Instructors 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 Senior Marketing Manager: Jennifer Jones Assistant Editor: Janine Tangney Associate Project Editor: Jill Clark Editorial Assistant: Amy Haines Copyright 2009 by Houghton Mifflin Harcourt Publishing Company. All rights reserved. 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 iii 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. iv CONTENTS 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 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........................................2 Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination ..........................9 Section 1.3 Applications of Systems of Linear Equations.....................................15 Review Exercises ..........................................................................................................21 2 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 1 2, a x a y 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 t = and obtain 2 4 02 42 .x tx tx t === So, you can describe the solution set as 2 x t = and , y t = where t is any real number. 9. Choosing y and z as the free variables, let y s = and , z t = and obtain 1 x s t + + = or 1 . x s t = So, you can describe the solution set as 1 , x s t = y s = and , z t = where s and t are any real numbers. 11. From Equation 2 you have 23. x = Substituting this value into Equation 1 produces 13 2 x = or 15. x = So, the system has exactly one solution: 15 x = and 23. x = 13. From Equation 3 you can conclude that 0. z = Substituting this value into Equation 2 produces 322 0 3.yy+ == Finally, by substituting 32y = and 0 z = into Equation 1, you obtain 32320 0.xx + == So, the system has exactly one solution: 3 32 2, , x y = = and 0. z = 15. Begin by rewriting the system in row-echelon form. The equations are interchanged. 1 21 2 32 05 2 0x xx x x+ =+ + = The first equation is multiplied by 12. 121 2 31205 2 0xxx x x+ =+ + = Adding 5 times the first equation to the second equation produces a new second equation. 1232121200xxxx+ == + The second equation is multiplied by 2. 1 22 31202 0x xx x+ = = To represent the solutions, choose 3x to be the free variable and represent it by the parameter t. Because 2 32 x x = and 1 212, x x = you can describe the solution set as 1 2, 2 , x t x t = =3, x t = where t is any real number. 17. 2 42x yx y+ = = Adding the first equation to the second equation produces a new second equation, 3 6, x = or 2. x = So, 0, y = and the solution is 2, x = 0. y = This is the point where the two lines intersect. x123342244 4x y = 22x + y = 4y Sect i on 1. 1 Int roduct i on t o Syst ems of Li near Equat i ons 3 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 19. 12 2 5x yx y = + = Adding 2 times the first equation to the second equation produces a new second equation. 10 7x 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 72 9x yx y =+ = Adding the first equation to 5 times the second equation produces a new second equation, 13 52, x = or 4. x = So, ( ) 2 4 9, y + = or 1, y = and the solution is: 4, x = 1. y = This is the point where the two lines intersect. 23. 2 55 11x yx y = = Subtracting the first equation from the second equation produces a new second equation, 3 6, x = or 2. x = So, ( ) 2 2 5, y = or 1, y = and the solution is: 2, x = 1. y = This is the point where the two lines intersect. 25. 3 114 32 12x yx y+ + = = Multiplying the first equation by 12 produces a new first equation. 3 4 72 12x yx y+ = = Adding the first equation to 4 times the second equation produces a new second equation, 11 55, x = or 5. x = So, ( ) 2 5 12, y = or 2, y = and the solution is: 5, x = 2. y = This is the point where the two lines intersect. 27. 0.05 0.03 0.070.07 0.02 0.16x yx y =+ = Multiplying the first equation by 200 and the second equation by 300 produces new equations. 10 6 1421 6 48x yx y =+ = Adding the first equation to the second equation produces a new second equation, 31 62, x = or 2. x = So, ( ) 10 2 6 14, y = or 1, y = and the solution is: 2, x = 1. y = This is the point where the two lines intersect. 2 2 4 6 8 10x2463x 5y = 72x + y = 9810y1 2 4 5 6 7 8 24624680.05x 0.03y = 0.070.07x + 0.02y = 0.16xyxy2 4 6 4 62x y = 55x y = 1124121 2 3 6 7 2468101246x + 34y 13= 1 +2x y = 12xyx1342 43422x + 2y = 5x y = 1y1 3 44 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 29. 14 63x yx y+ = = Adding 6 times the first equation to the second equation produces a new second equation, 529, x = or 185. x = So, 1853, y = or 35, y = and the solution is: 185, x = 35. 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 12 4, . x y = = (d) Adding 14 times the first equation to the second equation produces the new second equation 343 , y = or 14. y = So, 12, x = and the solution is 12, x =14. 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 942 , , 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 2201x xx == Now, using back-substitution you can conclude that the system has exactly one solution: 11 x = and 21. x = 39. Interchanging the two equations produces the system 2 120. 2 120u vu v+ =+ = Adding 2 times the first equation to the second equation produces a new second equation. 2 1203 120u vv+ = = Solving the second equation you have 40. v = Substituting this value into the first equation gives 80 120 u + = or 40. u = So, the system has exactly one solution: 40 u = and 40. v = 41. Dividing the first equation by 9 produces a new first equation. 1 13 91 2 15 5 3x yx y = + = Adding 15 times the first equation to the second equation produces a new second equation. 1 13 97 1415 45x yy = = Multiplying the second equation by 157produces a new second equation. 1 13 923x yy = = Now, using back-substitution, you can substitute 23y = into the first equation to obtain 2 19 9x + = or 13. x = So, you can conclude that the system has exactly one solution: 13x = and 23. y = 1 1 3 422468x y = 3x4 = 1 +y6xy43433x y = 36 + 2 = 1 x y43432x 8y = 3 12x + y = 043434x 8y = 90.8x 1.6y = 1.8 Sect i on 1. 1 Int roduct i on t o Syst ems of Li near Equat i ons 5 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 43. To begin, change the form of the first equation. 23 1 12 3 62 5x yx y+ = = Multiplying the first equation by 2 yields a new first equation. 23 23 32 5x yx y+ = = Subtracting the first equation from the second equation yields a new second equation. 23 23 38 83 3x yy+ = = Dividing the second equation by 83 yields a new second equation. 23 23 31x yy+ == 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 21 22.5 9.53 4 52x xx x = + = Adding 3 times the first equation to the second equation produces a new second equation. 1 222.5 9.511.5 80.5x xx = = Now, using back-substitution, you can conclude that the system has exactly one solution: 18 x = and 27. x = 47. Adding 2 times the first equation to the second equation yields a new second equation. 63 93 0x y zy zx z+ + = = = Adding 3 times the first equation to the third equation yields a new third equation. 63 93 4 18x y zy zy z+ + = = = Dividing the second equation by 3 yields a new second equation. 13633 4 18x y zyzy z+ + =+ = = Adding 3 times the second equation to the third equation yields a new third equation. 13633 9x y zy zz+ + =+ = = Dividing the third equation by 3 yields a new third equation. 13633x y zy zz+ + =+ == Now, using back-substitution you can conclude that the system has exactly one solution: 1, x = 2, y = and 3. z = 49. Dividing the first equation by 3 yields a new first equation. 12 31 2 31 2 32 4 13 3 32 32 3 6 8xx xx x xx x x + =+ = + = Subtracting the first equation from the second equation yields a new second equation. 12 32 31 2 32 4 13 3 35 10 83 3 32 3 6 8xx xx xx x x + = = + = Adding 2 times the first equation to the third equation yields a new third equation. 12 32 32 32 4 13 3 35 10 83 3 35 10 223 3 3xx xx xx 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. 6 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 51. Dividing the first equation by 2 yields a new first equation. 12 31 31 2 31 32 224 2 102 3 13 8xx xx xx x x+ =+ = + = Adding 4 times the first equation to the second equation produces a new second equation. 12 32 31 2 31 32 222 8 22 3 13 8xx xx xx x x+ = + = + = Adding 2 times the first equation to the third equation produces a new third equation. 