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Elementary Linear Algebra, 8th Edition Solution Manual

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    Complete Solutions Manual to Accompany Elementary Linear Algebra EIGHTH EDITION Ron Larson The Pennsylvania State University, State College, PA

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    CONTENTS Chapter 1 Introduction to Systems of Linear Equations........................................ 1 Chapter 2 Matrices ................................................................................................ 29 Chapter 3 Determinants......................................................................................... 76 Chapter 4 Vector Spaces ..................................................................................... 104 Chapter 5 Inner Product Spaces .......................................................................... 154 Chapter 6 Linear Transformations...................................................................... 206 Chapter 7 Eigenvalues and Eigenvectors ........................................................... 244

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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 .......................... 8 Section 1.3 Applications of Systems of Linear Equations ..................................... 14 Review Exercises .......................................................................................................... 22 Project Solutions .......................................................................................................... 27

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    2 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 2 x and 2 y 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 1 2 1 2 1 6 3 9 3 9 3 . x t x t x t − = = + = + So, you can describe the solution set as 1 6 3 x t = + and , y t = where t is any real number. 10. Choosing 2 x and 3 x as free variables, let 2 x s = and 3 x t = and obtain 1 12 24 36 12. x s t + − = 1 1 2 3 1 1 2 3 . x s t x s t + − = = − + So, you can describe the solution set as 1 1 2 3 , x s t = − + 3 2 , and , x t x s = = where s and t are any real number. 12. 3 2 2 3 x y x 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. 14. The two lines coincide. Multiplying the first equation by 2 produces a new first equation. 2 3 4 3 2 2 4 x y x y − = − + = − Adding 2 times the first equation to the second equation produces a new second equation. 2 3 2 0 0 x y − = = Choosing y t = as the free variable, you obtain 2 3 2. x t = + So, you can describe the solution set as 2 3 2 x t = + and , y t = where t is any real number. 16. 3 17 4 3 7 x y x 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. x 4 − 2 − 2 − 3 − 4 − 3 ( − 1, 1) y − x + 2 y = 3 x + 3 y = 2 − 2 x + y = − 4 x 2 1 3 − 3 4 − 2 − 2 − 3 − 4 4 y 4 3 x − y = 1 1 2 1 3 x y − 2 − 4 − 6 − 8 − 2 2 6 8 − x + 3 y = 17 4 x + 3 y = 7 ( − 2, 5)

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    Section 1.1 Introduction to Systems of Linear Equations 3 18. 5 21 6 5 21 x y x 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. 20. 1 2 4 2 3 2 5 x y x y − + + = − = Multiplying the first equation by 6 produces a new first equation. 3 2 23 2 5 x y x 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. 22. 0.2 0.5 27.8 0.3 0.4 68.7 x y x y − = − + = Multiplying the first equation by 40 and the second equation by 50 produces new equations. 8 20 1112 15 20 3435 x y x 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. 24. 2 1 2 3 6 3 4 4 x y x 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. 26. From Equation 2 you have 2 3. x = Substituting this value into Equation 1 produces 1 2 12 6 x − = or 1 9. x = So, the system has exactly one solution: 1 9 x = and 2 3. x = 28. From Equation 3 you have 2. z = Substituting this value into Equation 2 produces 3 2 11 y + = or 3. y = Finally, substituting 3 y = into Equation 1, you obtain 3 5 x − = or 8. x = So, the system has exactly one solution: 8, 3, and 2. x y z = = = 30. From the second equation you have 2 0. x = Substituting this value into Equation 1 produces 1 3 0. x x + = Choosing 3 x as the free variable, you have 3 x t = and obtain 1 0 x t + = or 1 . x t = − So, you can describe the solution set as 1 , x t = − 2 0, x = and 3 . x t = x y − 3 3 6 9 − 3 − 6 − 9 3 x − 5 y = 21 6 x + 5 y = 21 (6, − 3) x y 0.2 x − 0.5 y = − 27.8 0.3 x + 0.4 y = 68.7 50 150 − 50 100 150 250 (101, 96) x y − 1 2 3 4 5 6 − 2 1 2 3 4 5 4 x + y = 4 2 3 2 3 1 6 x + y = x y x − 2 y = 5 x − 1 2 y + 2 3 + = 4 − 3 6 12 3 6 9 12 (7, 1)

