row reduction as follows.

0 0 a b a b a b cb c d d ad bc a ⎡ ⎤ ⎡ ⎤ _⇒ _⎢ _⎥ _⇒ ⎡ ⎤ − ⎢ ⎥ _⎢ ₊ _⎥ ⎢ ₋ ⎥ ⎣ ⎦ _⎢_⎣ _⎥_⎦ ⎣ ⎦

So, ad − bc ≠ 0.Similarly, if c ≠ 0,interchange rows and proceed as above. So the original matrix is row equivalent to the identity if and only if ad −bc ≠ 0. 61. Form the augmented matrix for this system

2 1 0 1 2 0 λ λ − ⎡ ⎤ ⎢ ₋ ⎥ ⎣ ⎦

and reduce the system using elementary row operations.

2 1 2 0 1 2 0 2 1 0 0 4 3 0 λ λ λ λ λ − − ⎡ ⎤ _⇒ ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ₋ ₊ ⎥ ⎣ ⎦ ⎣ ⎦

To have a nontrivial solution you must have

(

)(

)

2 ₄ ₃ ₀ 1 3 0. λ λ λ λ − + = − − =

So, if 1λ = or 3,λ = the system will have nontrivial solutions.

63. To show that it is possible you need give only one example, such as 1 2 3 1 2 3 0 1 x x x x x x + + = + + =

which has fewer equations than variables and obviously has no solution.

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Section 1.3 Applications of Systems of Linear Equations 15 65. a b a c b d a c b d c d c d c d c d a b a b a b − − − − − − ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⇒ ⇒ ⇒ ⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

The rows have been interchanged. In general, the second and third elementary row operations can be used in this manner to interchange two rows of a matrix. So, the first elementary row operation is, in fact, redundant.

67. When a system of linear equations is inconsistent, the row-echelon form of the corresponding augmented matrix will have a row that is all zeros except for the last entry.

69. A matrix in reduced echelon form has zeros above each row’s leading 1, which may not be true for a matrix in row-echelon form.

Section 1.3 Applications of Systems of Linear Equations

1. (a) Because there are three points, choose a second- degree polynomial,

( )

2

0 1 2 .

p x = a + a x+ a x Then substitute x = 2, 3,and 4 into p x and equate the

( )

results to y = 5, 2,and 5, respectively.

( )

2 0 1 2 0 1 2 2 0 1 2 0 1 2 2 0 1 2 0 1 2 2 2 2 4 5 3 3 3 9 2 4 4 4 16 5 a a a a a a a a a a a a a a a a a a + + = + + = + + = + + = + + = + + =

Form the augmented matrix 1 2 4 5 1 3 9 2 1 4 16 5 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

and use Gauss-Jordan elimination to obtain the equivalent reduced row-echelon matrix

1 0 0 29 0 1 0 18 . 0 0 1 3 ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ So, _{p x}

( )

₌ ₂₉ ₋₁₈_x ₊_{3 .}_x2 (b)

3. (a) Because there are three points, choose a second- degree polynomial,

( )

2

0 1 2 .

p x = a + a x+ a x Then substitute x = 2, 3, and 5 into p x and equate the

( )

results to y = 4, 6,and 10, respectively.

( )

2 0 1 2 0 1 2 2 0 1 2 0 1 2 2 0 1 2 0 1 2 2 2 2 4 4 3 3 3 9 6 5 5 5 25 10 a a a a a a a a a a a a a a a a a a + + = + + = + + = + + = + + = + + =

Use Gauss-Jordan elimination on the augmented matrix for this system.

1 2 4 4 1 0 0 0 1 3 9 6 0 1 0 2 1 5 25 10 0 0 1 0 ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ _⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ So, p x

( )

= 2 .x (b) x 1 1 2 2 3 3 4 4 5 5 6 6 (2, 5) (4, 5) (3, 2) y x 2 1 2 3 4 5 4 6 8 10 (2, 4) (3, 6) (5, 10) y

www.elsolucionario.org

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5. (a) Using the translation z = x −2007,the points

(

z y are ,

)

(

−1, 5 , 0, 7 ,

) (

)

and

(

1, 12 . Because there are three points,

)

choose a second-degree polynomial

( )

2

0 1 2 .

p z = a + a z+ a z Then substitute z = −1, 0,and 1 into p z and equate the

( )

results to y = 5, 7,and 12 respectively.