12 32 32 31 32 222 8 24 16 4xx xx xx x+ = + = = Dividing the second equation by 2 yields a new second equation. 12 32 32 31 32 224 14 16 4xx xx xx x+ = = = Adding 4 times the second equation to the third equation produces a new third equation. 12 32 31 32 224 10 0xx xx x+ = = = Adding 12 times the second equation to the first equation produces a new first equation. 132 31 52 24 1xxx x+ = = Choosing 3x t = as the free variable, you can describe the solution as 15 12 2, x t = 24 1, x t = and 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 2 180 72x 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 62 3 127 4 5 22. 2 6x y z wy z wy z wy z+ + + = = + + = = Adding 7 times the second equation to the third, and 1 times the second equation to the fourth, produces 62 3 1218 26 1063 6.x y z wy z wz ww+ + + = = + == Using back-substitution, you find the original system has exactly one solution: 1, x = 0, y = 3, z = and 2. w = Answers may vary slightly for Exercises 5763. 57. Using a computer software program or graphing utility, you obtain 1 2 3 415, 40, 45, 75 x x x x = = = = 59. Using a computer software program or graphing utility, you obtain 1.2, 0.6, 2.4. x y z = = = 61. Using a computer software program or graphing utility, you obtain 1 2 31 4 15 5 2, , . x x x = = = 63. Using a computer software program or graphing utility, you obtain 6.8813, 163.3111, x y = = 210.2915, z = 59.2913. w = Sect i on 1. 1 Int roduct i on t o Syst ems of Li near Equat i ons 7 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 65. 0 x y z = = = is clearly a solution. Dividing the first equation by 4 yields a new first equation. 3 174 405 4 22 04 2 19 0xy zx y zx y z+ + =+ + =+ + = Adding 5 times the first equation to the second equation yields a new second equation. 3 174 41 34 4004 2 19 0xy zy zx y z+ + =+ =+ + = Adding 4 times the first equation to the third equation yields a new third equation. 3 174 41 34 4002 0xy zy zy z+ + =+ = + = Multiplying the second equation by 4 yields a new second equation. 3 174 403 02 0xy zy zy z+ + =+ = + = Adding the second equation to the third equation yields a new third equation. 3 174 403 05 0xy zy zz+ + =+ == Dividing the third equation by 5 yields a new third equation. 3 174 403 00xy zy zz+ + =+ == Now, using back-substitution, you can conclude that the system has exactly one solution: 0, x = 0, y = and 0. z = 67. 0 x y z = = = is clearly a solution. Dividing the first equation by 5 yields a new first equation. 15010 5 2 05 15 9 0x yzx y zx y z+ =+ + =+ = Adding 10 times the first equation to the second equation yields a new second equation. 1505 4 05 15 9 0x yzy zx y z+ = + =+ = Adding 5 times the first equation to the third equation yields a new third equation. 1505 4 010 8 0x yzy zy z+ = + = = Dividing the second equation by 5 yields a new second equation. 15450010 8 0x y zy zy z+ = = = Adding 10 times the second equation to the third equation yields a new third equation. 1545000 0x yzyz+ = == Adding 1 times the second equation to the first equation yields a new first equation. 354500xzyz+ = = Choosing z t = as the free variable you find the solution to be 35, x t = 45, 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 t = as the free variable, the solution is ,c bx ta a= , y t = where t is any real number. (b) False. For example, consider the system 1 2 31 2 312x x xx x x+ + =+ + = which is an inconsistent system. (c) False. A consistent system may have only one solution.8 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 71. Because 1x t = and 2 13 4 3 4, x t x = = one answer is the system 1 21 23 4. 3 4x xx x = + = Letting 2, x t = you get 14 4.3 3 3t tx += = + 73. Substituting 1 A x = and 1 B y = into the original system yields 12 12 7. 3 4 0A BA B =+ = Reduce this system to row-echelon form. Dividing the first equation by 12 yields a new first equation. 7123 4 0A BA B =+ = Adding 3 times the first equation to the second equation yields a new second equation. 712747A BB == Dividing the second equation by 7 yields a new second equation. 71214A BB == So, 1 4 A = and 1 3. B = Because 1 A x = and 1 , B y = the solution of the original system of equations is: 3 x = and 4. y = 75. Substituting 1 , A x = 1 , B y = and 1 , C z = into the original system yields 2 3 44 2 102 3 13 8.A B CA CA B C+ =+ = + = Reduce this system to row-echelon form. 1 32 222 8 24 16 4AB CB CB C+ = + = = 3 12 224 1A B CB C+ = = Letting t C = be the free variable, you have , C t = 4 1 B t = and ( ) 5.2tA += So, the solution to the original problem is 2,5xt=1,4 1yt=1, zt= where 15, , 0.4t 77. Reduce the system to row-echelon form. Dividing the first equation by cos yields a new second equation. ( ) ( )sin 1cos cossin cos 0x yx y + = + = Multiplying the first equation by sin and adding to the second equation yields a new second equation. sin 1cos cos1 sincos cosx yy + = = 2 2 2sin sin cos 1Because coscos cos cos ++ = = Multiplying the second equation by cos yields a new second equation. sin 1cos cossinx yy + = = Substituting sin y = into the first equation yields 2 2sin 1sincos cos1 sin coscos .cos cosxx + = = = = So, the solution of the original system of equations is: cos x = and sin . y = 79. For this system to have an infinite number of solutions both equations need to be equivalent. Multiply the second equation by 2. 4 62 2 6x kykx y+ = = So, when 2 k = the system will have an infinite number of solutions. 81. Reduce the system to row-echelon form. ( )201 0x kyk y+ = = 20, 1 0 0x kyk y+ = = 201 0 0,xk y = = If 21 0, k that is if 1, k the system will have exactly one solution. Sect i on 1. 2 Gaussi an El i mi nat i on and Gauss-Jordan El i mi nat i on 9 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 83. To begin, reduce the system to row-echelon form. ( )2 68 3 14x y kzk z+ + = = This system will have no solution if 8 3 0, k = that is, 83. k = 85. Reducing the system to row-echelon form, you have ( ) ( )( ) ( )231 1 11 1 1 3x y kzk y k zk y k z k+ + = + = + = ( ) ( )( )231 1 1. 2 3x y kzk y k zk k z k+ + = + = + = If 22 0, k k + = 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 35 6 13x yx y = = 4 314 28x yy = = 4 32x yy = = 52xy == 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. 110019922y xy 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 1s, 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 1202xx == So, the solution is: 10 x = and 22. x = xx 4y = 35 2 4 4 3432534514y = 28yxx 4y = 35 2 4 4 34325345y = 2yx2 4 3 2 143253145y = 2x = 5y4xx 4y = 35x 6y = 135 2 4 3 4245345y10 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 17. Because the matrix is in row-echelon form, convert back to a system of linear equations. 1 22 3332 11x xx xx = == Solve this system by back-substitution. ( )221 2 11xx == Substituting 21 x = into equation 1, ( )113 1. 2xx == So, the solution is: 1 22, 1, x x = = and 31. x = 19. Interchange the first and second rows. 1 1 1 02 1 1 30 1 2 1 Interchange the second and third rows. 1 1 1 00 1 2 12 1 1 3 Add 2 times the first row to the third row to produce a new third row. 1 1 1 00 1 2 10 3 3 3 Add 3 times the second row to the third row to produce a new third row. 1 1 1 00 1 2 10 0 9 0 Divide the third row by 9 to produce a new third row. 1 1 1 00 1 2 10 0 1 0 Convert back to a system of linear equations. 1 2 32 3302 10x x xx xx + =+ == Solve this system by back-substitution. ( ) 2 31 2 31 2 1 1 2 01 0 1x xx x x= = == = = So, the solution is: 11, x =21, x = and 30. x = 21. Because the matrix is in row-echelon form, convert back to a system of linear equations. 1 2 42 3 43 442 42 32 14x x xx x xx xx+ + =+ + =+ == Solve this system by back-substitution. ( )( )( )3 42 3 41 2 41 2 1 2 4 73 2 3 2 7 4 134 2 4 2 13 4 26x xx x xx x x= = = = = == = = So, the solution is: 126, x = 213, x =37, x = and 44. x = 23. The augmented matrix for this system is 1 2 7.2 1 8 Adding 2 times the first row to the second row yields a new second row. 1 2 70 3 6 Dividing the second row by 3 yields a new second row. 1 2 70 1 2 Converting back to a system of linear equations produces 2 72.x yy+ == Finally, using back-substitution you find that 3 x = and 2. y = 25. The augmented matrix for this system is 1 2 1.5.2 4 3 Gaussian elimination produces the following. 