Solution Manual for Elementary Linear Algebra, 7th Edition

CONTENTS Chapter 1 Systems of Linear Equations………………………………………………………… 1 Chapter 2 Matrices ……………………………………………………………………………………27 Chapter 3 Determinants……………………………………………………………………………..64 Chapter 4 Vector Spaces …………………………………………………………………………… 88 Chapter 5 Inner Product Spaces………………………………………………………………..133 Chapter 6 Linear Transformations…………………………………………………………….180 Chapter 7 Eigenvalues and Eigenvectors …………………………………………………..217 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 …………………………………………………………………………………………….21 Project Solutions……………………………………………………………………………………………..26 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 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 a1 x + a2 y = b. So, this equation is not linear in the variables x and y. 14. y 4 3 2 1 4. Because the terms x 2 and y 2 cannot be rewritten as ax + by for any real numbers a and b, the equation cannot be written in the form a1 x + a2 y = b. So, this equation is not linear in the variables x and y. 4 −2 −3 The two lines coincide. Multiplying the first equation by 2 produces a new first equation. 8. Choosing y as the free variable, let y = t and obtain x − 23 y = 3x − 12 t = 9 Adding 2 times the first equation to the second equation produces a new second equation. x = 3 + 16 t. x − 23 y = 2 So, you can describe the solution set as x = 3 + 16 t and y = t , where t is any real number. 0 = 0 Choosing y = t as the free variable, you obtain 10. Choosing x2 and x3 as free variables, let x3 = t and x2 = s and obtain 13×1 − 26 x + 39t = 13. x = 23 t + 2. So, you can describe the solution set as x = 23 t + 2 and y = t , where t is any real number. Dividing this equation by 13 you obtain x1 − 2s + 3t = 1 x1 = 1 + 2 s − 3t. So, you can describe the solution set as x1 = 1 + 2 s − 3t , x2 = s, and x3 = t , where t and s are any real numbers. −x + 2y = 3 4 16. 4x + 3y = 7 x −3 −2 x + 3y = 2 x + 3y = 2 −x + 2 y = 3 Adding the first equation to the second equation produces a new second equation, 5 y = 5 or y = 1. y 8 (− 2, 5) 6 − x + 3y = 17 2 −8 −6 −4 −2 −2 (−1, 1) −2 −3 −4 2 −2 x + 43 y = −4 3x = 9 + 12 t y x −4 −3 −2 6. Because the equation is in the form a1 x + a2 y = b, it is linear in the variables x and y. 12. 1 x − 13 y = 1 2 −2x + 43 y = −4 x − x + 3 y = 17 4x + 3 y = 7 Subtracting the first equation from the second equation produces a new second equation, 5 x = −10 or x = −2. So, 4( −2) + 3 y = 7 or y = 5, and the solution is: x = −2, y = 5. This is the point where the two lines intersect. So, x = 2 − 3 y = 2 − 3(1), and the solution is: x = −1, y = 1. This is the point where the two lines intersect. 2 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 1.1 18. 22. y 3 −3 Introduction to Systems of Linear Equations x −3 −6 −9 3 y 250 6x + 5y = 21 9 150 (6, − 3) 100 6 0.3x + 0.4y = 68.7 0.2x − 0.5y = −27.8 (101, 96) x − 5y = 21 x 50 −50 150 x − 5 y = 21 0.2 x − 0.5 y = −27.8 6 x + 5 y = 21 0.3x + 0.4 y = 68.7 Adding the first equation to the second equation produces a new second equation, 7 x = 42 or x = 6. Multiplying the first equation by 40 and the second equation by 50 produces new equations. So, 6 − 5 y = 21 or y = −3, and the solution is: x = 6, y = −3. This is the point where the two lines intersect. 8 x − 20 y = −1112 15 x + 20 y = 3435 Adding the first equation to the second equation produces a new second equation, 23 x = 2323 or x = 101. y 20. 12 9 6 3 −3 y+2 x−1 + =4 3 2 So, 8(101) − 20 y = −1112 or y = 96, and the solution x − 2y = 5 (7, 1) 6 3 is: x = 101, y = 96. This is the point where the two lines intersect. x 12 x −1 y + 2 + = 4 2 3 x − 2y = 5 Multiplying the first equation by 6 produces a new first equation. 3x + 2 y = 23 x − 2y = 5 Adding the first equation to the second equation produces a new second equation, 4 x = 28 or x = 7. So, 7 − 2 y = 5 or y = 1, and the solution is: x = 7, y = 1. This is the point where the two lines intersect. y 24. 5 4 3 2 1 −1 −2 2 x + 16 y = 23 3 4x + y = 4 x 2 3 4 5 6 2x + 1 y = 2 3 6 3 4x + y = 4 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 y = 4 − 4t. So, you can describe the solution as x = t and y = 4 − 4t , where t is any real number. 26. From Equation 2 you have x2 = 3. Substituting this value into Equation 1 produces 2 x1 − 12 = 6 or x1 = 9. So, the system has exactly one solution: x1 = 9 and x2 = 3. 28. From Equation 3 you conclude that z = 2. Substituting this value into Equation 2 produces 2 y + 2 = 6 or y = 2. Finally, substituting y = 2 and z = 2 into Equation 1, you obtain x − 2 = 4 or x = 6. So, the system has exactly one solution: x = 6, y = 2, and z = 2. 