( )

2 0 1 2 0 1 2 2 0 1 2 0 2 0 1 2 0 1 2 1 1 5 0 0 7 1 9 1 12 a a a a a a a a a a a a a a a + − + − = − + = + + = = + + = + + =

7 2 3 2 1 1 1 5 1 0 0 7 1 0 0 7 0 1 0 1 1 1 12 0 0 1 ⎡ ⎤ − ⎡ ⎤ ⎢ ⎥ ⎢ _{⎥ ⇒ ⎢} ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ So,

( )

7 3 2 2 2 7 . p z = + z + z

Letting z = x− 2007,you have

( )

7

(

)

3

(

)

2

2 2

7 2007 2007 .

p x = + x − + x−

(b)

7. Choose a fourth-degree polynomial and substitute 1, 2, 3,

x = and 4 into

( )

2 3 4

0 1 2 3 4 .

p x = a + a x+ a x + a x + a x However, when you substitute x = into 3 p x and equate it to

( )

2

y = and y = you get the contradictory equations 3

0 1 2 3 4 0 1 2 3 4 3 9 27 81 2 3 9 27 81 3 a a a a a a a a a a + + + + = + + + + =

and must conclude that the system containing these two equations will have no solution. Also, y is not a function of x because the x-value of 3 is repeated. By similar reasoning, you cannot choose

( )

2 3 4

0 1 2 3 4

p y = b + b y+b y + b y + b y because 1

y = corresponds to both x = and 2.1 x =

9. Letting

( )

2

0 1 2 ,

p x = a + a x+ a x substitute x = 0, 2, and 4 into p x and equate the results to

( )

y = 1, 3,and 5, respectively.

( )

2 0 1 2 0 2 0 1 2 0 1 2 2 0 1 2 0 1 2 0 0 1 2 2 2 4 3 4 4 4 16 5 a a a a a a a a a a a a a a a a + + = = + + = + + = + + = + + =

1 0 0 1 1 0 0 1 1 2 4 3 0 1 0 1 1 4 16 5 0 0 1 0 ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ _⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ So, p x

( )

= +1 x.

The graphs of y =1 p x

( )

= 1 1

(

+ x

)

and that of the

function 7 1 2

15 15

1

y = − x+ x are shown below. z 3 9 12 −1 1 (2006) (2007) (2008) (−1, 5) (0, 7) (1, 12) y 3 4 5 (0, 1) 2, 4, 1 ₁ 3 y =₁₅x2−₁₅7x+ 1

)

₎

)

₎

y y = 1 1 + x

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Section 1.3 Applications of Systems of Linear Equations 17

11. To begin, substitute x = − and 11 x = into

( )

2 3

0 1 2 3

p x = a +a x+ a x + a x and equate the results to 2y = and 2,y = − respectively. 0 1 2 3 0 1 2 3 2 2 a a a a a a a a − + − = + + + = −

Then, differentiate p, yielding

( )

2 1 2 2 3 3 .

p x′ = a + a x + a x Substitute 1x = − and 1x = into p x′

( )

and equate the results to 0.

1 2 3 1 2 3 2 3 0 2 3 0 a a a a a a − + = + + =

Combining these four equations into one system and forming the augmented matrix, you obtain

1 1 1 1 2 1 1 1 1 2 . 0 1 2 3 0 0 1 2 3 0 − − ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ − ⎥ ⎢ ⎥ ⎣ ⎦

Use Gauss-Jordan elimination to find the equivalent reduced row-echelon matrix 1 0 0 0 0 0 1 0 0 3 . 0 0 1 0 0 0 0 0 1 1 ⎡ ⎤ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

So, _{p x}

( )

_{= −}₃_x ₊ _x3_._{The graph of}_y ₌ _{p x}

( )

_{is shown below.}

13. (a) Because you are given three points, choose a second-degree polynomial,

( )

2

0 1 2 .

p x = a + a x+ a x Because the x-values are large, use the translation z = x−1990 to obtain

(

−10, 227 , 0, 249 , 10, 281 .

) (

)

Substituting the given points into

( )

p x produces the following system of linear equations.