1 2 1.52 4 3 321 22 4 3 321 20 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. Sect i on 1. 2 Gaussi an El i mi nat i on and Gauss-Jordan El i mi nat i on 11 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 27. The augmented matrix for this system is 3 5 223 4 4 .4 8 32 Dividing the first row by 3 yields a new first row. 5 223 313 4 44 8 32 Adding 3 times the first row to the second row yields a new second row. 5 223 310 9 184 8 32 Adding 4 times the first row to the third row yields a new third row. 5 223 38 43 310 9 180 Dividing the second row by 9 yields a new second row. 5 223 38 43 310 1 20 Adding 43times the second row to the third row yields a new third row. 5 223 310 1 20 0 0 Converting back to a system of linear equations produces 5 223 32.x yy == Finally, using back-substitution you find that the solution is: 4 x = and 2. y = 29. The augmented matrix for this system is 1 0 3 23 1 2 5 .2 2 1 4 Gaussian elimination produces the following. 1 0 3 2 1 0 3 2 1 0 3 23 1 2 5 0 1 7 11 0 1 7 112 2 1 4 0 2 7 8 0 0 7 14 Back substitution now yields ( )( )32 31 3211 7 11 7 2 32 3 2 3 2 4.xx xx x== = = = + = + = So, the solution is: 1 2 34, 3, and 2. x x x = = = 31. The augmented matrix for this system is 1 1 5 31 0 2 1 .2 1 1 0 Subtracting the first row from the second row yields a new second row. 1 1 5 30 1 3 22 1 1 0 Adding 2 times the first row to the third row yields a new third row. 1 1 5 30 1 3 20 3 9 6 Multiplying the second row by 1 yields a new second row. 1 1 5 30 1 3 20 3 9 6 Adding 3 times the second row to the third row yields a new third row. 1 1 5 30 1 3 20 0 0 0 Adding 1 times the second row to the first row yields a new first row. 1 0 2 10 1 3 20 0 0 0 Converting back to a system of linear equations produces 1 32 32 13 2.x xx x = = Finally, choosing 3x t = as the free variable you can describe the solution as 11 2 , x t = +22 3 , x t = + and 3, x t = where t is any real number. 12 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 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. 7 112 41 3 53 9 5 28 30 Adding 3 times the first row to the second row yields a new second row. 7 112 41 274 21 3 50 0 13 Multiplying the second row by 4 yields a new second row. 7 112 41 3 50 0 1 52 54 Converting back to a system of linear equations produces 7 114 23 552 54.x y z wz w+ = = Choosing y s = and w t = as the free variables you can describe the solution as 100 3 96 , x s 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 61 1 4 2.2 5 20 101 2 8 4 Gaussian elimination produces the following. 3 3 12 6 1 1 4 21 1 4 2 1 1 4 22 5 20 10 2 5 20 101 2 8 4 1 2 8 41 1 4 2 1 1 4 20 0 0 0 0 3 12 60 3 12 6 0 0 0 00 3 12 6 0 0 0 01 1 4 2 1 0 0 00 1 4 2 0 1 4 20 0 0 0 0 0 0 00 0 0 0 0 0 0 0 Letting z t = be the free variable, the solution is: 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 23.53610 1 0 4.1111 18.5444 .0 0 1 2.1389 7.4306 Letting 4x t = be the free variable, you obtain 123423.5361 0.527818.5444 4.11117.43062 2.1389, where is any real number.x tx tx tx t t= += += += 39. Using a computer software program or graphing utility, the augmented matrix reduces to 1 0 0 0 0 20 1 0 0 0 2. 0 0 1 0 0 30 0 0 1 0 50 0 0 0 1 1 So, the solution is: 1 2 3 42, 2, 3, 5, x x x x = = = = and 51. x = 41. Using a computer software program or graphing utility, the augmented matrix reduces to 1 0 0 0 0 0 10 1 0 0 0 0 20 0 1 0 0 0 6.0 0 0 1 0 0 30 0 0 0 1 0 40 0 0 0 0 1 3 So, the solution is: 1 2 31, 2, 6, x x x = = = 4 53, 4, x x = = and 63. x = 43. The corresponding system of equations is 12 3000 0xx x =+ == Letting 3x t = be the free variable, the solution is: 1 2 30, , , x x t x t = = = where t is any real number. 45. The corresponding system of equations is 1 43000 0x xx + === Letting 4x t = and 3x s = be the free variables, the solution is: 1, x t = 2, x s =30, x =4. x t = Sect i on 1. 2 Gaussi an El i mi nat i on and Gauss-Jordan El i mi nat i on 13 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 47. (a) Because there are two rows and three columns, there are two equations and two variables. (b) Gaussian elimination produces the following. 1 21 2 1 273 4 1 0 4 3 7 0 14 3kk kkk + + The system is consistent if 4 3 0, k + or if 43. 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 01 2 0 1 2 073 4 1 0 0 4 3 7 0 0 1 04 3kk kkk + + Notice that 4 3 0, k + or 43. k But if 43 is substituted for k in the original matrix. Guassian elimination produces the following: 4 43 31 2 0 1 2 03 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 20 1 1 2.1 0 1 20 a b c Then use Gauss-Jordan elimination as follows. 1 1 0 2 1 1 0 20 1 1 2 0 1 1 20 1 1 0 0 1 1 00 0 21 1 0 2 1 1 0 20 1 1 2 0 1 1 20 0 2 2 0 0 2 20 2 0 0 21 1 0 2 1 1 0 20 1 1 2 0 1 1 20 0 1 1 0 0 1 10 0 2 0 0 0a b c b a c ab a c a a b c ba b c b a b c + + + + 1 1 0 2 1 0 0 10 1 0 1 0 1 0 10 0 1 1 0 0 1 10 0 0 0 0 0 a b c a b c + + + + Converting back to a system of linear equations 1110 .xyza b c==== + + The system (a) will have a unique solution if 0, a b c + + = (b) will have no solution if 0, a b c + + and (c) cannot have an infinite number of solutions. 14 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 51. Solve each pair of equations by Gaussian elimination as follows. (a) Equations 1 and 2: 5 86 35 86 31 0 4 2 5 160 1 1 1 0 0 8 53 68 53 6,,x ty tz t= = += (b) Equations 1 and 3: 36 1114 1413 4014 141 0 4 2 5 160 1 1 3 2 6 18 117 1420 137 14,,x ty tz t= = += (c) Equations 2 and 3: 1 1 0 0 1 0 1 31 3 2 6 0 1 1 3 3 ,3 ,x ty tz 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 01 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 0k These represent all possible 2 2 reduced row-echelon matrices. 57. (a) True. In the notation , m n m is the number of rows of the matrix. So, a 6 3 matrix has six rows. (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 00 1 0 0 10 0 1 0 20 0 0 1 3 which gives the solution 1 2 30, 1, 2, x x x = = = 4and 3. x = (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 0a ba b a bcbc d ad bc da + So, 0. ad bc Similarly, if 0, c interchange rows and proceed as above. So the original matrix is row equivalent to the identity if and only if 0. ad bc 61. Form the augmented matrix for this system 2 1 01 2 0 and reduce the system using elementary row operations. 21 2 0 1 2 02 1 0 0 4 3 0 + To have a nontrivial solution you must have ( )( )24 3 01 3 0. + = = So, if 1 = or 3, = the system will have nontrivial solutions. 63. To show that it is possible you need give only one example, such as 1 2 31 2 301x x xx x x+ + =+ + = which has fewer equations than variables and obviously has no solution. Sect i on 1. 3 Appl i cat i ons of Syst ems of Li near Equat i ons 15 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 65. a b a c b d a c b d c d c dc 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 row-echelon form has zeros above each rows 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, ( )20 1 2. p x a a x a x = + + Then substitute 2, 3, x = and 4 into ( ) p x and equate the results to 5, 2, y = and 5, respectively. ( ) ( )( ) ( )( ) ( )20 1 2 0 1 220 1 2 0 1 220 1 2 0 1 22 2 2 4 53 3 3 9 24 4 4 16 5a a a a a aa a a a a aa a a a a a+ + = + + =+ + = + + =+ + = + + = Form the augmented matrix 1 2 4 51 3 9 21 4 16 5 and use Gauss-Jordan elimination to obtain the equivalent reduced row-echelon matrix 1 0 0 290 1 0 18 .0 0 1 3 So, ( )229 18 3 . p x x x = + (b) 3. (a) Because there are three points, choose a second- degree polynomial, ( )20 1 2. p x a a x a x = + + Then substitute 2, 3, x = and 5 into ( ) p x and equate the results to 4, 6, y = and 10, respectively. ( ) ( )( ) ( )( ) ( )20 1 2 0 1 220 1 2 0 1 220 1 2 0 1 22 2 2 4 43 3 3 9 65 5 5 25 10a a a a a aa a a a a aa a a a a a+ + = + + =+ + = + + =+ + = + + = Use Gauss-Jordan elimination on the augmented matrix for this system. 1 2 4 4 1 0 0 01 3 9 6 0 1 0 21 5 25 10 0 0 1 0 So, ( ) 2 . p x x = (b) x112233445566(2, 5) (4, 5)(3, 2)yx21 2 3 4 546810(2, 4)(3, 6)(5, 10)y16 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 5. (a) Using the translation 2007, z x = the points ( ) , z y are ( ) ( ) 1, 5 , 0, 7 , and ( ) 1, 12 . Because there are three points, choose a second-degree polynomial ( )20 1 2. p z a a z a z = + + Then substitute 1, 0, z = and 1 into ( ) p z and equate the results to 5, 7, y = and 12 respectively. ( ) ( )( ) ( )( ) ( )20 1 2 0 1 220 1 2 020 1 2 0 1 21 1 50 0 71 9 1 12a a a a a aa a a aa a a a a+ + = + =+ + = =+ + = + + = Use Gauss-Jordan elimination on the augmented matrix for this system. 