30. From the second equation you have x2 = 0. Substituting this value into Equation 1 produces x1 + x3 = 0. Choosing x3 as the free variable, you have x3 = t and obtain x1 + t = 0 or x1 = −t. So, you can describe the solution set as x1 = −t , x2 = 0, and x3 = t. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 4 Chapter 1 Systems of Linear Equations 38. Adding −2 times the first equation to the second equation produces a new second equation. −8x + 10y = 14 32. (a) 4 4x − 5y = 3 −6 0 = 10 Because the second equation is a false statement, the original system of equations has no solution. −4 (b) This system is inconsistent, because you see two parallel lines on the graph of the system. 34. (a) 3x + 2 y = 2 6 40. Adding −6 times the first equation to the second equation produces a new second equation. x1 − 2 x2 = 0 1 x + 13 y = 0 2 2 14 x2 = 0 9x − 4y = 5 Now, using back-substitution, the system has exactly one solution: x1 = 0 and x2 = 0. −3 3 42. Multiplying the first equation by 32 produces a new first equation. −2 (b) Two lines corresponding to two equations intersect at a point, so this system is consistent. (c) The solution is approximately x = 13 and y = − 12 . (d) Adding −18 times the second equation to the first equation, you obtain −10 y = 5 or y = − 12 . Substituting y = − 12 into the first equation, you obtain 9 x = 3 or x = 13. The solution is: x = 13 and y = − 12 . (e) The solutions in (c) and (d) are the same. 36. (a) 2 −3 −5.3x + 2.1y = 1.25 3 x1 + 14 x2 = 0 4 x1 + x2 = 0 Adding −4 times the first equation to the second equation produces a new second equation. x1 + 14 x2 = 0 0 = 0 Choosing x2 = t as the free variable, you obtain x1 = − 14 t. So you can describe the solution set as x1 = − 14 t and x2 = t , where t is any real number. 44. To begin, change the form of the first equation. x1 x 5 + 2 = − 3 2 6 3×1 − x2 = − 2 Multiplying the first equation by 3 yields a new first equation. 15.9x − 6.3y = −3.75 −2 (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 25 . So let x = t , then the solution set is y = 53 x + 42 21 25 , where t is any real number. x = t , y = 53 t + 42 21 (d) Adding 3 times the first equation to the second equation you obtain −5.3x + 2.1 y = 1.25 0 = 0. Choosing x = t as the free variable, you obtain 2.1y = 5.3t + 1.25 or 21 y = 53t + 12.5 or 25 . So, you can describe the solution set y = 53 t + 42 21 25 , where t is any real as x = t , y = 53 t + 42 21 number. (e) The solutions in (c) and (d) are the same. 3 5 x2 = − 2 2 3×1 − x2 = − 2 x1 + Adding –3 times the first equation to the second equation produces a new second equation. 3 5 x2 = − 2 2 11 11 − x2 = 2 2 x1 + Multiplying the second equation by − 2 yields a new 11 second equation. x1 + 3 5 x2 = − 2 2 x2 = −1 Now, using back-substitution, the system has exactly one solution: x1 = −1 and x2 = −1. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 1.1 46. Multiplying the first equation by 20 and the second equation by 100 produces a new system. Introduction to Systems of Linear Equations 50. Interchanging the first and third equations yields a new system. x1 − 0.6 x2 = 4.2 x1 − 11×2 + 4 x3 = 3 7 x1 + 2 x1 + 4 x2 − 2 x2 = 17 Adding −7 times the first equation to the second equation produces a new second equation. x1 − 0.6 x2 = 4.2 6.2 x2 = −12.4 Now, using back-substitution, the system has exactly one solution: x1 = 3 and x2 = −2. 48. Adding the first equation to the second equation yields a new second equation. x+ y + z = 2 Adding −2 times the first equation to the second equation yields a new second equation. x1 − 11×2 + 4 x3 = 3 26 x2 − 9 x3 = 1 5 x1 − 3 x2 + 2 x3 = 3 Adding −5 times the first equation to the third equation yields a new third equation. x1 − 11×2 + 4 x3 = 3 26 x2 − 9 x3 = 1 Adding −4 times the first equation to the third equation yields a new third equation. x+ y + z = 2 4 y + 3z = 10 −3 y − 4 z = −4 Dividing the second equation by 4 yields a new second equation. x+ y + y + 5 2 Adding 3 times the second equation to the third equation yields a new third equation. y + z = 2 3z 4 − 74 z = 52 = 72 Multiplying the third equation by − 74 yields a new third equation. x+ y + 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. z = 2 y + 34 z = Now, using back-substitution the system has exactly one solution: x = 0, y = 4, and z = −2. 13 −2 x2 − 15 x3 = −45 −2 x2 − 15 x3 = −45 The third equation can be disregarded because it is the same as the second one. Choosing x3 as a free variable and letting x3 = t , you can describe the solution as x1 = 13 − 4t − 15 x2 = 45 t 2 2 x3 = t , where t is any real number. 