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    4 Chapter 1 Systems of Linear Equations 32. (a) (b) This system is inconsistent, because you see two parallel lines on the graph of the system. 34. (a) (b) Two lines corresponding to two equations intersect at a point, so this system is consistent. (c) The solution is approximately 1 3 x = and 1 2 . y = − (d) Adding 18 − times the second equation to the first equation, you obtain 10 5 y − = or 1 2 . y = − Substituting 1 2 y = − into the first equation, you obtain 9 3 x = or 1 3 . x = The solution is: 1 3 x = and 1 2 . y = − (e) The solutions in (c) and (d) are the same. 36. (a) (b) Because the lines coincide, the system is consistent. (c) All solutions of this system lie on the line 1 2 7 . y x = + So, let , x t = then the solution set is 1 2 , 7 , x t y t = = + where t is any real number. (d) Adding 3 times the first equation to the second equation you obtain 44.1 6.3 3.15 0 0. x y − + = = Choosing x t = as a free variable, you obtain 14.7 2.1 1.05 t y − + = or 147 21 105 t y − + = or 1 2 7 . y t = + So, you can describe the solution set as 1 2 , 7 , x t y t = = + where t is any real number. (e) The solutions in (c) and (d) are the same. 38. Adding 2 − times the first equation to the second equation produces a new second equation. 3 2 2 0 10 x 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 2 2 2 0 14 0 x x x − = = Now, using back-substitution, the system has exactly one solution: 1 0 x = and 2 0. x = 42. Multiplying the first equation by 3 2 produces a new first equation. 1 2 1 2 1 4 0 4 0 x x x x + = + = Adding 4 − times the first equation to the second equation produces a new second equation. 1 2 1 4 0 0 0 x x + = = Choosing 2 x t = as the free variable, you obtain 1 1 4 . x t = − So you can describe the solution set as 1 1 4 x t = − and 2 , x t = where t is any real number. 44. To begin, change the form of the first equation. 1 2 1 2 5 3 2 6 3 2 x x x x + = − − = − Multiplying the first equation by 3 yields a new first equation. 1 2 1 2 3 5 2 2 3 2 x x x x + = − − = − Adding –3 times the first equation to the second equation produces a new second equation. 1 2 2 3 5 2 2 11 11 2 2 x x x + = − − = Multiplying the second equation by 2 11 − yields a new second equation. 1 2 2 3 5 2 2 1 x x x + = − = − Now, using back-substitution, the system has exactly one solution: 1 2 1 and 1. x x = − = − 4 − 4 − 6 6 4 x − 5 y = 3 − 8 x + 10 y = 14 2 − 2 − 3 3 9 x − 4 y = 5 1 2 1 3 x + y = 0 − 3 − 2 3 2 44.1 x − 6.3 y = − 3.15 − 14.7 x + 2.1 y = 1.05

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    Section 1.1 Introduction to Systems of Linear Equations 5 46. Multiplying the first equation by 20 and the second equation by 100 produces a new system. 1 2 1 2 0.6 4.2 7 2 17 x x x x − = + = Adding 7 − times the first equation to the second equation produces a new second equation. 1 2 2 0.6 4.2 6.2 12.4 x x x − = = − Now, using back-substitution, the system has exactly one solution: 1 3 x = and 2 2. x = − 48. Adding the first equation to the second equation yields a new second equation. 2 4 3 10 4 4 x y z y z x y + + = + = + = Adding 4 − times the first equation to the third equation yields a new third equation. 2 4 3 10 3 4 4 x y z y z y z + + = + = − − = − Dividing the second equation by 4 yields a new second equation. 3 5 4 2 2 3 4 4 x y z y z y z + + = + = − − = − Adding 3 times the second equation to the third equation yields a new third equation. 3 5 4 2 7 7 4 2 2 x y z y z z + + = + = − = Multiplying the third equation by 4 7 − yields a new third equation. 3 5 4 2 2 2 x y z y z z + + = + = = − 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 3 1 2 3 1 2 3 11 4 3 2 4 7 5 3 2 3 x x x x x x x x x − + = + − = − + = Adding 2 − times the first equation to the second equation yields a new second equation. 1 2 3 2 3 1 2 3 11 4 3 26 9 1 5 3 2 3 x x x x x x x x − + = − = − + = Adding 5 − times the first equation to the third equation yields a new third equation. 1 2 3 2 3 2 3 11 4 3 26 9 1 52 18 12 x x x x x x 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 3 2 3 2 3 4 13 2 15 45 2 15 45 x x x x x x + = − − = − − − = − The third equation can be disregarded because it is the same as the second one. Choosing 3 x as a free variable and letting 3 , x t = you can describe the solution as 1 2 45 15 2 2 13 4 x t x 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 3 2 3 2 5 2 8 16 8 x x x x x − + = − = − Dividing the second equation by 8 yields a new second equation. 1 2 3 2 3 2 5 2 2 1 x x x x x − + = − = − Adding 2 times the second equation to the first equation yields a new first equation. 1 3 2 3 0 2 1 x x x x + = − = − Letting 3 x t = be the free variable, you can describe the solution as 1 , x t = − 2 2 1, x t = − and 3 , x t = where t is any real number.