72321 1 1 5 1 0 0 71 0 0 7 0 1 01 1 1 12 0 0 1 So, ( )2 7 32 27 . p z z z = + + Letting 2007, z x = you have ( ) ( ) ( )27 32 27 2007 2007 . p x x x = + + (b) 7. Choose a fourth-degree polynomial and substitute 1, 2, 3, x = and 4 into ( )2 3 40 1 2 3 4. p x a a x a x a x a x = + + + + However, when you substitute 3 x = into ( ) p x and equate it to 2 y = and 3 y = you get the contradictory equations 0 1 2 3 40 1 2 3 43 9 27 81 23 9 27 81 3a a a a aa 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 40 1 2 3 4p y b b y b y b y b y = + + + + because 1 y = corresponds to both 1 x = and 2. x = 9. Letting ( )20 1 2, p x a a x a x = + + substitute 0, 2, x = and 4 into ( ) p x and equate the results to 1, 3, y = and 5, respectively. ( ) ( )( ) ( )( ) ( )20 1 2 020 1 2 0 1 220 1 2 0 1 20 0 12 2 2 4 34 4 4 16 5a a a aa a a a a aa a a a a a+ + = =+ + = + + =+ + = + + = Use Gauss-Jordan elimination on the augmented matrix for this system. 1 0 0 1 1 0 0 11 2 4 3 0 1 0 11 4 16 5 0 0 1 0 So, ( ) 1 . p x x = + The graphs of ( ) ( ) 1 1 1 y p x x = = + and that of the function 2 7 115 151 y x x = + are shown below. z39121 1(2006) (2008) (2007)(1, 5)(0, 7)(1, 12)y1 2 43345x(0, 1)2,4,1135y = + 1 x27x15 15))))y1y =11 + x Sect i on 1. 3 Appl i cat i ons of Syst ems of Li near Equat i ons 17 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 11. To begin, substitute 1 x = and 1 x = into ( )2 30 1 2 3p x a a x a x a x = + + + and equate the results to 2 y = and 2, y = respectively. 0 1 2 30 1 2 322a a a aa a a a + =+ + + = Then, differentiate p, yielding ( )21 2 32 3 . p x a a x a x = + + Substitute 1 x = and 1 x = into ( ) p x and equate the results to 0. 1 2 31 2 32 3 02 3 0a a aa a a + =+ + = Combining these four equations into one system and forming the augmented matrix, you obtain 1 1 1 1 21 1 1 1 2.0 1 2 3 00 1 2 3 0 Use Gauss-Jordan elimination to find the equivalent reduced row-echelon matrix 1 0 0 0 00 1 0 0 3.0 0 1 0 00 0 0 1 1 So, ( )33 . p x x x = + The graph of ( ) y p x = is shown below. 13. (a) Because you are given three points, choose a second-degree polynomial, ( )20 1 2. p x a a x a x = + + Because the x-values are large, use the translation 1990 z x = to obtain ( ) ( ) ( ) 10, 227 , 0, 249 , 10, 281 . Substituting the given points into ( ) p x produces the following system of linear equations. ( ) ( )( ) ( )( ) ( )20 1 2 0 1 220 1 2 020 1 2 0 1 210 10 10 100 2270 0 24910 10 10 100 281a a a a a aa a a aa a a a a a+ + = + =+ + = =+ + = + + = Form the augmented matrix 1 10 100 2271 0 0 2491 10 100 281 and use Gauss-Jordan elimination to obtain the equivalent reduced row-echelon matrix 1 0 0 2490 1 0 2.7 .0 0 1 0.05 So, ( )2249 2.7 0.05 p z z z = + + and ( ) ( ) ( )2249 2.7 1990 0.05 1990 . p x x x = + + To predict the population in 2010 and 2020, substitute these values into ( ). p x ( ) ( ) ( )22010 249 2.7 20 0.05 20 323 million p = + + = ( ) ( ) ( )22020 249 2.7 30 0.05 30 375 million p = + + = 1 221x(1, 2)(1, 2)y118 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 15. (a) Letting 2000 z x = the four points are ( ) 1, 10,003 , ( ) 3, 10,526 , ( ) 5, 12,715 , and ( ) 7, 14,410 . Let ( )2 30 1 2 3. p z a a z a z a z = + + + ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2 30 1 2 3 0 1 2 32 30 1 2 3 0 1 2 32 30 1 2 3 0 1 2 32 30 1 2 3 0 1 2 31 1 1 10,0033 3 3 3 9 27 10,5265 5 5 5 25 125 12,7157 7 7 7 49 343 14,410a a a a a a a aa a a a a a a aa a a a a a a aa 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 11,041.251 3 9 27 10,526 0 1 0 0 1606.51 5 25 125 12,715 0 0 1 0 613.251 7 49 343 14,410 0 0 0 1 45 So, ( )2 311,041.25 1606.5 613.25 45 . p z z z z = + + Letting 2000, z x = ( ) ( ) ( ) ( )2 311,041.25 1606.5 2000 613.25 2000 45 2000 . p x x x x = + + Because the actual net profit increased each year from 2000 2007 except 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, ( )20 1 2, p x a a x a x = + + substitute 0, ,2x = and into ( ) p x and equate the results to 0, 1, and 0, y = respectively. 0220 120 1 2012 40aa a aa a a =+ + =+ + = Then form the augmented matrix, 221 0 0 01 12 41 0 and use Gauss-Jordan elimination to obtain the equivalent reduced row-echelon matrix 21 0 0 040 1 0.40 0 1 So, ( ) ( )2 22 24 4 4. p x x x x x = = Furthermore, sin3 222243 3 34 2 80.889.9 9p = = = Note that sin 3 0.866 = to three significant digits. 19. (i) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )20 1 2 0 1 220 1 2 020 1 2 0 1 21 1 10 0 01 1 1p a a a a a ap a a a ap a a a a a a = + + = += + + == + + = + + (ii) 0 1 200 1 2000a a aaa a a + ==+ + = (iii) From the augmented matrix 1 1 1 01 0 0 01 1 1 0 and use Gauss-Jordan elimination to obtain the equivalent reduced row-echelon matrix 1 0 0 00 1 0 0 .0 0 1 0 So, 0 10, 0, a a = = and 20. a = Sect i on 1. 3 Appl i cat i ons of Syst ems of Li near Equat i ons 19 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 21. (a) Each of the networks six junctions gives rise to a linear equation as shown below. input = output 1 31 2 42 53 64 7 65 7600500600500x xx x xx xx xx x xx 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 01 1 0 1 0 0 0 0 0 1 0 0 0 0 1 00 1 0 0 1 0 0 500 0 0 1 0 0 1 0 6000 0 1 0 0 1 0 600 0 0 0 1 0 1 1 00 0 0 1 0 1 1 0 0 0 0 0 1 0 1 5000 0 0 0 1 0 1 500 0 0 0 0 0 0 0 0 Letting 7x t = and 6x s = be the free variables, you have 123456600500x sx tx sx s tx tx s=== = = = 7, x t = where s and t are any real numbers. (b) If 6 70, x x = = then the solution is 10, x =20, x =3600, x =40, x =5500, x =60, x =70. x = (c) If 51000 x = and 60, x = then the solution is 10, x =2500, x = 3600, x =4500, x =51000, x =60, x = and 7500. x = 23. (a) Each of the networks four junctions gives rise to a linear equation, as shown below. input = output 2 14 23 41 3200100200100x xx xx xx x+ == += ++ = Rearranging these equations and forming the augmented matrix, you obtain 1 1 0 0 2000 1 0 1 100.0 0 1 1 2001 0 1 0 100 Gauss-Jordan elimination produces the matrix 1 0 0 1 1000 1 0 1 100.0 0 1 1 2000 0 0 0 0 Letting 4, x t = you have 1100 , x t = +2100 , x t = +3200 , x t = + and 4, x t = where t is a real number. (b) When 40, x t = = then 1100, x =2100, x = 3200. x = (c) When 4100, x t = = then 1 2 3200, 0, 300. x x x = = =20 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 25. Applying Kirchoffs first law to either junction produces 1 3 2I I I + = and applying the second law to the two paths produces 1 1 2 2 1 22 2 3 3 2 34 3 33 4R I R I I IR I R I I I+ = + =+ = + = Rearrange these equations, form the augmented matrix, and use Gauss-Jordan elimination. 1 1 1 0 1 0 0 04 3 0 3 0 1 0 10 3 1 4 0 0 1 1 So, 1 20, 1, I I = = and 31. I = 27. (a) To find the general solution, let A have a volts and B have b volts. Applying Kirchoffs first law to either junction produces 1 3 2I I I + = and applying the second law to the two paths produces 1 1 2 2 1 22 2 3 3 2 322 3R I R I I I aR I R I I b+ = + =+ = + = Rearrange these three equations and form the augmented matrix. 1 1 1 01 2 00 2 4ab Gauss-Jordan elimination produces the matrix ( )( )( )1 0 0 3 70 1 0 4 14 .0 0 1 3 2 14a ba bb a + When 5 a = and 8, b = then 1 2 31, 2, 1. I I I = = = (b) When 2 a = and 6, b = then 1 2 30, 1, 1. I I I = = = 29. ( ) ( ) ( )22 241 11 1 1x A B Cx xx x x= + + ++ + ( ) ( )( ) ( )( ) ( )222 2 22 24 1 1 1 14 24 2x A x B x x C xx Ax Ax A Bx B Cx Cx A B x A C x A B C= + + + + = + + + + = + + + + So, 42 0. 0A BA CA B C+ =+ = = Use Gauss-Jordan elimination to solve the system. 1 1 0 4 1 0 0 12 0 1 0 0 1 0 31 1 1 0 0 0 1 2 The solution is: 1, 3, and 2 A B C = = = So, ( ) ( ) ( )22 24 1 3 2.1 11 1 1xx xx x x= + ++ + 31. ( )( ) ( )22 2202 22 2 2x A B Cx xx x x = + =+ + ( ) ( )( ) ( )( ) ( )222 2 22 220 2 2 2 220 4 4 4 220 4 4 4 2x A x B x x C xx Ax Ax A Bx B Cx Cx A B x A C x A B C = + + + + = + + + + = + + + + + So, 14 04 4 2 20.A BA CA B C+ = + = + = Use Gauss-Jordan elimination to solve the system. 1 1 0 1 1 0 0 14 0 1 0 0 1 0 24 4 2 20 0 0 1 4 The solution is: 1, 2, and 4. A B C = = = So, ( )( ) ( )22 220 1 2 4.2 22 2 2xx xx x x = ++ + 33. Use Gauss-Jordan elimination to solve the system 2 0 1 0 1 0 0 20 2 1 0 0 1 0 21 1 0 4 0 0 1 4 So, 2 x = 2 y = and 4. = Revi ew Exerci ses f or Chapt er 1 21 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 35. Let 1x = number of touchdowns, 2x = number of extra-point kicks, and 3x = number of field goals. 