54. Adding −3 times the first equation to the second equation produces a new second equation. x1 − 2 x2 + 5 x3 = 2 8 x2 − 16 x3 = −8 5 2 z = −2 + 4 x3 = x1 z = 2 3z = 4 −3 y − 4 z = −4 x + y + 52 x2 − 18 x3 = −12 = 4 y x3 = 7 5 x1 − 3 x2 + 2 x3 = 3 4 y + 3 z = 10 4x + 5 Dividing the second equation by 8 yields a new second equation. x1 − 2 x2 + 5 x3 = 2 x2 − 2 x3 = −1 Adding 2 times the second equation to the first equation yields a new first equation. x1 + x3 = 0 x2 − 2 x3 = −1 Letting x3 = t be the free variable, you can describe the solution as x1 = −t , x2 = 2t − 1, and x3 = t , where t is any real number. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 6 Chapter 1 Systems of Linear Equations 56. Adding −2 times the first equation to the fourth equation, yields + 3×4 = 4 x1 2 x2 − x3 − 8 x + 3 y + 3 z = 0. Multiplying the fourth equation by −1, and interchanging it with the second equation, yields + 3 x4 = 4 Adding −4 times the first equation to the second equation, and −8 times the first equation to the third, yields x + − 2 x4 = 2 x2 − x3 − 1 x4 = 0. Adding −3 times the second equation to the third, and −2 times the second equation to the fourth, produces x1 + 3 x4 = 4 x2 − 4 x3 + 6 x4 = 7 x3 − 13 x4 = −6. Dividing the third equation by 12 yields + 3 x4 = 4 x2 − 4 x3 + 6 x4 = 3 x3 − 5x 3 4 = − 23 7 x3 − 13 x4 = −6. Adding −7 times the third equation to the fourth yields x1 −9 y + 3 z = 0. Adding −3 times the second equation to the third equation yields x + = 0 3y 2 −3 y − z = 0 6 z = 0. 3 12 x3 − 20 x4 = −8 x1 = 0 3y 2 −3 y − z = 0 x2 − 4 x3 + 6 x4 = 3 3×2 = 0 4x + 3y − z = 0 1 − x2 + 4 x3 − 6 x4 = −3. x1 Dividing the first equation by 2 produces x + 32 y x4 = 0 − 2 x4 = 3×2 62. x = y = z = 0 is clearly a solution. Using back-substitution, you conclude there is exactly one solution: x = y = z = 0. 64. x = y = z = 0 is clearly a solution. Dividing the first equation by 12 yields 5 y + 1 z = 0 x + 12 12 12 x + 4 y − z = 0. Adding −12 times the first equation to the second yields 5 y + 1 z = 0 x + 12 12 + 3 x4 = 4 x2 − 4 x3 + 6 x4 = 3 Letting z = t be the free variable, you can describe the x3 − 53 x4 = − 32 solution as x = 34 t , y = −2t , and z = t , where t is any 4x 3 4 = 4 3. Using back-substitution, the original system has exactly one solution: x1 = 1, x2 = 1, x3 = 1, and x4 = 1. Answers may vary slightly for Exercises 58–60. 58. Using a computer software program or graphing utility, you obtain x = 0.8, y = 1.2, z = −2.4. 60. Using a computer software program or graphing utility, you obtain x = 6.8813, y = −163.3111, z = −210.2915, w = −59.2913. − y − 2 z = 0. real number. 66. Let x = the speed of the plane that leaves first and y = the speed of the plane that leaves second. y − x = 80 Equation 1 2 x + 32 y = 3200 −2 x + 2 y = 2x + 3 y 2 7 y 2 Equation 2 160 = 3200 = 3360 y = 960 960 − x = 80 x = 880 Solution: First plane: 880 kilometers per hour; second plane: 960 kilometers per hour © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 1.1 68. (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 of the text, and will be proved later in Theorem 2.5. (b) True. See definition on page 6 of the text. (c) False. Consider the following system of three linear equations with two variables. 2 x + y = −3 Introduction to Systems of Linear Equations 72. Substituting A = −6 x − 3 y = 9 = x 1. 2 A + 3B = 0 3A − 4B = − 25. 6 Reduce the system to row-echelon form. 8 A + 12 B = 0 9 A − 12 B = − 25 2 0 8 A + 12 B = x1 − x2 + x3 = 3 Letting x3 = t and x2 = s be the free variables, you can describe the solution as x1 = 3 + s − t , x2 = s, and x3 = t , where t and s are any real numbers. 2A + = − So, A = − 25 25 1 and B = . Because A = and x 34 51 1 , the solution of the original system of equations y 34 51 is: x = − and y = . 25 25 1 1 1 , B = , and C = into the original system yields y z x B − 2C = 5 3A − 4B 2A + 25 2 17 A B = − x1 + x2 − x3 = −3 74. Substituting A = 1 1 and B = into the original system y x yields The solution to this system is: x = 1, y = −5. 70. Because x1 = t and x2 = s, you can write x3 = 3 + s − t = 3 + x2 − x1. One system could be 7 = −1 B + 3C = 0. Reduce the system to row-echelon form. 2 A + B − 2C = 5 3A − 4B = −1 5C = −5 3A − 4B = −1 −11B + 6C = −17 5C = −5 So, C = −1. Using back-substitution, −11B + 6( −1) = −17, or B = 1 and 3 A − 4(1) = −1, or A = 1. Because A = 1 x, B = 1 y , and C = 1 z , the solution of the original system of equations is: x = 1, y = 1, and z = −1. 76. Multiplying the first equation by sin θ and the second by cos θ produces (sin θ cos θ ) x + (sin 2 θ ) y = sin θ −(sin θ cos θ ) x + (cos 2 θ ) y = cos θ . Adding these two equations yields (sin 2 θ + cos2 θ ) y = sin θ + cos θ y = sin θ + cos θ . So, (cos θ ) x + (sin θ ) y = (cos θ ) x + sin θ (sin θ + cos θ ) = 1 and x = (1 − sin 2 θ − sin θ cos θ ) = (cos2 θ − sin θ cos θ ) = cos θ − sin θ . cos θ cos θ Finally, the solution is x = cos θ − sin θ and y = cos θ + sin θ . © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 8 Chapter 1 Systems of Linear Equations 78. Interchange the two equations and row reduce. x − 32 y = −6 kx + y = 4 86. If c1 = c2 = c3 = 0, then the system is consistent because x = y = 0 is a solution. 