( )

2 0 1 2 0 1 2 2 0 1 2 0 2 0 1 2 0 1 2 10 10 10 100 227 0 0 249 10 10 10 100 281 a a a a a a a a a a a a a a a a + − + − = − + = + + = = + + = + + =

Form the augmented matrix

1 10 100 227 1 0 0 249 1 10 100 281 − ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

1 0 0 249 0 1 0 2.7 . 0 0 1 0.05 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

So, _{p z}

( )

₌ ₂₄₉₊ _2.7_z ₊ _0.05_z2_and_{p x}

( )

₌ ₂₄₉₊ _2.7

(

_x ₋₁₉₉₀

)

₊ _0.05

(

_x₋_{1990 .}

)

2 _{To predict the population in 2010} and 2020, substitute these values into p x

( )

.

p

(

2010

)

= 249 + 2.7 20

( )

+0.05 20

( )

2 = 323 million

(

₂₀₂₀

)

₂₄₉ _{2.7 30}

( )

_{0.05 30}

( )

2 _{375 million} p = + + = −1 2 −2 −1 x (−1, 2) (1, −2) y 1

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15. (a) Letting z = x− 2000the four points are

(

1, 10,003 ,

)

(

3, 10,526 ,

)

(

5, 12,715 , and

)

(

7, 14,410 .

)

Let

( )

2 3 0 1 2 3 . p z = a + a z+ a z + a z

( )

2 3 0 1 2 3 0 1 2 3 2 3 0 1 2 3 0 1 2 3 2 3 0 1 2 3 0 1 2 3 2 3 0 1 2 3 0 1 2 3 1 1 1 10,003 3 3 3 3 9 27 10,526 5 5 5 5 25 125 12,715 7 7 7 7 49 343 14,410 a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a + + + = + + + = + + + = + + + = + + + = + + + = + + + = + + + =

(b) Use Gauss-Jordan elimination to solve the system. 1 1 1 1 10,003 1 0 0 0 11,041.25 1 3 9 27 10,526 0 1 0 0 1606.5 1 5 25 125 12,715 0 0 1 0 613.25 1 7 49 343 14,410 0 0 0 1 45 ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ _⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ − ⎥ ⎣ ⎦ ⎣ ⎦ So, _{p z}

( )

₌_11,041.25₊_1606.5_z ₊ _613.25_z2 ₋_{45 .}_z3 Letting z = x− 2000, _{p x}

( )

₌ _11,041.25₊_1606.5

(

_x₋ ₂₀₀₀

)

₊_613.25

(

_x₋ ₂₀₀₀

)

2 ₋ ₄₅

(

_x₋ _{2000 .}

)

3

Because the actual net profit increased each year from 2000 −2007except for 2006 and the predicted values decrease each year after 2008, the solution does not produce a reasonable model for predicting future net profits.

17. Choosing a second-degree polynomial approximation,

( )

2

0 1 2 ,

p x = a + a x+ a x substitute 0, , 2 x = π and

π into p x and equate the results to

( )

y = 0, 1, and 0, respectively. 0 2 2 0 1 2 0 1 2 0 1 2 4 0 a a a a a a a π π π π = + + = + + =

Then form the augmented matrix,

2 2 1 0 0 0 1 1 2 4 1 0 π π π π ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

2 1 0 0 0 4 0 1 0 _. 4 0 0 1 π π ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ So,

( )

2

(

2

)

2 2 4 4 4 _. p x x x πx x π π π = − = − Furthermore, sin 3 π ≈ 2 2 2 2 4 3 3 3 4 2 8 _0.889. 9 9 p π π π π π π π ⎡ ⎤ ⎛ ⎞ ₌ _⎢ ⎛ ⎞₋⎛ ⎞ _⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎢_⎣ ⎝ ⎠ ⎝ ⎠ ⎥_⎦ ⎡ ⎤ = _⎢ _⎥ = ≈ ⎣ ⎦ 19. (i)

( )

2 0 1 2 0 1 2 2 0 1 2 0 2 0 1 2 0 1 2 1 1 1 0 0 0 1 1 1 p a a a a a a p a a a a p a a a a a a − = + − + − = − + = + + = = + + = + + (ii) 0 1 2 0 0 1 2 0 0 0 a a a a a a a − + = = + + =

(iii) From the augmented matrix 1 1 1 0 1 0 0 0 1 1 1 0 − ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

1 0 0 0 0 1 0 0 . 0 0 1 0 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ So, a0 = 0,a1 = 0,and a2 = 0.