1 2 31 2 32 3136 3 460x x xx x xx x+ + =+ + = = Use Gauss-Jordan elimination to solve the system. 1 1 1 13 1 0 0 56 1 3 46 0 1 0 40 1 1 0 0 0 1 4 Because 15, x =24, x = and 34, x = there were 5 touchdowns, 4 extra-point kicks, and 4 field goals. Review Exercises for Chapter 1 1. 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. 3. Because the equation is in the form 1 2, a x a y b + = it is 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. Because the equation is in the form 1 2, a x a y b + = it is linear in the variables x and y. 9. Choosing y and z as the free variables and letting y s = and , z t = you have 3 1 14 2 24 2 6 14 1 2 6.x s tx s tx s t + = = += + So, the solution set can be described as 3 1 14 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. 123 32 21 1 2 1 0 1 1 2 1 1 20 1 0 1 3 1 0 0 4 6 Converting back to a linear system, the solution is: 12x = and 32. y = 13. Rearrange the equations as shown below. 42 3 0x yx y = = Row reduce the augmented matrix for this system. 1 1 4 1 1 4 1 1 4 1 0 122 3 0 0 1 8 0 1 8 0 1 8 Converting back to a linear system, the solution is: 12 x = and 8. y = 15. Row reduce the augmented matrix for this system. 1 1 0 1 1 0 1 1 0 1 0 02 1 0 0 1 0 0 1 0 0 1 0 Converting back to a linear system, the solution is: 0 x = and 0. y = 22 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 17. The augmented matrix for this system is 1 1 91 1 1 which is equivalent to the reduced row-echelon matrix 1 1 0.0 0 1 Because the second row corresponds to 0 1, = which is a false statement, you can conclude that the system has no solution. 19. Multiplying both equations by 100 and forming the augmented matrix produces 20 30 14.40 50 20 Use Gauss-Jordan elimination as shown below. 3 7 1 3 7 3 72 2 10 2 10 2 104 45 51 0 1 1 10 1 0 1 40 50 20 0 10 8 So, the solution is: 112x = and 245. x = 21. Expanding the second equation, 3 2 0, x y + = the augmented matrix for this system is 1 12 303 2 0 which is equivalent to the reduced row-echelon matrix 1 0 0. 0 1 0 So, the solution is: 0 x = and 0. y = 23. Because the matrix has 2 rows and 3 columns, it has size 2 3. 25. This matrix has the characteristic stair step pattern of leading 1s so it is in row-echelon form. However, the leading 1 in row three of column four has 1s above it, so the matrix is not in reduced row-echelon form. 27. Because the first row begins with 1, this matrix is not in row-echelon form. 29. This matrix corresponds to the system 1 232 00.x xx+ == Choosing 2x t = as the free variable you can describe the solution as 12 , x t = 2, x t = and 30, x = where t is a real number. 31. The augmented matrix for this system is 1 1 2 12 3 1 25 4 2 4 which is equivalent to the reduced row-echelon matrix 1 0 0 20 1 0 3 .0 0 1 3 So, the solution is: 2, 3, x y = = and 3. z = 33. Use Gauss-Jordan elimination on the augmented matrix. 12132 3 3 3 1 0 06 6 12 13 0 1 012 9 1 2 0 0 1 1 So, 12, x =13, y = and 1. z = 35. The augmented matrix for this system is 1 2 1 62 3 0 71 3 3 11 which is equivalent to the reduced row-echelon matrix 1 0 3 40 1 2 5 .0 0 0 0 Choosing z t = as the free variable you find that the solution set can be described by 4 3 , x t = + 5 2 , y t = + and , z t = where t is a real number. Revi ew Exerci ses f or Chapt er 1 23 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 37. Use the Gauss-Jordan elimination on the augmented matrix for this system. 322 1 2 4 1 0 22 2 0 5 0 1 2 12 1 6 2 0 0 0 0 So, the solution is: 322 , x t = 1 2 , y t = + , z t = where t is any real number. 39. The augmented matrix for this system is 2 1 1 2 15 2 1 3 01 3 2 2 13 2 3 5 12 which is equivalent to the reduced row-echelon matrix 1 0 0 0 10 1 0 0 40 0 1 0 30 0 0 1 2 So, the solution is: 1 21, 4, x x = =33, x = and 42. x = 41. Using a graphing utility, the augmented matrix reduces to 1 0 0 00 1 4 20 0 0 00 0 0 0 Choosing z t = as the free variable, you find that the solution set can be described by 0, x = 4 , y z t = and , z t = where t is a real number. 43. Using a graphing utility, the augmented matrix reduces to 1 0 0 0 10 1 0 0 00 0 1 0 40 0 0 1 2 So, the solution is: 1, 0, x y = = 4, z = and 2. w = 45. Using a graphing utility, the augmented matrix reduces to 1 0 2 0 00 1 1 0 00 0 0 1 0 Choosing z t = as the free variable you find that the solution set can be described by 2 , x t = , y t = , z t = and 0, w = where t is a real number. 47. Use Guass-Jordan elimination on the augmented matrix. 1 2 8 0 1 0 0 03 2 0 0 0 1 0 01 1 7 0 0 0 1 0 So, the solution is 1 2 30. x x x = = = 49. The augmented matrix for this system is 2 8 4 03 10 7 00 10 5 0 which is equivalent to the reduced row-echelon matrix 121 0 4 00 1 00 0 0 0 Choosing 3x t = as the free variable, you find that the solution set can be described by 1 2124 , , x t x t = = and 3, x t = where t is a real number. 51. Forming the augmented matrix 1 01 1kk and using Gauss-Jordan elimination, you obtain 2222211 1 1 01 1 1 11, 1 0.1 0 0 1 0 10 111kk kkkkk k k kkk So, the system is inconsistent if 1. k = 53. Row reduce the augmented matrix. ( ) ( )1 2 3 1 2 30 2 9 3 9 b a a a b (a) There will be no solution if 2 0 b a = and 9 3 0. a That is if 2 b a = and 3. a = (b) There will be exactly one solution if 2 . b a (c) There will be an infinite number of solutions if 2 b a = and 3. a = That is, if 3 a = and 6. b = 24 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 55. You can show that two matrices of the same size are row equivalent if they both row reduce to the same matrix. The two given matrices are row equivalent because each is row equivalent to the identity matrix. 57. Adding a multiple of row one to each row yields the following matrix. ( )( )( ) ( ) ( )( )1 2 30 2 10 2 4 2 10 1 2 1 1 1nn n n nn n n nn n n n n n n """# # # #" Every row below row two is a multiple of row two. Therefore, reduce these rows to zeros. ( )1 2 30 2 10 0 0 00 0 0 0nn n n n """# # # " # Dividing row two by n yields a new second row. 1 2 30 1 2 10 0 0 00 0 0 0nn """# # # #" Adding 2 times row two to row one yields a new first row. 1 0 1 20 1 2 10 0 0 00 0 0 0nn """# # # #" This matrix is in reduced row-echelon form. 59. (a) False. See page 3, following Example 2. (b) True. See page 5, Example 4(b). 61. (a) Let 1x = number of three-point baskets, 2x = number of two-point baskets, and 3x = number of one-point free throws. 1 2 31 22 33 2 593 01x x xx xx x+ + = = = (b) Use Gauss-Jordan elimination to solve the system. 3 2 1 59 1 0 0 53 1 0 0 0 1 0 150 1 1 1 0 0 1 14 Because 5, x = 15, y = and 14, z = there were 5 three-point baskets, 15 two-point baskets, and 14 one-point free throws. 63. ( )( ) ( )22 23 3 22 22 2 2x x A B Cx xx x x = + ++ + ( ) ( )( ) ( )( ) ( )222 2 22 23 3 2 2 2 2 23 3 2 4 4 4 23 3 2 4 4 4 2x x A x B x x C xx x Ax Ax A Bx B Cx Cx x A B x A C x A B C = + + + + = + + + + = + + + + + So, 34 3. 4 4 2 2A BA CA B C+ = + = + = Use Gauss-Jordan elimination to solve the system. 1 1 0 3 1 0 0 14 0 1 3 0 1 0 24 4 2 2 0 0 1 1 The solution is: 1, 2, A B = = and 1. C = So, ( )( ) ( )22 23 3 2 1 2 1.2 22 2 2x xx xx x x = + ++ + Revi ew Exerci ses f or Chapt er 1 25 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 65. (a) Because there are three points, choose a second degree polynomial, ( )20 1 2. p x a a x a x = + + By substituting the values at each point into this equation you obtain the system 0 1 20 1 20 1 22 4 53 9 0. 