88. Multiplying the first equation by c, and the second by a, produces x − 32 y = −6 acx + bcy = ec ( 32 k + 1) y = 4 + 6k acx + day = af . = − 23 , there will be an infinite number of So, if k Subtracting the second equation from the first yields acx + bcy = ec solutions. (ad − bc) y = af − ec. 80. Reduce the system. So, there is a unique solution if ad − bc ≠ 0. x + ky = 2 (1 − k ) y = 4 − 2k 2 If k = ±1, there will be no solution. 82. Interchange the first two equations and row reduce. x + y + z = 0 ky + 2kz = 4k −3 y − z = z = 0 −3 −2 −1 −2 x 1 4 5 The two lines coincide. 2x − 3y = 7 0 = 0 y + 2z = 4 5 z = 13. Letting y = t , x = Because this system has exactly one solution, the answer is all k ≠ 0. 84. Reducing the system to row-echelon form produces x + 3 2 1 −4 −5 1 If k = 0, then there is an infinite number of solutions. Otherwise, x + y + y 90. 5y + z = 0 y − 2z = 0 (a − 10) y + (b − 2) z = c x + 5y + z = 0 y − 2z = 0 (2a + b − 22) z = c. 7 + 3t . 2 The graph does not change. 92. 21x − 20 y = 0 13x − 12 y = 120 Subtracting 5 times the second equation from 3 times the first equation produces a new first equation, −2 x = −600, or x = 300. So, 21(300) − 20 y = 0 or y = 315, and the solution is: x = 300, y = 315. The graphs are misleading because they appear to be parallel, but they actually intersect at (300, 315). So, you see that (a) if 2a + b − 22 ≠ 0, then there is exactly one solution. (b) if 2a + b − 22 = 0 and c = 0, then there is an infinite number of solutions. (c) if 2a + b − 22 = 0 and c ≠ 0, there is no solution. Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination 2. Because the matrix has 4 rows and 1 column, it has size 4 × 1. 6. Because the matrix has 1 row and 5 columns, it has size 1 × 5. 4. Because the matrix has 1 row and 1 column, it has size 1 × 1. ⎡ 3 −1 −4⎤ ⎡3 −1 −4⎤ 8. ⎢ ⎥ ⇒ ⎢ ⎥ ⎣−4 3 7⎦ ⎣5 0 −5⎦ Add 3 times Row 1 to Row 2. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination 3 −2⎤ ⎡−1 −2 ⎡−1 −2 3 −2⎤ ⎢ ⎥ ⎢ ⎥ 10. ⎢ 2 − 5 1 − 7⎥ ⇒ ⎢ 0 − 9 7 −11⎥ ⎢ 5 ⎢ 0 − 6 8 − 4⎥ 4 −7 6⎥⎦ ⎣ ⎣ ⎦ Add 2 times Row 1 to Row 2. Add 5 times Row 1 to Row 3. 12. Because the matrix is in reduced row-echelon form, you can convert back to a system of linear equations x1 = 2 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. 8⎤ ⎡ 2 6 16⎤ ⎡ 1 3 ⎡ 1 3 8⎤ ⎢ ⎥ ⇒ ⎢ ⎥ ⇒ ⎢ ⎥ ⎣−2 −6 −16⎦ ⎣−2 −6 −16⎦ ⎣0 0 0⎦ x2 = 3. 14. Because the matrix is in row-echelon form, you can convert back to a system of linear equations x1 + 2 x2 + x3 = 0 x3 = −1. Using back-substitution, you have x3 = −1. Letting x2 = t be the free variable, you can describe the solution as x1 = 1 − 2t , x2 = t , and x3 = −1, where t is any real number. 16. Gaussian elimination produces the following. ⎡2 1 1 0⎤ ⎡ 1 0 1 0⎤ ⎢ ⎥ ⎢ ⎥ 1 − 2 1 − 2 ⇒ ⎢ ⎥ ⎢ 1 −2 1 −2⎥ ⎢ 1 0 1 0⎥ ⎢2 1 1 0⎥ ⎣ ⎦ ⎣ ⎦ Converting back to a system of linear equations, you have x + 3 y = 8. Choosing y = t as the free variable, you can describe the solution as x = 8 − 3t 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 − 12 − 20 ⎡2 −1 −0.1⎤ ⎢ ⎥ ⇒ ⎢ 8⎥ 2 ⎥ ⎣ 3 2 1.6⎦ 5⎦ ⎣⎢3 ⎡ 1 − 12 ⇒ ⎢ 7 2 ⎣⎢0 ⎡ 1 0 1 0⎤ ⎡ 1 0 1 0⎤ ⎢ ⎥ ⎢ ⎥ ⇒ ⎢0 −2 0 −2⎥ ⇒ ⎢0 1 0 1⎥ ⎢2 1 1 0⎥ ⎢2 1 1 0⎥ ⎣ ⎦ ⎣ ⎦ ⎡ 1 0 1 0⎤ ⎡ 1 0 1 0⎤ ⎢ ⎥ ⎢ ⎥ ⇒ ⎢0 1 0 1⎥ ⇒ ⎢0 1 0 1⎥ ⎢0 1 −1 0⎥ ⎢0 0 1 1⎥ ⎣ ⎦ ⎣ ⎦ Because the matrix is in row-echelon form, convert back to a system of linear equations. + x3 = 0 x1 x2 9 = 1 x3 = 1 By back-substitution, x1 = − x3 = −1. So, the solution is: x1 = −1, x2 = 1, and x3 = 1. 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. 1⎤ − 20 ⎥ 7 ⎥ 4⎦ 1⎤ ⎡ 1 − 12 − 20 ⎡1 0 ⇒ ⎢ ⎥ ⇒ ⎢ 1 1 ⎢⎣0 ⎢⎣0 1 2⎥ ⎦ 1⎤ 5 ⎥ 1 2⎥ ⎦ Converting back to a system of equations, the solution is: x = 15 and y = 12 . 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⎥ ⎣ ⎦ ⎣ ⎦ 0⎤ ⎡ 1 2 0⎤ ⎡1 2 ⎢ ⎥ ⎢ ⎥ 1 −6⎥ ⇒ ⎢0 1 −6⎥ ⇒ ⎢0 ⎢0 −8 8⎥ ⎢0 0 −40⎥ ⎣ ⎦ ⎣ ⎦ Because the third row corresponds to the equation 0 = − 40, you conclude that the system has no solution. 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. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 10 Chapter 1 Systems of Linear Equations 32. The augmented matrix for this system is ⎡2 −1 3 24⎤ ⎢ ⎥ ⎢0 2 −1 14⎥. ⎢7 −5 0 6⎥ ⎣ ⎦ Gaussian elimination produces the following. 