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Section 1.3 Applications of Systems of Linear Equations 19

21. (a) Each of the network’s six junctions gives rise to a linear equation as shown below. input = output 1 3 1 2 4 2 5 3 6 4 7 6 5 7 600 500 600 500 x x x x x x x x x x x x x x = + = + + = + = + = = +

Rearrange these equations, form the augmented matrix, and use Gauss-Jordan elimination.

1 0 1 0 0 0 0 600 1 0 0 0 0 1 0 0 1 1 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 1 0 0 500 0 0 1 0 0 1 0 600 0 0 1 0 0 1 0 600 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 0 1 0 1 500 0 0 0 0 1 0 1 500 0 0 0 0 0 0 0 0 − ⎡ ⎤ ⎡ ⎤ ⎢ ₋ ₋ ⎥ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⇒ ⎢ ⎥ ⎢ ⎥ ⎢ − ⎥ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

Letting x7 = and t x6 = be the free variables, you have s 1 2 3 4 5 6 600 500 x s x t x s x s t x t x s = = = − = − = − = 7 ,

x = t where s and t are any real numbers.

(b) If x6 = x7 = 0,then the solution is x1 = 0, x2 = 0, x3 = 600, x4 = 0, x5 = 500, x6 = 0, x7 = 0.

(c) If x5 = 1000and x6 = 0,then the solution is x1 = 0, x2 = −500, x3 = 600, x4 = 500, x5 = 1000, x6 = 0,and 7 500.

x = −

23. (a) Each of the network’s four junctions gives rise to a linear equation, as shown below. input = output 2 1 4 2 3 4 1 3 200 100 200 100 x x x x x x x x + = = + = + + =

Rearranging these equations and forming the augmented matrix, you obtain

1 1 0 0 200 0 1 0 1 100 . 0 0 1 1 200 1 0 1 0 100 − ⎡ ⎤ ⎢ ₋ ₋ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ − − ⎥ ⎣ ⎦

Gauss-Jordan elimination produces the matrix 1 0 0 1 100 0 1 0 1 100 . 0 0 1 1 200 0 0 0 0 0 − ⎡ ⎤ ⎢ ₋ ₋ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

Letting x4 = t,you have x1 = 100+ t, x2 = −100 +t, x3 = 200+ and t, x4 = t,where t is a real number. (b) When x₄ = t = 0,then x₁ =100, x2 = −100, x3 = 200.

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25. Applying Kirchoff’s first law to either junction produces

1 3 2

I + I = I

and applying the second law to the two paths produces

1 1 2 2 1 2 2 2 3 3 2 3 4 3 3 3 4 R I R I I I R I R I I I + = + = + = + =

Rearrange these equations, form the augmented matrix, and use Gauss-Jordan elimination.

1 1 1 0 1 0 0 0 4 3 0 3 0 1 0 1 0 3 1 4 0 0 1 1 − ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ _⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ So, I1 = 0, I2 = 1,and I3 = 1.

27. (a) To find the general solution, let A have a volts and B have b volts. Applying Kirchoff’s first law to either

junction produces

1 3 2

I + I = I

and applying the second law to the two paths produces 1 1 2 2 1 2 2 2 3 3 2 3 2 2 3 R I R I I I a R I R I I b + = + = + = + =

Rearrange these three equations and form the augmented matrix. 1 1 1 0 1 2 0 0 2 4 a b − ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

Gauss-Jordan elimination produces the matrix

(

)

(

)

(

)

1 0 0 3 7 0 1 0 4 14 . 0 0 1 3 2 14 a b a b b a ⎡ − ⎤ ⎢ ₊ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎣ ⎦

When 5a = and 8,b = then 1 1, 2 2, 3 1.

I = I = I =

(b) When a = 2and 6,b = then 1 0, 2 1, 3 1. I = I = I = 29.

(

) (

)

(

)

2 2 2 4 1 1 1 1 1 x A B C x x x + x− = − + + + x+

(

)

(

)(

)

(

)

(

)

(

)

2 2 2 2 2 2 2 4 1 1 1 1 4 2 4 2 x A x B x x C x x Ax Ax A Bx B Cx C x A B x A C x A B C = + + + − + − = + + + − + − = + + + + − − So, 4 2 0 . 0 A B A C A B C + = + = − − =

Use Gauss-Jordan elimination to solve the system.