4 16 20a a aa a aa a a+ + =+ + =+ + = Forming the augmented matrix 1 2 4 51 3 9 01 4 16 20 and using Gauss-Jordan elimination you obtain 13522521 0 0 900 1 0 .0 0 1 So, ( )2 135 252 290 . p x x x = + (b) 67. Establish the first year as 0 x = and substitute the values at each point into ( )20 1 2p x a a x a x = + + to obtain the system 00 1 20 1 250602 4 75.aa a aa a a=+ + =+ + = Forming the augmented matrix 1 0 0 501 1 1 601 2 4 75 and using Gauss-Jordan elimination, you obtain 152521 0 0 500 1 0 .0 0 1 So, ( )2 15 52 250 . p x x x = + + To predict the sales in the fourth year, evaluate ( ) p x when 3. x = ( ) ( ) ( )215 52 23 50 3 3 $95. p = + + = 69. (a) There are three points: (0, 80), (4, 68), and (80, 30). Because you are given three points, choose a second-degree polynomial, ( )20 1 2. p x a a x a x = + + Substituting the given points into ( ) p x produces the following system of linear equations. ( ) ( )( ) ( )( ) ( )20 1 2 020 1 2 020 1 2 00 04 480 80a a a aa a a aa a a a+ + =+ + =+ + =1 21 2804 16 6880 6400 30a aa a=+ + =+ + = (b) Form the augmented matrix 1 0 0 801 4 16 681 80 6400 30 and use Gauss-Jordan elimination to obtain the equivalent reduced row-echelon matrix 2581321 0 0 800 1 0 .0 0 1 So, ( )2 25 18 3280 . p x x x = + (c) The graphing utility gives 080, a =1258, a = and 2132. a = In other words ( )2 25 18 3280 . p x x x = + (d) The results to (b) and (c) are the same. (e) There is precisely one polynomial function of degree 1 n (or less) that fits n distinct points. 71. Applying Kirchoffs first law to either junction produces 1 3 2I I I + = and applying the second law to the two paths produces 1 1 2 2 1 22 2 3 3 2 33 4 34 2 2.R I R I I IR I R I I I+ = + =+ = + = Rearrange these equations, form the augmented matrix, and use Gauss-Jordan elimination. 5136131131 1 1 0 1 0 03 4 0 3 0 1 00 4 2 2 0 0 1 So, the solution is 1513, I =2613, I = and 3113. I = 1 2 3 4 5510152025x(2, 5)(3, 0)(4, 20)y 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........................................2 Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination ..........................8 Section 1.3 Applications of Systems of Linear Equations.....................................13 Review Exercises ..........................................................................................................19 Project Solutions...........................................................................................................24 2 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 2. Because the term xy cannot be rewritten as ax by + for any real numbers a and b, the equation cannot be written in the form 1 2. a x a y b + = So this equation is not linear in the variables x and y. 4. Because the terms 2x and 2y cannot be rewritten as ax by + for any real numbers a and b, the equation cannot be written in the form 1 2. a x a y b + = So this equation is not linear in the variables x and y. 6. Because the equation is in the form 1 2, a x a y b + = it is linear in the variables x and y. 8. Choosing y as the free variable, let y t = and obtain 1212163 93 93 .x tx tx t == += + So, you can describe the solution set as 163 x t = + and , y t = where t is any real number. 10. Choosing 2x and 3x as free variables, let 3x t = and 2x s = and obtain 113 26 39 13. x x t + = Dividing this equation by 13 you obtain 112 3 11 2 3 .x s tx s t + == + So, you can describe the solution set as 11 2 3 , x s t = + 2, x s = and 3, x t = where t and s are any real numbers. 12. From Equation 2 you have 23. x = Substituting this value into Equation 1 produces 12 12 6 x = or 19. x = So, the system has exactly one solution: 19 x = and 23. x = 14. From Equation 3 you conclude that 2. z = Substituting this value into Equation 2 produces 2 2 6 y + = or 2. y = Finally, substituting 2 y = and 2 z = into Equation 1, you obtain 2 4 x = or 6. x = So, the system has exactly one solution: 6, 2, x y = = and 2. z = 16. From the second equation you have 20. x = Substituting this value into Equation 1 produces 1 30. x x + = Choosing 3x as the free variable, you have 3x t = and obtain 10 x t + = or 1. x t = So, you can describe the solution set as 1, x t = 20, x = and 3. x t = 18. 3 22 3x yx y+ = + = Adding the first equation to the second equation produces a new second equation, 5 5, y = or 1. y = So, ( ) 2 3 2 3 1 , x y = = and the solution is: 1, x = 1. y = This is the point where the two lines intersect. 20. The two lines coincide. Multiplying the first equation by 2 produces a new first equation. 234322 4x yx y = + = Adding 2 times the first equation to the second equation produces a new second equation. 2320 0x y == Choosing y t = as the free variable, you obtain 232. x t = + So, you can describe the solution set as232 x t = + and , y t = where t is any real number. x422343(1, 1)yx + 2y = 3x + 3y = 22x + y = 4x2133422 3 4 4y43x y = 11213 Sect i on 1. 1 Int roduct i on t o Syst ems of Li near Equat i ons 3 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 22. 3 174 3 7x yx y + =+ = Subtracting the first equation from the second equation produces a new second equation, 5 10, x = or 2. x = So, ( ) 4 2 3 7, y + = or 5, y = and the solution is: 2, x = 5. y = This is the point where the two lines intersect. 24. 5 216 5 21x yx y =+ = Adding the first equation to the second equation produces a new second equation, 7 42, x = or 6. x = So, 6 5 21, y = or 3, y = and the solution is: 6, x = 3. y = This is the point where the two lines intersect. 26. 1 242 32 5x yx y ++ = = Multiplying the first equation by 6 produces a new first equation. 3 2 232 5x yx y+ = = Adding the first equation to the second equation produces a new second equation, 4 28, x = or 7. x = So, 7 2 5, y = or 1, y = and the solution is: 7, x = 1. y = This is the point where the two lines intersect. 28. 0.2 0.5 27.80.3 0.4 68.7x yx y = + = Multiplying the first equation by 40 and the second equation by 50 produces new equations. 8 20 111215 20 3435x yx y = + = Adding the first equation to the second equation produces a new second equation, 23 2323, x = or 101. x = So, ( ) 8 101 20 1112, y = or 96, y = and the solution is: 101, x = 96. y = This is the point where the two lines intersect. 30. 2 1 23 6 34 4x yx y+ =+ = Adding 6 times the first equation to the second equation produces a new second equation, 0 0. = Choosing x t = as the free variable, you obtain 4 4 . y t = So, you can describe the solution as x t = and 4 4 , y t = where t is any real number. 32. (a) (b) This system is inconsistent, because you see two parallel lines on the graph of the system. (c) Because the system is inconsistent, you cannot approximate the solution. (d) Adding 2 times the first equation to the second you obtain 0 8, = which is a false statement. This system has no solution. (e) You obtained the same answer both geometrically and algebraically. 446 64x 5y = 38x + 10y = 14xy2 4 6 82268x + 3y = 174x + 3y = 7xy3 3 6 93693x 5y = 216x + 5y = 21xyx 2y = 5x 12y + 23+ = 43 6 1236912xy1 2 3 4 5 62123454x + y = 4232316x + y =xy0.2x 0.5y = 27.80.3x + 0.4y = 68.750 10015020050100150250(101, 96)4 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 34. (a) (b) Two lines corresponding to two equations intersect at a point, so this system has a unique solution. (c) 13, x 12y (d) Adding 18 times the second equation to the first equation you obtain 10 5 y = or 12. y = Substituting 12y = into the first equation you obtain 9 3 x = or 13. x = The solution is: 13x = and 12. y = (e) You obtained the same answer both geometrically and algebraically. 36. (a) (b) Because each equation has the same line as a graph, there are infinitely many solutions. (c) All solutions of this system lie on the line 53 2521 42. y x = + So let , x t = then the solution set is , x t =53 2521 42y t = + for any real number t. (d) Adding 3 times the first equation to the second equation you obtain 5.3 2.1 1.250 0.x y + == Choosing x t = as the free variable, you obtain 2.1 5.3 1.25 y t = + or 21 53 12.5 y t = + or 53 2521 42. y t = + So, you can describe the solution set as , x t =53 2521 42y t = + for any real number t. (e) You obtained the same answer both geometrically and algebraically. 38. Adding 2 times the first equation to the second equation produces a new second equation. 3 2 20 10x y + == Because the second equation is a false statement, the original system of equations has no solution. 40. Adding 6 times the first equation to the second equation produces a new second equation. 1 222 014 0x xx == Now, using back-substitution, the system has exactly one solution: 10 x = and 20. x = 42. Multiplying the first equation by 32produces a new first equation. 1 21 21404 0x xx x+ =+ = Adding 4 times the first equation to the second equation produces a new second equation. 1 21400 0x x + == Choosing 2x t = as the free variable, you obtain 114. x t = So you can describe the solution set as 114x t = and 2, x t = where t is any real number. 44. To begin, change the form of the first equation. 