3 12⎤ 3 3 ⎡1 − 1 ⎡1 − 1 ⎡1 − 1 12⎤ ⎡2 −1 3 24⎤ 2 2 2 2 2 2 ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ 1 2 −1 14⎥ ⇒ ⎢0 2 −1 14⎥ ⇒ ⎢0 1 −2 ⎢0 2 −1 14⎥ ⇒ ⎢0 ⎢7 −5 0 6⎥ ⎢0 − 3 − 21 −78⎥ ⎢0 ⎢7 −5 0 6⎥ 0 − 45 ⎣ ⎦ 2 2 4 ⎣ ⎦ ⎣ ⎦ ⎣ 12⎤ ⎥ 7⎥ ⎥ − 135 2 ⎦ Back-substitution now yields x3 = 6 x2 = 7 + 12 x3 = 7 + 12 (6) = 10 x1 = 12 − 32 x3 + 12 x2 = 12 − 23 (6) + 12 (10) = 8. So, the solution is: x1 = 8, x2 = 10, and x3 = 6. 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. 3⎤ ⎡ 1 1 −5 ⎢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 x1 − 2 x3 = 1 x2 − 3 x3 = 2. Finally, choosing x3 = t as the free variable, you can describe the solution as x1 = 1 + 2t , x2 = 2 + 3t , and x3 = t , where t is any real number. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination 11 36 The augmented matrix for this system is 1 8⎤ ⎡ 1 2 ⎢ ⎥. ⎣−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. 38. The augmented matrix for this system is ⎡2 ⎢ ⎢3 ⎢1 ⎢ ⎣⎢5 1 −1 2 −6⎤ ⎥ 1 1⎥ . 5 2 6 −3⎥ ⎥ 2 −1 −1 3⎥⎦ 4 0 Gaussian elimination produces the following. ⎡1 ⎢ ⎢3 ⎢2 ⎢ ⎢⎣5 6 −3⎤ 5 2 6 −3⎤ 5 2 6 −3⎤ ⎡1 ⎡1 ⎥ ⎢ ⎥ ⎢ 6 17 − 10 ⎥ 1 1⎥ 0 − 11 − 6 − 17 10 0 1 ⎥ ⇒ ⎢ 11 11 11 ⎥ ⇒ ⎢ ⎢0 −9 −5 −10 0⎥ ⎢0 −9 −5 −10 1 −1 2 −6⎥ 0⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣0 −23 −11 −31 18⎥⎦ ⎢⎣0 −23 −11 −31 18⎥⎦ 2 −1 −1 3⎥⎦ 5 2 4 0 ⎡1 ⎢ ⎢0 ⇒ ⎢ 0 ⎢ ⎢⎣0 ⎡1 ⎢ ⎢0 ⇒ ⎢ 0 ⎢ ⎢⎣0 5 2 6 1 6 11 1 − 11 17 11 17 11 43 11 50 11 0 0 −3⎤ ⎡1 ⎥ ⎢ 10 − 11 ⎥ ⎢0 ⇒ ⎢0 1 − 43 90⎥ ⎥ ⎢ 781 1562 − 11 ⎥⎦ 0 ⎢⎣0 11 5 2 6 1 6 11 17 11 0 0 −3⎤ ⎡1 ⎥ ⎢ 10 − 11 ⎥ ⎢0 ⇒ 90 ⎥ ⎢0 − 11 ⎥ ⎢ ⎥ ⎢⎣0 − 32 11 ⎦ −3⎤ ⎥ − 10 1 11 ⎥ 0 1 − 43 90⎥ ⎥ 50 − 32 ⎥ 0 17 11 11 4⎦ 5 2 6 6 11 17 11 −3⎤ ⎥ − 10 11 ⎥ 1 − 43 90⎥ ⎥ 0 1 −2⎥⎦ 5 2 6 1 6 11 17 11 0 0 Back-substitution now yields w = −2 z = 90 + 43w = 90 + 43( −2) = 4 6 z − 17 w = − 10 − 6 4 − 17 −2 = 0 − 11 y = − 10 ( ) 11 ( ) 11 11 11 ( ) 11 ( ) . x = −3 − 5 y − 2 z − 6w = −3 − 5(0) − 2( 4) − 6( −2) = 1. So, the solution is: x = 1, y = 0, z = 4, and w = −2. 40. Using a computer software program or graphing utility, you obtain x1 = 1 x2 = −1 x3 = 2 x4 = 0 x5 = −2 x6 = 1. 42. The corresponding equations are x1 = 0 x2 + x3 = 0. Choosing x4 = t and x3 = t as the free variables, you can describe the solution as x1 = 0, x2 = − s, x3 = s, and x4 = t , where s and t are any real numbers. 44. The corresponding equations are all 0 = 0. So, there are three free variables. So, x1 = t , x2 = s, and x3 = r , where t , s, and r are any real numbers. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 12 Chapter 1 Systems of Linear Equations 46. x = number of $1 bills y = number of $5 bills z = number of $10 bills w = number of $20 bills x + 5 y + 10 z + 20w = 95 x + y + z + w = 26 y − 4z x − 2y = 0 = −1 ⎡ 1 5 10 20 95⎤ ⎡1 ⎢ ⎥ ⎢ ⎢ 1 1 1 1 26⎥ ⇒ ⎢0 ⎢0 ⎢0 1 −4 0 0⎥ ⎢ ⎥ ⎢ ⎢⎣0 ⎣⎢ 1 −2 0 0 −1⎥⎦ 0 0 0 15⎤ ⎥ 1 0 0 8⎥ 0 1 0 2⎥ ⎥ 0 0 1 1⎥⎦ x = 15 y = 8 z = 2 w =1 The server has 15 $1 bills, 8 $5 bills, 2 $10 bills, and one $20 bill. 48. (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. 3 ⎤ ⎡ 2 −1 3⎤ ⎡2 −1 ⎢ ⎥ ⎢ ⎥ ⎢−4 2 k ⎥ ⇒ ⎢0 0 k + 6⎥ ⎢ 4 −2 6⎥ ⎢0 0 0 ⎥⎦ ⎣ ⎦ ⎣ This system is consistent if and only if k + 6 = 0, so k = −6. 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 − 1 2 2 ⎢ ⎥ 0 k + 6⎥ ⎢0 ⎢0 0 0 ⎥⎦ ⎣ 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 ⎢ ⎢0 ⎢1 ⎢ ⎣⎢a 1 0 0⎤ ⎡1 ⎢ ⎥ 0 1 1 0⎥ ⇒ ⎢ ⎢ ⎥ 0 0 1 0 ⎢ ⎥ b c 0⎦⎥ ⎣⎢0 1 1 −1 (b − a ) 0 0⎤ ⎡1 ⎥ ⎢ 1 0⎥ 0 ⇒ ⎢ ⎥ ⎢ 1 0 0 ⎥ ⎢ c 0⎦⎥ ⎣⎢0 1 0 1 1 0 2 0 ( a − b + c) 0⎤ ⎡1 ⎥ ⎢ 0⎥ 0 ⇒ ⎢ ⎥ ⎢ 0 0 ⎥ ⎢ 0⎦⎥ ⎣⎢0 1 0 0⎤ ⎥ 1 1 0⎥ 0 1 0⎥ ⎥ 0 0 0⎦⎥ 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. 54. Use Gauss-Jordan elimination as follows. 3⎤ ⎡1 2 3⎤ ⎡1 2 ⎡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⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination 13 56. Begin by finding all possible first rows [0 0 0], [0 0 1], [0 1 0], [0 1 a], [1 0 0], [1 0 a], [1 a b], [1 a 0], 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 a⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢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⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡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 a⎥ , ⎢0 0 0⎥ ⎢0 0 0⎥ ⎢0 0 1⎥ ⎢0 0 0⎥ ⎢0 0 0⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡1 a 0⎤ ⎡1 a 0⎤ ⎡1 a b⎤ ⎡1 0 a⎤ ⎡1 0 a⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢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⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 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 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 c = 0 is not an elementary row operation. (This would change the system by eliminating the equation corresponding to this row.) ⎡1 2⎤ ⎡1 0⎤ 60. No, the row-echelon form is not unique. For instance, ⎢ ⎥ and ⎢ ⎥. The reduced row-echelon form is unique. ⎣0 1⎦ ⎣0 1⎦ 62. First, you need a ≠ 0 or c ≠ 0. If a ≠ 0, then you have b ⎡a ⎤ b ⎤ ⎡a b ⎤ ⎢ ⎥ ⇒ ⎡a ⇒ cb ⎢ ⎥ ⎢ ⎥. ⎢ ⎥ +b 0 − ⎣c d ⎦ ⎣0 ad − bc⎦ ⎢⎣ ⎥ a ⎦ So, ad − bc = 0 and b = 0, which implies that d = 0. If c ≠ 0, then you interchange rows and proceed. d ⎡c ⎤ d ⎤ ⎡a b ⎤ ⎢ ⎥ ⇒ ⎡c ⇒ ad ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ − bc⎦ c d 0 ad − + 0 b ⎣ ⎦ ⎣ c ⎣⎢ ⎦⎥ ⎡1 0⎤ ⎡a b ⎤ Again, ad − bc = 0 and d = 0, which implies that b = 0. In conclusion, ⎢ ⎥ if and only if ⎥ is row-equivalent to ⎢ c d ⎣0 0⎦ ⎣ ⎦ b = d = 0, and a ≠ 0 or c ≠ 0. 