1 1 0 4 1 0 0 1 2 0 1 0 0 1 0 3 1 1 1 0 0 0 1 2 ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ _⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ₋ ₋ ⎥ ⎢ ₋ ⎥ ⎣ ⎦ ⎣ ⎦

The solution is: 1,A = B = 3, andC = − 2 So,

(

) (

)

(

)

2 2 2 4 1 3 2 _. 1 1 1 1 1 x x x x + x− = − + + − x + 31.

(

)(

)

(

)

2 2 2 20 2 2 2 2 2 x A B C x x x x x − ₌ ₊ ₌ + − + − −

(

)

(

)(

)

(

)

(

)

(

)

2 2 2 2 2 2 2 20 2 2 2 2 20 4 4 4 2 20 4 4 4 2 x A x B x x C x x Ax Ax A Bx B Cx C x A B x A C x A B C − = − + + − + + − = − + + − + + − = + + − + + − + So, 1 4 0 4 4 2 20. A B A C A B C + = − − + = − + =

1 1 0 1 1 0 0 1 4 0 1 0 0 1 0 2 4 4 2 20 0 0 1 4 − ⎡ ⎤ ⎡ ⎤ ⎢₋ ⎥ _⇒ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

The solution is: 1,A = B = −2, and C = 4. So,

(

)(

)

(

)

2 2 2 20 1 2 4 _. 2 2 2 2 2 x x x x x x − ₌ ₋ ₊ + − + − −

33. Use Gauss-Jordan elimination to solve the system

2 0 1 0 1 0 0 2 0 2 1 0 0 1 0 2 1 1 0 4 0 0 1 4 ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ _⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎣ ⎦ ⎣ ⎦ So, 2x = y = and 2 λ = − 4.

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Review Exercises for Chapter 1 21

35. Let x1 = number of touchdowns, x2 = number of extra-point kicks, and x3 = number of field goals.

1 2 3 1 2 3 2 3 13 6 3 46 0 x x x x x x x x + + = + + = − =

1 1 1 13 1 0 0 5 6 1 3 46 0 1 0 4 0 1 1 0 0 0 1 4 ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ _⇒ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

Because x1 = 5, x2 = 4,and x3 = 4,there were 5 touchdowns, 4 extra-point kicks, and 4 field goals.

Review Exercises for Chapter 1

1 2 ,

a x+ a y = b it is not linear in the variables x and y. 3. Because the equation is in the form a x1 + a y2 = b,it is

linear in the variables x and y.

1 2 ,

a x+ a y = b it is not linear in the variables x and y. 7. Because the equation is in the form a x1 + a y2 = b,it is

linear in the variables x and y.

9. Choosing y and z as the free variables and letting y = and ,s z = t you have

3 1 1 4 2 2 4 2 6 1 4 1 2 6 . x s t x s t x s t − + − = − = − + = − + −

So, the solution set can be described as 1 1 3 4 2 2 ,

x= − + s− t

, ,

y = s z = t where s and t are real numbers.

11. Row reduce the augmented matrix for this system.

1 2 3 3 2 2 1 1 2 1 0 1 1 2 1 1 2 0 1 0 1 3 1 0 0 4 6 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ _⇒ ⎡ ⎤ _⇒ _{⇒ ⎢} ⎥ ⎢ ⎥ ⎢ ₋ ⎥ ⎢ ₋ ₋ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Converting back to a linear system, the solution is: 1 2

x = and 3

2.

y = 13. Rearrange the equations as shown below.

4 2 3 0 x y x y − = − − =

Row reduce the augmented matrix for this system.

1 1 4 1 1 4 1 1 4 1 0 12 2 3 0 0 1 8 0 1 8 0 1 8 − − − − − − − ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⇒ ⇒ ⇒ ⎢ ₋ ⎥ ⎢ ₋ ⎥ ⎢ ₋ ⎥ ⎢ ₋ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Converting back to a linear system, the solution is: x = − and 8.12 y = − 15. Row reduce the augmented matrix for this system.

1 1 0 1 1 0 1 1 0 1 0 0

2 1 0 0 1 0 0 1 0 0 1 0

⎡ ⎤ _⇒ ⎡ ⎤ _⇒ ⎡ ⎤ _⇒ ⎡ ⎤

⎢ ⎥ ⎢ ₋ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Converting back to a linear system, the solution is: x = 0and 0.y =

Elementary Linear Algebra - Larson - 6th Ed

Instructor’s Solutions Manual for