1 21 274 3 122 12x xx x+ = = Multiplying the first equation by 4 yields a new first equation. 1 21 27 43 32 12x xx x+ = = Adding 2 times the first equation to the second equation produces a new second equation. 1 227 43 311 223 3x xx+ = = Multiplying the second equation by 311 yields a new second equation. 1 227 43 32x xx+ == Now, using back-substitution, the system has exactly one solution: 15 x = and 22. x = 46. Multiplying the first equation by 20 and the second equation by 100 produces a new system. 1 21 20.6 4.27 2 17x xx x =+ = Adding 7 times the first equation to the second equation produces a new second equation. 1 220.6 4.26.2 12.4x xx == Now, using back-substitution, the system has exactly one solution: 13 x = and 22. x = 223 39x 4y = 51213x + y = 05.3x + 2.1y = 1.2515.9x 6.3y = 3.753 322 Sect i on 1. 1 Int roduct i on t o Syst ems of Li near Equat i ons 5 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 48. Adding the first equation to the second equation yields a new second equation. 24 3 104 4x y zy zx y+ + =+ =+ = Adding 4 times the first equation to the third equation yields a new third equation. 24 3 103 4 4x y zy zy z+ + =+ = = Dividing the second equation by 4 yields a new second equation. 3 54 223 4 4x y zy zy z+ + =+ = = Adding 3 times the second equation to the third equation yields a new third equation. 3 54 27 74 22 x y zy zz+ + =+ = = Multiplying the third equation by 47 yields a new third equation. 3 54 222x y zy zz+ + =+ == Now, using back-substitution, the system has exactly one solution: 0, 4, and 2. x y z = = = 50. Interchanging the first and third equations yields a new system. 1 2 31 2 31 2 311 4 32 4 75 3 2 3x x xx x xx x x + =+ = + = Adding 2 times the first equation to the second equation yields a new second equation. 1 2 32 31 2 311 4 326 9 15 3 2 3x x xx xx x x + = = + = Adding 5 times the first equation to the third equation yields a new third equation. 1 2 32 32 311 4 326 9 152 18 12x x xx xx x + = = = At this point you realize that Equations 2 and 3 cannot both be satisfied. So, the original system of equations has no solution. 52. Adding 4 times the first equation to the second equation and adding 2 times the first equation to the third equation produces new second and third equations. 1 32 32 34 132 15 452 15 45x xx xx x+ = = = The third equation can be disregarded because it is the same as the second one. Choosing 3x as a free variable and letting 3, x t = you can describe the solution as 1245 152 213 4 x tx t= = 3, x t = where t is any real number. 54. Adding 3 times the first equation to the second equation produces a new second equation. 1 2 32 32 5 28 16 8x x xx x + = = Dividing the second equation by 8 yields 1 2 32 32 5 22 1.x x xx x + = = Adding 2 times the second equation to the first equation yields 1 32 30. 2 1x xx x+ = = Letting 3x t = be the free variable, you can describe the solution as 1, x t = 22 1, x t = and 3, x t = where t is any real number. 6 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 56. Adding 2 times the first equation to the fourth equation, yields 1 42 3 42 42 3 43 42 03 2 14 6 3.x xx x xx xx x x+ = = = + = Multiplying the fourth equation by 1, and interchanging it with the second equation, yields 1 42 3 42 42 3 43 44 6 33 2 1. 2 0x xx x xx xx x x+ = + = = = Adding 3 times the second equation to the third, and 2 times the second equation to the fourth, produces 1 42 3 43 43 43 44 6 312 20 8. 7 13 6x xx x xx xx x+ = + = = = Dividing the third equation by 12 yields 1 42 3 43 43 45 23 33 44 6 3. 7 13 6x xx x xx xx x+ = + = = = Adding 7 times the third equation to the fourth yields 1 42 3 43 445 23 34 43 33 44 6 3.x xx x xx xx+ = + = = = Using back-substitution, the original system has exactly one solution: 11, x =21, x =31 x = and 41. x = Answers may vary slightly for Exercises 5864. 58. Using a computer software program or graphing utility, you obtain 10, x = 20, y = 40, z = 12. w = 60. Using a computer software program or graphing utility, you obtain 0.8, x = 1.2, y = 2.4. z = 62. Using a computer software program or graphing utility, you obtain 10.6, x =20.5, x = 30.8. x = 64. Using a computer software program or graphing utility, you obtain 6.8813, x = 163.3111, y = 210.2915, z = 59.2913. w = 66. 0 x y z = = = is clearly a solution. Dividing the first equation by 2 produces 3204 3 0. 8 3 3 0x yx y zx y z+ =+ =+ + = Adding 4 times the first equation to the second equation, and 8 times the first equation to the third, yields 3203 0. 9 3 0x yy zy z+ = = + = Adding 3 times the second equation to the third equation yields 3203 06 0.x yy zz+ = == Using back-substitution you conclude there is exactly one solution: 0. x y z = = = 68. 0 x y z = = = is clearly a solution. Dividing the first equation by 12 yields 5 112 120. 12 4 0x y zx y z+ + =+ = Adding 12 times the first equation one to the second yields 5 112 1202 0.x y zy z+ + = = Letting z t = be the free variable, you can describe the solution as 34, x t = 2 , y t = , z t = where t is any real number. 70. (a) False. Any system of linear equations is either consistent which means it has a unique solution, or infinitely many solutions; or inconsistent, i.e., it has no solution. This result is stated on page 6 of the text, and will be proved later in Theorem 2.5. (b) True. See definition on page 7 of the text. (c) False. Consider the following system of three linear equations with two variables. 2 36 3 91.x yx yx+ = == The solution to this system is: 1, 5. x y = = 72. Because 1x t = and 2, x s = you can write 3 2 13 3 . x s t x x = + = + One system could be 1 2 31 2 333x x xx x x + = + = Letting 3x t = and 2x s = be the free variables, you can describe the solution as 13 , x s t = + 2, x s = 3, x t = where t and s are any real numbers. Sect i on 1. 1 Int roduct i on t o Syst ems of Li near Equat i ons 7 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 74. Substituting 1Ax= and 1By= into the original system yields 2 3 0. 253 46A BA B+ = = Reduce the system to row-echelon form. 8 12 0259 1228 12 025172A BA BA BA+ = = + == So, 2534A = and 25.51B = Because 1Ax= and 1, By= the solution of the original system of equations is: 3425x = and 51.25y = 76. Substituting 1, Ax=1, By= and 1Cz= into the original system yields 2 2 53 4 12 3 0.A B CA BA B C+ = = + + = Reduce the system to row-echelon form. 2 2 53 4 15 5A B CA BC+ = = = 3 4 111 6 175 5A BB CC = + = = So, 1. C = Using back-substitution, ( ) 11 6 1 17, B + = or 1 B = and ( ) 3 4 1 1, A = or 1. A = Because 1 , A x = 1 , B y = and 1 , C z = the solution of the original system of equations is: 1, x = 1, y = and 1. z = 78. Multiplying the first equation by sin and the second by cos produces ( ) ( )( ) ( )22sin cos sin sinsin cos cos cosx yx y + = + = Adding these two equations yields ( )2 2sin cos sin cossin cos .yy + = += + So, ( ) ( ) ( ) ( ) cos sin cos sin sin cos 1 x y x + = + + = and ( ) ( )2 21 sin sin cos cos sin coscos sin .cos cosx = = = Finally, the solution is cos sin x = and cos sin . y = + 80. Interchange the two equations and row reduce. ( )3232326461 4 6x ykx yx yk y k = + = = + = + So, if 23, k = there will be an infinite number of solutions. 82. Reduce the system. ( )221 4 2x kyk y k+ = = If 1, k = there will be no solution. 84. Interchange the first two equations and row reduce 02 43 1x y zky kz ky z+ + =+ = = If 0, k = then there is an infinite number of solutions. Otherwise, 02 45 13.x y zy zz+ + =+ == Because this system has exactly one solution, the answer is all 0. k 8 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 86. Reducing the system to row-echelon form produces ( ) ( )5 02 010 2x y zy za y b z c+ + = = + = ( )5 02 02 22x y zy za b z c+ + = =+ = So, you see that (a) if 2 22 0, a b + then there is exactly one solution. (b) if 2 22 0 a b + = and 0, c = then there is an infinite number of solutions. (c) if 2 22 0 a b + = and 0, c there is no solution. 88. If 1 2 30, c c c = = = then the system is consistent because 0 x y = = is a solution. 90. Multiplying the first equation by c, and the second by a, produces .acx bcy ecacx day af+ =+ = Subtracting the second equation from the first yields ( ) .acx bcy ecad bc y af ec+ = = So, there is a unique solution if 0. ad bc 92. The two lines coincide. 