64. Row reduce the augmented matrix for this system. λ 0⎤ ⎡1 λ 0⎤ ⎡λ − 1 2 0⎤ ⎡ 1 ⎥ ⎢ ⎥ ⇒ ⎢ ⎥ ⇒ ⎢ 2 λ 0⎦ ⎢⎣0 ( −λ + λ + 2) 0⎥⎦ ⎣ 1 ⎣λ − 1 2 0⎦ To have a nontrivial solution you must have λ2 − λ − 2 = 0 (λ − 2)(λ + 1) = 0. So, if λ = −1 or λ = 2, the system will have nontrivial solutions. 66. Answers will vary. Sample answer: Because the third row consists of all zeros, choose a third equation that is a multiple of one of the other two equations. x + 3 z = −2 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. y + 4z = 1 2 y + 8z = 2 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 14 Chapter 1 Systems of Linear Equations Section 1.3 Applications of Systems of Linear Equations 2. (a) Because there are three points, choose a second-degree polynomial, p ( x) = a0 + a1 x + a2 x 2 . Then substitute x = 0, 2, and 4 into p ( x) and equate the results to y = 0, − 2, and 0, respectively. a0 + a1 (0) + a2 (0) = a0 2 = 0 a0 + a1 ( 2) + a2 ( 2) = a0 + 2a1 + 4a2 = − 2 2 a0 + a1 ( 4) + a2 ( 4) = a0 + 4a1 + 16a2 = 0 2 Use Gauss-Jordan elimination on the augmented matrix for this system. ⎡1 0 0 0⎤ 0⎤ ⎡1 0 0 ⎢ ⎥ ⎢ ⎥ 1 2 4 − 2 ⇒ 0 1 0 − 2⎥ ⎢ ⎢ ⎥ ⎢0 0 1 1⎥ ⎢1 4 16 0⎥⎦ ⎣ ⎣ 2⎦ So, p ( x) = − 2 x + 12 x 2 . y (b) 4 (0, 0) −2 (4, 0) x 2 −2 4 6 (2, −2) −4 4. (a) Because there are three points, choose a second-degree polynomial, p( x) = a0 + a1 x + a2 x 2 . Then substitute x = 2, 3, and 4 into p( x) and equate the results to y = 4, 4, and 4, respectively. a0 + a1 ( 2) + a2 ( 2) = a0 + 2a1 + 4a2 = 4 2 a0 + a1 (3) + a2 (3) = a0 + 3a1 + 9a2 = 4 2 a0 + a1 ( 4) + a2 ( 4) = a0 + 4a1 + 16a2 = 4 2 Use Gauss-Jordan elimination on the augmented matrix for this system. ⎡1 2 4 4⎤ ⎡1 0 0 4⎤ ⎢ ⎥ ⎢ ⎥ ⎢1 3 9 4⎥ ⇒ ⎢0 1 0 0⎥ ⎢1 4 16 4⎥ ⎢0 0 1 0⎥ ⎣ ⎦ ⎣ ⎦ So, p( x) = 4. y (b) 5 (2, 4) (4, 4) (3, 4) 3 2 1 x 1 2 3 4 5 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 1.3 Applications of Systems of Linear Equations 15 6. (a) Because there are four points, choose a third-degree polynomial, p ( x) = a0 + a1 x + a2 x 2 + a3 x3 . Then substitute x = 0, 1, 2, and 3 into p ( x) and equate the results to y = 42, 0, − 40, and − 72, respectively. a0 + a1 (0) + a2 (0) + a3 (0) = a0 2 a0 + a1 (1) + a2 (1) 3 + a3 (1) 2 3 = 42 = a0 + a1 + a2 + a3 a0 + a1 ( 2) + a2 ( 2) + a3 ( 2) = a0 + 2a1 + 4a2 + 8a3 2 3 = 0 = − 40 a0 + a1 (3) + a2 (3) + a3 (3) = a0 + 3a1 + 9a2 + 27 a3 = − 72 2 2 Use Gauss-Jordan elimination on the augmented matrix for this system. ⎡1 ⎢ ⎢1 ⎢1 ⎢ ⎣⎢1 42⎤ ⎡1 ⎥ ⎢ 0⎥ 0 ⇒ ⎢ ⎥ ⎢ 2 4 8 − 40 0 ⎥ ⎢ 3 9 27 − 72⎥⎦ ⎣⎢0 0 0 0 1 1 1 42⎤ ⎥ 1 0 0 − 41⎥ 0 1 0 − 2⎥ ⎥ 0 0 1 1⎦⎥ 0 0 0 So, p ( x) = 42 − 41x − 2 x 2 + x3 . y (b) 60 30 (0, 42) (1, 0) x −4 −2 −60 −90 4 6 8 10 (3, −72) (2, −40) 8. (a) Because there are five points, choose a fourth-degree polynomial, p ( x) = a0 + a1 x + a2 x 2 + a3 x3 + a4 x 4 . Then substitute x = − 4, 0, 4, 6, and 8 into p ( x) and equate the results to y = 18, 1, 0, 28, and 135, respectively. a0 + a1 ( − 4) + a2 ( − 4) + a3 (− 4) + a4 (− 4) = a0 − 4a1 + 16a2 − 2 3 a0 + a1 (0) + a2 ( 0 ) 2 a0 + a1 ( 4) + a2 ( 4) 2 + a3 ( 4) a0 + a1 (6) + a2 ( 6 ) 2 + a3 (6) a0 + a1 (8) + a2 (8) 2 + a3 (8) 4 + a3 (0) 3 3 3 3 + a4 (8) 64a3 + 256a4 = 18 + a 4 ( 0) 4 = a0 + a4 ( 4) 4 = a0 + 4a1 + 16a2 + 64a3 + 256a4 = 0 + a4 ( 6) 4 = a0 + 6a1 + 36a2 + 216a3 + 1296a4 = 28 4 = a0 + 8a1 + 64a2 + 512a3 + 4096a4 = 135 =1 Use Gauss-Jordan elimination on the augmented matrix for this system. ⎡1 ⎡1 − 4 16 − 64 256 18⎤ ⎢ ⎢ ⎥ 0 0 0 0 1⎥ ⎢0 ⎢1 ⎢1 4 16 64 256 0⎥ ⇒ ⎢0 ⎢ ⎢ ⎥ ⎢0 6 36 216 1296 28⎥ ⎢1 ⎢ ⎢ ⎥ 8 64 512 4096 135⎦ ⎣1 ⎣⎢0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1⎤ 3⎥ 4⎥ − 12 ⎥ ⎥ 3⎥ − 16 1⎥ ⎥ 16 ⎦ 3 3 1 x 4 = 1 16 + 12 x − 8 x 2 − 3 x 3 + x 4 . So, p ( x) = 1 + 34 x − 12 x 2 − 16 x + 16 ) 16 ( y (b) (8, 135) 120 80 (0, 1) (6, 28) 40 (4, 0) (− 4, 18) −8 −4 x 4 8 12 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 16 Chapter 1 Systems of Linear Equations 10. Assume that the equation of the ellipse is x 2 + ax + by 2 + cy − d = 0. Because each of the given points lies on the ellipse, you have the following linear equations. (− 5) + a(− 5) + b(1) + c(1) − d = − 5a + b + c − d + 25 = 0 2 2 (−3) + a(− 3) + b(2) + c( 2) − d = − 3a + 4b + 2c − d + 9 = 0 2 2 (−1) + a(−1) + b(1) + c(1) − d = − a + b + c − d + 1 = 0 2 2 − d + 9 = 0 ( − 3) + a(− 3) + b(0) + c(0) − d = − 3a 2 2 Use Gauss-Jordan elimination on the system. ⎡− 5 ⎢ ⎢− 3 ⎢ −1 ⎢ ⎣⎢− 3 1 1 −1 − 25⎤ ⎡1 ⎥ ⎢ 4 2 −1 − 9⎥ 0 ⇒ ⎢ ⎢0 1 1 −1 −1⎥ ⎥ ⎢ 0 0 −1 − 9⎥⎦ ⎢⎣0 6⎤ ⎥ 1 0 0 4⎥ 0 1 0 − 8⎥ ⎥ 0 0 1 − 9⎥⎦ 0 0 0 ( x + 3) + ( y − 1) 2 So, the equation of the ellipse is x 2 + 6 x + 4 y 2 − 8 y + 9 = 0 or 4 1 2 = 1. 