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    6 Chapter 1 Systems of Linear Equations 56. Adding 3 times the first equation to the fourth equation yields 1 4 2 3 4 2 4 2 3 4 2 1 4 2 0 2 3 6 7. x x x x x x x x x x − + = − − = − = − + + = Interchanging the second equation with the third equation yields 1 4 2 4 2 3 4 2 3 4 2 1 0 4 2 2 3 6 7. x x x x x x x x x x − + = − = − − = − + + = Adding 4 − times the second equation to the third equation, and adding 2 − times the second equation to the fourth equation yields 1 4 2 4 3 4 3 4 2 1 0 3 2 3 4 7. x x x x x x x x − + = − = − + = + = Adding 3 times the second equation to the third equation yields 1 4 2 4 3 4 4 2 1 0 3 2 13 13. x x x x x x x − + = − = − + = = Using back-substitution, the original system has exactly one solution: 1 2 3 4 1, 1, 1, and 1. x x x x = = = = Answers may vary slightly for Exercises 58–62. 58. Using a software program or graphing utility, you obtain 0.8, x = 1.2, y = 2.4. z = − 60. Using a software program or graphing utility, you obtain 10, 20, 40, 12. x y z w = = − = = − 62. Using a software program or graphing utility, you obtain 6.8813, 163.3111, x y = = − 210.2915, z = − 59.2913. w = − 64. 0 x y z = = = is clearly a solution. Dividing the first equation by 2 produces 3 2 0 4 3 0 . 8 3 3 0 x y x y z x y z + = + − = + + = Adding 4 − times the first equation to the second equation, and 8 − times the first equation to the third, yields 3 2 0 3 0 . 9 3 0 x y y z y z + = − − = − + = Adding 3 − times the second equation to the third equation yields 3 2 0 3 0 6 0. x y y z z + = − − = = Using back-substitution, you conclude there is exactly one solution: 0. x y z = = = 66. 0 x y z = = = is clearly a solution. Dividing the second equation by 2 yields a new second equation. 1 2 16 3 0 8 0 x y z x y z + + = + − = Adding 3 − times the second equation to the first equation produces a new first equation. 5 2 1 2 8 0 8 0 x z x y z − + = + − = Letting z t = be the free variable, you can describe the solution as 5 16 , 2 , x t y t = = − and , z t = where t is any real number. 68. Let x = the speed of the plane that leaves first and y = the speed of the plane that leaves second. 3 2 80 Equation 1 2 3200 Equation 2 y x x y − = + = 3 2 7 2 2 2 160 2 3200 3360 960 x y x y y y − + = + = = = 960 80 880 x x − = = Solution: First plane: 880 kilometers per hour; second plane: 960 kilometers per hour