2 3 70 0x y == Letting , y t =7 3.2tx += The graph does not change. 94. 21 20 013 12 120x yx y = = Subtracting 5 times the second equation from 3 times the first equation produces a new first equation, 2 600, x = or 300. x = So, ( ) 21 300 20 0, y = or 315, y = and the solution is: 300, x = 315. y = The graphs are misleading because they appear to be parallel, but they actually intersect at ( ) 300, 315 . Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination 2. Because the matrix has 2 rows and 4 columns, it has size 2 4. 4. Because the matrix has 1 row and 5 columns, it has size 1 5. 6. Because the matrix has 1 row and 1 column, it has size 1 1. 8. Because the matrix has 4 rows and 1 column, it has size 4 1. 10. Because the leading 1 in the first row is not farther to the left than the leading 1 in the second row, the matrix is not in row-echelon form. 12. The matrix satisfies all three conditions in the definition of row-echelon form. However, because the third column does not have zeros above the leading 1 in the third row, the matrix is not in reduced row-echelon form. 14. 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 four) has zeros elsewhere, the matrix is in reduced row-echelon form. 16. Because the matrix is in reduced row-echelon form, you can convert back to a system of linear equations 1223.xx == 18. Because the matrix is in row-echelon form, you can convert back to a system of linear equations 1 2 332 0. 1x x xx+ + == Using back-substitution, you have 31. x = Letting 2x t = be the free variable, you can describe the solution as 11 2 , x t = 2x t = and 31, x = where t is any real number. x2112451 4 5 2 33y Sect i on 1. 2 Gaussi an El i mi nat i on and Gauss-Jordan El i mi nat i on 9 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 20. Gaussian elimination produces the following. 2 1 1 0 1 0 1 01 2 1 2 1 2 1 21 0 1 0 2 1 1 01 0 1 0 1 0 1 00 2 0 2 0 1 0 12 1 1 0 2 1 1 01 0 1 0 1 0 1 00 1 0 1 0 1 0 10 1 1 0 0 0 1 1 Because the matrix is in row-echelon form, convert back to a system of linear equations. 1 323011x xxx+ === By back-substitution 1 31. x x = = So, the solution is: 1 21, 1, x x = = and 31. x = 22. Because the fourth row of this matrix corresponds to the equation 0 2, = there is no solution to the linear system. 24. The augmented matrix for this system is 2 6 16.2 6 16 Use Gauss-Jordan elimination as follows. 2 6 16 1 3 8 1 3 82 6 16 2 6 16 0 0 0 Converting back to system of linear equations, you have 3 8. x y + = Choosing y t = as the free variable, you can describe the solution as 8 3 x t = and , y t = where t is any real number. 26. The augmented matrix for this system is 2 1 0.1.3 2 1.6 Gaussian elimination produces the following. 1 12 20851 12 207 72 41 1 12 20 51 12 21 2 1 0.13 2 3 2 1.6101 1 00 1 0 1 Converting back to a system of equations, the solution is: 15x = and 12. y = 28. The augmented matrix for this system is 1 2 01 1 6 .3 2 8 Gaussian elimination produces the following. 1 2 0 1 2 01 1 6 0 1 63 2 8 0 8 81 2 0 1 2 00 1 6 0 1 60 8 8 0 0 40 Because the third row corresponds to the equation 0 40, = you conclude that the system has no solution. 30. The augmented matrix for this system is 2 1 3 240 2 1 14 .7 5 0 6 Gaussian elimination produces the following. 3 12 23 12 23 212 23 12 21245 1354 22 1 3 24 1 120 2 1 14 0 2 1 147 5 0 6 7 5 0 61 120 2 1 140 781 120 1 70 0 Back-substitution now yields ( )( ) ( )32 31 3 21 12 23 3 1 12 2 2 267 7 6 1012 12 6 10 8.xx xx x x== + = + == + = + = So, the solution is: 1 28, 10 x x = = and 36. x = 10 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 32. The augmented matrix for this system is 2 0 3 34 3 7 5 .8 9 15 10 Gaussian elimination produces the following. 3 32 23 3 3 32 2 2 21 13 32 0 3 3 1 04 3 7 5 4 3 7 58 9 15 10 8 9 15 101 0 1 00 3 1 1 0 10 9 3 20 0 0 1 Because the third row corresponds to the equation 0 1, = there is no solution to the original system. 34. The augmented matrix for this system is 1 2 1 8.3 6 3 21 Gaussian elimination produces the following matrix. 1 2 1 80 0 0 3 Because the second row corresponds to the equation 0 3, = there is no solution to the original system. 36. The augmented matrix for this system is 2 1 1 2 63 4 0 1 11 5 2 6 35 2 1 1 3 Gaussian elimination produces the following. 6 17 1011 11 111 5 2 6 3 1 5 2 6 3 1 5 2 6 33 4 0 1 1 0 11 6 17 10 0 12 1 1 2 6 0 9 5 10 0 0 9 5 10 05 2 1 1 3 0 23 11 31 18 0 23 11 31 18 6 17 10 6 17 1011 11 11 11 11 1143 90 111 11 1117 50 32 17 50 3211 11 11 11 11 46 17 10 6 17 1011 11 11 11 11 1781 156211 111 5 2 6 3 1 5 2 6 30 1 0 10 0 0 0 1 43 900 0 0 01 5 2 6 3 1 5 2 6 30 1 0 10 0 1 43 900 0 0 10 0 1 43 900 0 0 1 2 Back-substitution now yields ( )( ) ( ) ( ) ( )( ) ( ) ( )10 6 17 10 6 1711 11 11 11 11 11290 43 90 43 2 44 2 0.3 5 2 6 3 5 0 2 4 6 2 1.wz wy z wx y z w= = + = + == = == = = So, the solution is: 1, 0, 4 x y z = = = and 2. w = Sect i on 1. 2 Gaussi an El i mi nat i on and Gauss-Jordan El i mi nat i on 11 Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 38. Using a computer software program or graphing utility, you obtain 14.362932.756928.6356.xyz=== 40. Using a computer software program or graphing utility, you obtain 1234521341.xxxxx== === 42. Using a computer software program or graphing utility, you obtain 123456112021.xxxxxx== === = 44. The corresponding equations are 12 30. 0xx x =+ = Choosing 4x t = and 3x t = as the free variables, you can describe the solution as 10, x =2, x s = 3, x s = and 4, x t = where s and t are any real numbers. 46. The corresponding equations 0 0 = and 3 free variables. So, 1, x t =2, x s =3, x r = where , , t s r are any real numbers. 48. (a) If A is the augmented matrix of a system of linear equations, then number of equations in this system is three (because it is equal to the number of rows of the augmented matrix). Number of variables is two because it is equal to number of columns of the augmented matrix minus one. (b) Using Gaussian elimination on the augmented matrix of a system, you have the following. 2 1 34 24 2 6k 2 1 30 0 60 0 0k + This system is consistent if and only if 6 0, k + = so 6. k = (c) If A is the coefficient matrix of a system of linear equations, then the number of equations is three, because it is equal the number of rows of the coefficient matrix. The number of variables is also three, because it is equal to the number of columns of the coefficient matrix. (d) Using Gaussian elimination on A you obtain the following coefficient matrix of an equivalent system. 3 12 210 0 60 0 0k + Because the homogeneous system is always consistent, the homogeneous system with the coefficient matrix A is consistent for any value of k. 50. Using Gaussian elimination on the augmented matrix, you have the following. ( )( )1 1 0 0 1 1 0 00 1 1 0 0 1 1 00 1 1 0 1 0 1 00 0 01 1 0 00 1 1 00 0 2 00 0 01 1 0 00 1 1 00 0 1 00 0 0 0b a c a b ca b c + From this row reduced matrix you see that the original system has a unique solution. 52. Because the system composed of equations 1 and 2 is consistent, but has a free variable, this system must have an infinite number of solutions. 12 Chapt er 1 Syst ems of Li near Equat i ons Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. 54. Use Guass-Jordan elimination as follows. 1 2 3 1 2 3 1 2 3 1 0 14 5 6 0 3 6 0 1 2 0 1 27 8 9 0 6 12 0 0 0 0 0 0 56. Begin by finding all possible first rows [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 0 0 0 , 0 0 1 , 0 1 0 , 0 1 , 1 0 0 , 1 0 , 1 , 1 0 , a a a b a where a and b are nonzero real numbers. For each of these examine the possible remaining rows. 0 0 0 0 0 1 0 1 0 0 1 0 0 10 0 0 , 0 0 0 , 0 0 0 , 0 0 1 , 0 0 0 ,0 0 0 0 0 0 0 0 0 0 0 0 0 0 0a 1 0 0 1 0 0 1 0 0 1 0 0 1 0 00 0 0 , 0 1 0 , 0 1 0 , 0 0 1 , 0 1 ,0 0 0 0 0 0 0 0 1 0 0 0 0 0 0a 1 0 1 0 1 1 0 1 00 0 0 , 0 0 1 , 0 0 0 , 0 0 0 , 0 1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0a a a b a a 58. (a) False. A 4 7 matrix has 4 rows and 7 columns. (b) True. Reduced row echelon form of a given matrix is unique while row echelon form is not. (See also exercise 64 of this section.) (c) True. See Theorem 1.1 on page 25. (d) False. Multiplying a row by a nonzero constant is one of the elementary row operations. However, multiplying a row of a matrix by a constant 0 c = is not an elementary row operation. (This would ch