12. (a) Because there are four points, choose a third-degree polynomial, p ( x ) = a0 + a1 x + a2 x 2 + a3 x3 . Then substitute x = 1, 1.189, 1.316, and 1.414 into p ( x) and equate the results to y = 1, 1.587, 2.080, and 2.520, respectively. a0 + a1 (1) + a2 (1) + a3 (1) 2 3 = a0 + a1 + a2 + a3 = 1 a0 + a1 (1.189) + a2 (1.189) + a3 (1.189) ≈ a0 + 1.189a1 + 1.414a2 + 1.681a3 = 1.587 2 3 a0 + a1 (1.316) + a2 (1.316) + a3 (1.316) ≈ a0 + 1.316a1 + 1.732a2 + 2.279a3 = 2.080 2 3 a0 + a1 (1.414) + a2 (1.414) + a3 (1.414) ≈ a0 + 1.414a1 + 1.999a2 + 2.827 a3 = 2.520 2 3 Use Gauss-Jordan elimination on the augmented matrix for this system. ⎡1 ⎢ ⎢1 ⎢1 ⎢ ⎣⎢1 1 1 1 ⎤ ⎡1 ⎥ ⎢ 1.189 1.414 1.681 1.587 ⎥ 0 ⇒ ⎢ ⎥ ⎢ 1.316 1.732 2.279 2.080 0 ⎥ ⎢ 1.414 1.999 2.827 2.520⎦⎥ ⎣⎢0 1 0 0 0 − 0.095⎤ ⎥ 1 0 0 0.103⎥ 0 1 0 0.405⎥ ⎥ 0 0 1 0.587⎦⎥ So, p ( x) ≈ − 0.095 + 0.103 x + 0.405 x 2 + 0.587 x3 . y (b) 4 3 2 1 −3 −2 (1.414, 2.520) (1.316, 2.080) (1.189, 1.587) (1, 1) x −1 −2 −3 −4 1 2 3 14. Choosing a second-degree polynomial approximation p( x) = a0 + a1 x + a2 x 2 , substitute x = 1, 2, and 4 into p( x) and equate the results to y = 0, 1, and 2, respectively. a0 + a1 + a2 = 0 a0 + 2a1 + 4a2 = 1 a0 + 4a1 + 16a2 = 2 The solution to this system is a0 = − 43 , a1 = 32 , and a2 = − 16 . So, p( x) = − 43 + 23 x − 16 x 2 . Finally, to estimate log 2 3, calculate p(3) = − 43 + 23 (3) − 16 (3) = 53 . 2 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 1.3 Applications of Systems of Linear Equations 17 16. Assume that the equation of the ellipse is ax 2 + cy 2 + dx + ey + f = 0. Because each of the given points lies on the ellipse, you have the following linear equations. a( − 5) + c(1) + d (−5) + e(1) + f = 25a + c − 5d + e + f = 0 a( − 3) + c( 2) + d ( −3) + e( 2) + f = 9a + 4c − 3d + 2e + f = 0 a( −1) + c(1) + d ( −1) + e(1) + f = d + e + f = 0 2 2 2 2 2 2 a + c − a( − 3) + c(0) + d ( −3) + e(0) + f = 9a 2 2 − 3d + f = 0 Use Gauss-Jordan elimination on the system. ⎡25 ⎢ ⎢ 9 ⎢ 1 ⎢ ⎢⎣ 9 ⎡1 1 1 0⎤ ⎢ ⎥ 4 −3 2 1 0⎥ ⎢0 ⇒ ⎢ ⎥ 1 −1 1 1 0 ⎢0 ⎥ ⎢0 0 −3 0 1 0⎥⎦ ⎣ 1 −5 0 0 0 − 19 1 0 0 0 1 0 0 0 1 − 94 − 23 8 9 0⎤ ⎥ 0⎥ 0⎥⎥ 0⎥⎦ Letting f = t be the free variable, you have a = 19 t , c = 94 t , d = 32 t , e = − 98 t , and f = t , where t is any real number. 18. (a) Letting z = x − 1940 , the four data points are (0, 132), (1, 151), ( 2, 179), and (3, 203). 10 Let p ( z ) = a0 + a1 z + a2 z 2 + a3 z 3 . Substituting the points into p ( z ) produces the following system of linear equations. a0 + a1 (0) + a2 (0) + a3 (0) = a0 2 3 = 132 a0 + a1 (1) + a2 (1) + a3 (1) = a0 + a1 + a2 + a3 = 151 a0 + a1 ( 2) + a2 ( 2) + a3 ( 2) = a0 + 2a1 + 4a2 + 8a3 = 179 2 2 3 3 a0 + a1 (3) + a2 (3) + a3 (3) = a0 + 3a1 + 9a2 + 27 a3 = 203 2 3 Form the augmented matrix ⎡1 ⎢ ⎢1 ⎢1 ⎢ ⎢⎣1 132 ⎤ ⎥ 1 1 1 151 ⎥ 2 4 8 179 ⎥ ⎥ 3 9 27 203⎥⎦ 0 0 0 and use Gauss-Jordan elimination to obtain the equivalent reduced row-echelon matrix ⎡1 ⎢ ⎢0 ⎢ ⎢ ⎢0 ⎢ ⎢0 ⎣ 132⎤ ⎥ 61⎥ 1 0 0 6⎥ ⎥. 0 1 0 11⎥ 13 ⎥ 0 0 1 − ⎥ 6⎦ 0 0 0 So, the cubic polynomial is p ( z ) = 132 + 61 13 3 z + 11z 2 − z . 6 6 2 Because z = 3 x − 1940 61⎛ x − 1940 ⎞ 13 ⎛ x − 1940 ⎞ ⎛ x − 1940 ⎞ , p ( x) = 132 + ⎜ ⎟ + 11⎜ ⎟ − ⎜ ⎟ . 10 6⎝ 10 10 6⎝ 10 ⎠ ⎝ ⎠ ⎠ (b) To estimate the population in 1980, let x = 1980. p (1980) = 132 + 61 13 3 2 ( 4) + 11( 4) − ( 4) = 210 million, which is 6 6 less than the actual population of 227 million. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 18 Chapter 1 Systems of Linear Equations 20. (a) Letting z = x − 2000, the five points are ( 2, 244.5), (3, 256.3), ( 4, 285.2), (5, 312.4), and (6, 345.0). Let p ( z ) = a0 + a1 z + a2 z 2 + a3 z 3 + a4 z 4 . a0 + a1 ( 2) + a2 ( 2) + a3 ( 2) + a4 ( 2) = a0 + 2a1 + 4a2 + 8a3 + 16a4 = 244.5 a0 + a1 (3) + a2 (3) + a3 (3) + a4 (3) = a0 + 3a1 + 9a2 + 27 a3 + 81a4 = 256.3 2 3 2 4 3 4 a0 + a1 ( 4) + a2 ( 4) + a3 ( 4) + a4 ( 4) = a0 + 4a1 + 16a2 + 64a3 2 3 4 + 256a4 = 285.2 a0 + a1 (5) + a2 (5) + a3 (5) + a4 (5) = a0 + 5a1 + 25a2 + 125a3 + 625a4 = 312.4 2 3 4 a0 + a1 (6) + a2 (6) + a3 (6) + a4 (6) = a0 + 6a1 + 36a2 + 216a3 + 1296a4 = 345.0 2 3 4 (b) Use Gauss-Jordan elimination to solve the system. ⎡1 ⎢ ⎢1 ⎢1 ⎢ ⎢1 ⎢ ⎣1 2 3 4 5 6 ⎡ ⎢1 ⎢ ⎢ 4 8 16 244.5⎤ ⎢0 ⎥ 9 27 81 256.3⎥ ⎢ 16 64 256 285.2⎥ ⇒ ⎢⎢0 ⎥ ⎢ 25 125 625 312.4⎥ ⎢0 ⎥ 36 216 1296 345.0⎦ ⎢ ⎢ ⎢0 ⎣⎢ 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 4769 ⎤ 10 ⎥ ⎥ 33,433 ⎥ − 120 ⎥ ⎥ 27,209 ⎥ 240 ⎥ ⎥ 2189 ⎥ − 120 ⎥ ⎥ 259 ⎥ 240 ⎦⎥ 4769 33,433 27,209 2 2189 3 259 4 − z + z − z + z . 