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    Section 1.1 Introduction to Systems of Linear Equations 7 70. (a) False. Any system of linear equations is either consistent, which means it has a unique solution, or infinitely many solutions; or inconsistent, which means it has no solution. This result is stated on page 5, and will be proved later in Theorem 2.5. (b) True. See definition on page 6. (c) False. Consider the following system of three linear equations with two variables. 2 3 6 3 9 1. x y x y x + = − − − = = The solution to this system is: 1, 5. x y = = − 72. Because 1 x t = and 2 , x s = you can write 3 2 1 3 3 . x s t x x = + − = + − One system could be 1 2 3 1 2 3 3 3 x x x x x x − + = − + − = − Letting 3 x t = and 2 x s = be the free variables, you can describe the solution as 1 3 , x s t = + − 2 , x s = and 3 , x t = where t and s are any real numbers. 74. Substituting 1 A x = and 1 B y = into the original system yields 3 2 1 17 2 3 . 6 A B A B + = − − = − Reduce the system to row-echelon form. 27 18 9 12 18 17 A B A B + = − − = − 27 18 9 39 26 A B A + = − = − Using back substitution, 2 3 A = − and 1 . 2 B = Because 1 A x = and 1 , B y = the solution of the original system of equations is: 3 2 x = − and 2. y = 76. Substituting 1 , A x = 1 , B y = and 1 C z = into the original system yields 2 2 5 . 3 4 1 2 3 0 A B C A B A B C + − = − = − + + = Reduce the system to row-echelon form. 2 2 5 3 4 1 5 5 A B C A B C + − = − = − = − 3 4 1 11 6 17 5 5 A B B C C − = − − + = − = − 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 θ and the second by cos θ produces ( ) ( ) ( ) ( ) 2 2 sin cos sin sin sin cos cos cos . x y x y θ θ θ θ θ θ θ θ + = − + = Adding these two equations yields ( ) 2 2 sin cos sin cos sin cos . y y θ θ θ θ θ θ + = + = + So, ( ) ( ) ( ) ( ) cos sin cos sin sin cos 1 x y x θ θ θ θ θ θ + = + + = and ( ) ( ) 2 2 1 sin sin cos cos sin cos cos sin . cos cos x θ θ θ θ θ θ θ θ θ θ − − − = = = − Finally, the solution is cos sin x θ θ = − and cos sin . y θ θ = +

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    8 Chapter 1 Systems of Linear Equations 80. Reduce the system to row-echelon form. ( ) 2 0 1 0 x ky k y + = − = 2 0 0, 1 0 x ky y k + = = − ≠ 2 0 0, 1 0 x y k = = − ≠ If 2 1 0, k − ≠ that is if 1, k ≠ ± the system will have exactly one solution. 82. 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 3 0, k − = that is, 8 3 . k = 84. Reduce the system to row-echelon form. ( ) 16 4 3 0 kx y k x + = + = The system will have an infinite number of solutions when 3 4 4 3 0 . k k + =  = − 86. Reducing the system to row-echelon form produces ( ) ( ) 5 0 2 0 10 2 x y z y z a y b z c + + = − = − + − = ( ) 5 0 2 0 2 22 . x y z y z a 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 3 0, 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 ec acx day af + = + = Subtracting the second equation from the first yields ( ) . acx bcy ec ad bc y af ec + = − = − So, there is a unique solution if 0. ad bc − ≠ 92. The two lines coincide. 2 3 7 0 0 x y − = = Letting , y t = 7 3 . 2 t x + = The graph does not change. 94. 21 20 0 13 12 120 x y x 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 4 rows and 1 column, it has size 4 1. × 4. Because the matrix has 1 row and 1 column, it has size 1 1. × 6. Because the matrix has 1 row and 5 columns, it has size 1 5. × 8. 3 1 4 3 1 4 4 3 7 5 0 5 − − − −          − −     Add 3 times Row 1 to Row 2. 10. 1 2 3 2 1 2 3 2 2 5 1 7 0 9 7 11 5 4 7 6 0 6 8 4 − − − − − −         − −  − −         − − −     Add 2 times Row 1 to Row 2. Then add 5 times Row 1 to Row 3. x 2 1 − 1 − 2 − 4 − 5 1 4 5 − 2 − 3 3 y

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    Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination 9 12. Because the matrix is in reduced row-echelon form, you can convert back to a system of linear equations 1 2 2 3. x x = = 14. Because the matrix is in row-echelon form, you can convert back to a system of linear equations 1 2 3 3 2 0 . 1 x x x x + + = = − Using back-substitution, you have 3 1. x = − Letting 2 x t = be the free variable, you can describe the solution as 1 1 2 , x t = − 2 , x t = and 3 1, x = − where t is any real number. 16. Gaussian elimination produces the following. 3 1 1 5 1 0 1 2 1 2 1 0 1 2 1 0 1 0 1 2 3 1 1 5 1 0 1 2 1 0 1 2 0 2 0 2 0 2 0 2 3 1 1 2 0 1 2 1 1 0 1 2 1 0 1 2 0 1 2 1 0 1 2 1 0 2 0 2 0 0 4 4 1 0 1 2 0 1 2 1 0 0 1 1 −                  −              −  −         − − − −                       − − −                Because the matrix is in row-echelon form, convert back to a system of linear equations. 1 3 2 3 3 2 2 1 1 x x x x x + = + = = By back-substitution, 1 2 3 1, 1, and 1. x x x = = − = 18. Because the fourth row of this matrix corresponds to the equation 0 2, = there is no solution to the linear system. 20. 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. 22. 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. 24. 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. 26. 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 8 2 6 16 2 6 16 0 0 0               − − − − − −       Converting back to a 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. 28. The augmented matrix for this system is 2 1 0.1 . 3 2 1.6 − −       Gaussian elimination produces the following. 1 1 2 20 8 5 1 1 2 20 7 7 2 4 1 1 1 2 20 5 1 1 2 2 1 2 1 0.1 3 2 3 2 1.6 1 0 1 1 0 0 1 0 1   − − − −                − −            − −               Converting back to a system of equations, the solution is: 1 5 x = and 1 2 . y = 30. The augmented matrix for this system is 1 2 0 1 1 6 . 3 2 8         −   Gaussian elimination produces the following. 1 2 0 1 2 0 1 1 6 0 1 6 3 2 8 0 8 8 1 2 0 1 2 0 0 1 6 0 1 6 0 8 8 0 0 40          −         − −              −  −         − −     Because the third row corresponds to the equation 0 40, = − you can conclude that the system has no solution.

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    10 Chapter 1 Systems of Linear Equations 32. The augmented matrix for this system is 3 2 3 22 0 3 1 24 . 6 7 0 22 −     −     − −   Gaussian elimination produces the following. 2 22 3 3 1 3 2 22 3 3 1 3 2 22 3 3 1 3 2 22 3 3 1 3 1 1 3 2 3 22 0 3 1 24 0 1 8 6 7 0 22 6 7 0 22 1 1 0 1 8 0 3 6 66 1 1 0 1 8 0 0 7 42 1 1 0 1 8 0 0 1 6   − −         −  −       − − − −         −      −   − − −       −      −   − −       −      −       Back-substitution now yields ( ) ( ) ( ) 3 2 3 1 2 3 1 1 3 3 22 2 22 2 3 3 3 3 6 8 8 6 10 10 6 8. x x x x x x = = + = + = = + − = + − = So, the solution is: 1 2 3 8, 10, and 6. x x x = = = 34. 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 2 − times the first row to the third row 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 1 − times the second row to the first row yields a new first row. 1 0 2 1 0 1 3 2 0 0 0 0 −     −       Converting back to a system of linear equations produces 1 3 2 3 2 1 3 2. x x x x − = − = Finally, choosing 3 x t = as the free variable, you can describe the solution as 1 1 2 , x t = + 2 2 3 , x t = + and 3 , x t = where t is any real number. 36. The augmented matrix for this system is 1 2 1 8 . 3 6 3 21     − − − −   Gaussian elimination produces the following matrix. 1 2 1 8 0 0 0 3       Because the second row corresponds to the equation 0 3, = there is no solution to the original system.

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    Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination 11 38. The augmented matrix for this system is 2 1 1 2 6 3 4 0 1 1 . 1 5 2 6 3 5 2 1 1 3 − −         −    − −    Gaussian elimination produces the following. 6 17 10 11 11 11 1 5 2 6 3 1 5 2 6 3 1 5 2 6 3 3 4 0 1 1 0 11 6 17 10 0 1 2 1 1 2 6 0 9 5 10 0 0 9 5 10 0 5 2 1 1 3 0 23 11 31 18 0 23 11 31 18 − − −             − − − −               − − − − − − − −        − −   − − −   − − −        6 17 10 6 17 10 11 11 11 11 11 11 43 90 1 11 11 11 17 50 32 17 50 32 11 11 11 11 11 4 6 17 10 6 17 10 11 11 11 11 11 1 781 1562 11 11 1 5 2 6 3 1 5 2 6 3 0 1 0 1 0 0 0 0 1 43 90 0 0 0 0 1 5 2 6 3 1 5 2 6 3 0 1 0 1 0 0 1 43 90 0 0 0 − −         − −           − − −         − −     −   −   − −       −     −   1 0 0 1 43 90 0 0 0 1 2         −    −    Back-substitution now yields ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 10 6 17 10 6 17 11 11 11 11 11 11 2 90 43 90 43 2 4 4 2 0 . 3 5 2 6 3 5 0 2 4 6 2 1. w z w y z w x y z w = − = + = + − = = − − − = − − − − = = − − − − = − − − − − = So, the solution is: 1, 0, 4, x y z = = = and 2. w = − 40. Using a software program or graphing utility, the augmented matrix reduces to 1 0 0 0 0 2 0 1 0 0 0 1 . 0 0 1 0 0 3 0 0 0 1 0 4 0 0 0 0 1 1     −             So, the solution is: 1 2 3 4 5 2, 1, 3, 4, and 1. x x x x x = = − = = = 42. Using a computer software program or graphing utility, you obtain 1 2 3 4 5 6 1 1 2 0 2 1. x x x x x x = = − = = = − = 44. The corresponding equations are 1 2 3 0 . 0 x x x = + = Choosing 4 x t = and 3 x t = as the free variables, you can describe the solution as 1 0, x = 2 , x s = − 3 , x s = and 4 , x t = where s and t are any real numbers. 46. The corresponding equations are all 0 0. = So, there are three free variables. So, 1 , x t = 2 , x s = and 3 , x r = where , , and t s r are any real numbers.

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    12 Chapter 1 Systems of Linear Equations 48. number of $1 bills number of $5 bills number of $10 bills number of $20 bills x y z w = = = = 5 10 20 95 26 4 0 2 1 x y z w x y z w y z x y + + + = + + + = − = − = − 1 5 10 20 95 1 0 0 0 15 1 1 1 1 26 0 1 0 0 8 0 1 4 0 0 0 0 1 0 2 1 2 0 0 1 0 0 0 1 1                  −      − −        15 8 2 1 x y z w = = = = The server has 15 $1 bills, 8 $5 bills, 2 $10 bills, and one $20 bill. 50. (a) If A is the augmented matrix of a system of linear equations, then the number of equations in this system is three (because it is equal to the number of rows of the augmented matrix). The number of variables is two because it is equal to the 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 3 4 2 4 2 6 k −     −      −   2 1 3 0 0 6 0 0 0 k −     +       This system is consistent if and only if 6 0, k + = so 6. k = − If A is the coefficient matrix of a system of linear equations, then the number of equations is three, because it is equal to 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. Using Gaussian elimination on A you obtain the following coefficient matrix of an equivalent system. 3 1 2 2 1 0 0 6 0 0 0 k   −   +       Because the homogeneous system is always consistent, the homogeneous system with the coefficient matrix A is consistent for any value of k . 52. Using Gaussian elimination on the augmented matrix, you have the following. ( ) ( ) 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 2 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 b a c a b c a b c                                    −         − − +                 From this row reduced matrix you see that the original system has a unique solution. 54. 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. 56. Use Gauss-Jordan elimination as follows. 1 2 3 1 2 3 1 2 3 1 0 1 4 5 6 0 3 6 0 1 2 0 1 2 7 8 9 0 6 12 0 0 0 0 0 0 −                  − −                   − −        

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    Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination 13 58. 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 1 0 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 0 a                                                   1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 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 0 a                                                   1 0 1 0 1 1 0 1 0 0 0 0 , 0 0 1 , 0 0 0 , 0 0 0 , 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 a a a b a a                                                   60. (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 21. (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 change the system by eliminating the equation corresponding to this row.) 62. No, the row-echelon form is not unique. For instance, 1 2 0 1       and 1 0 . 0 1       The reduced row-echelon form is unique. 64. First, you need 0 a ≠ or 0. c ≠ If 0, a ≠ then you have . 0 0 a b a b a b cb c d ad bc b a                 − − +         So, 0 ad bc − = and 0, b = which implies that 0. d = If 0, c ≠ then you interchange rows and proceed. 0 0 c d a b c d ad c d ad bc b c                 − − +         Again, 0 ad bc − = and 0, d = which implies that 0. b = In conclusion, a b c d       is row-equivalent to 1 0 0 0       if and only if 0, b d = = and 0 a ≠ or 0. c ≠ 66. Row reduce the augmented matrix for this system. 2 2 9 5 0 1 0 1 0 1 0 2 9 5 0 0 2 9 5 0 λ λ λ λ λ λ λ + − − −               − + − + −       To have a nontrivial solution you must have the following. ( )( ) 2 2 9 5 0 5 2 1 0 λ λ λ λ + − = + − = So, if 5 λ = − or 1 2 , λ = the system will have nontrivial solutions. 68. A matrix is in reduced row-echelon form if every column that has a leading 1 has zeros in every position above and below its leading 1. A matrix in row-echelon form may have any real numbers above the leading 1’s. 70. (a) 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. (b) When a system of linear equations has infinitely many solutions, the row-echelon form of the corresponding augmented matrix will have a row that consists entirely of zeros or more than one column with no leading 1’s. The last column will not contain a leading 1.

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