10 120 240 120 240 So, p ( z ) = Letting z = x − 2000, p ( x) = ( x − 2000) + 3 4769 33,433 27,209 2189 2 − ( x − 2000) + ( x − 2000) − 10 120 240 120 259 4 ( x − 2000) . 240 To determine the reasonableness of the model for years after 2006, compare the predicted values for 2007–2009 to the actual values. x 2007 2008 2009 p ( x) 416 584.3 934.7 Actual 374.5 401.2 405.0 The predicted values are not close to the actual values. 22. Let p1 ( x) = a0 + a1 x + a2 x 2 + p2 ( x) = b0 + b1x + b2 x 2 + + an −1x n −1 and + bn −1 x n −1 be two different polynomials that pass through the n given points. The polynomial p1 ( x) − p2 ( x) = ( a0 − b0 ) + ( a1 − b1 ) x + ( a2 − b2 ) x 2 + + ( an −1 − bn −1 ) x n −1 is zero for these n values of x. So, a0 = b0 , a1 = b1 , a2 = b2 , …, an −1 = bn −1. Therefore, there is only one polynomial function of degree n − 1 (or less) whose graph passes through n points in the plane with distinct x-coordinates. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 1.3 Applications of Systems of Linear Equations 24. Let p( x) = a0 + a1x + a2 x 2 be the equation of the parabola. Because the parabola passes through the points (0, 1) and 26. Choose a fourth-degree polynomial and substitute x = 1, 2, 3, and 4 into ( 12 , 12 ), you have a0 + a1 (0) + a2 (0) = a0 2 () = 1 ( ) = a + 12 a + 14 a = 12 . a0 + a1 12 + a2 12 2 0 1 2 Because p( x) has a horizontal tangent at 19 ( 12 , 12 ), the derivative of p( x), p′( x) = a1 + 2a2 x, equals zero when x = 12 . So, you have a third linear equation () a1 + 2a2 12 = a1 + a2 = 0. Use Gauss-Jordan elimination on the augmented matrix for this linear system. p( x) = a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 . However, when you substitute x = 3 into p( x) and equate it to y = 2 and y = 3 you get the contradictory equations a0 + 3a1 + 9a2 + 27 a3 + 81a4 = 2 a0 + 3a1 + 9a2 + 27 a3 + 81a4 = 3 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 p( y ) = b0 + b1 y + b2 y 2 + b3 y 3 + b4 y 4 because y = 1 corresponds to both x = 1 and x = 2. 1⎤ ⎡ 1 0 0 1⎤ ⎡1 0 0 ⎢ 1 1 1⎥ ⎢ ⎥ 1 ⇒ 0 1 0 − 2 ⎢ ⎢ ⎥ 2 4 2⎥ ⎢0 1 1 0⎥ ⎢0 0 1 2⎥ ⎣ ⎦ ⎣ ⎦ So, p( x) = 1 − 2 x + 2 x 2 . y 1 1 2 (0, 1) ( 12, 12 ) x 1 2 1 28. (a) Each of the network’s four junctions gives rise to a linear equation as shown below. input = output 300 = x1 + x2 x1 + x3 = x4 + 150 x2 + 200 = x3 + x5 x4 + x5 = 350 Rearrange these equations, form the augmented matrix, and use Gauss-Jordan elimination. ⎡ 1 1 0 0 0 300⎤ ⎡ 1 0 1 0 1 500⎤ ⎢ 1 0 1 −1 0 150⎥ ⎢ ⎥ ⎢ ⎥ ⇒ ⎢0 1 −1 0 −1 −200⎥ ⎢0 1 −1 0 −1 −200⎥ ⎢0 0 0 1 1 350⎥ ⎢ ⎥ ⎢ ⎥ 0⎦ ⎣0 0 0 1 1 350⎦ ⎣0 0 0 0 0 Letting x5 = t and x3 = s be the free variables, you have x1 = 500 − s − t x2 = −200 + s + t x3 = s x4 = 350 − t x5 = t , where t and s are any real numbers. (b) If x2 = 200 and x3 = 50, then you have s = 50 and t = 350. So, the solution is: x1 = 100, x2 = 200, x3 = 50, x4 = 0, and x5 = 350. (c) If x2 = 150 and x3 = 0, then you have s = 0 and t = 350. So, the solution is: x1 = 150, x2 = 150, x3 = 0, x4 = 0, and x5 = 350. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 20 Chapter 1 Systems of Linear Equations 30. (a) Each of the network’s four junctions gives rise to a linear equation as shown below. input = output 400 + x2 = x1 x1 + x3 = x4 + 600 300 = x2 + x3 + x5 x4 + x5 = 100 Rearrange these equations, form the augmented matrix, and use Gauss-Jordan elimination. ⎡ 1 −1 0 0 0 400⎤ ⎡ 1 0 1 0 1 700⎤ ⎢ 1 0 1 −1 0 600⎥ ⎢0 1 1 0 1 300⎥ ⎢ ⎥ ⇒ ⎢ ⎥ ⎢0 1 1 0 1 300⎥ ⎢0 0 0 1 1 100⎥ ⎢0 0 0 1 1 100⎥ ⎢0 0 0 0 0 0⎥⎦ ⎣ ⎦ ⎣ Letting x5 = t and x3 = s be the free variables, you can describe the solution as x1 = 700 − s − t x2 = 300 − s − t x3 = s x4 = 100 − t x5 = t , where t and s are any real numbers. (b) If x3 = 0 and x5 = 100, then the solution is: x1 = 600, x2 = 200, x3 = 0, x4 = 0, and x5 = 100. (c) If x3 = x5 = 100, then the solution is: x1 = 500, x2 = 100, x3 = 100, x4 = 0, and x5 = 100. 32. Applying Kirchoff’s first law to three of the four junctions produces I1 + I 3 = I 2 I1 + I 4 = I 2 I3 + I 6 = I5 and applying the second law to the three paths produces R1I1 + R2 I 2 = 3I1 + 2 I 2 = 14 R2 I 2 + R4 I 4 + R5 I 5 + R3 I 3 = 2 I 2 + 2 I 4 + I 5 + 4 I 3 = 25 R5 I 5 + R6 I 6 = I5 + I 6 = 8. Rearrange these equations, form the augmented matrix, and use Gauss-Jordan elimination. ⎡ 1 −1 ⎢ ⎢ 1 −1 ⎢0 0 ⎢ ⎢3 2 ⎢ ⎢0 2 ⎢0 0 ⎣ 1 0 0 0 0 0 0 1 1 0 −1 1 0 0 0 0 4 2 1 0 0 0 1 1 0⎤ ⎡1 ⎢ ⎥ 0⎥ ⎢0 ⎢0 ⎥ 0 ⎥ ⇒ ⎢ ⎢0 14⎥ ⎢ ⎥ 25⎥ ⎢0 ⎢0 ⎥ 8⎦ ⎣ 0 0 0 0 0 2⎤ ⎥ 1 0 0 0 0 4⎥ 0 1 0 0 0 2⎥ ⎥ 0 0 1 0 0 2⎥ ⎥ 0 0 0 1 0 5⎥ 0 0 0 0 1 3⎥⎦ So, the solution is: I1 = 2, I 2 = 4, I 3 = 2, I 4 = 2, I 5 = 5, and I 6 = 3. 34. (a) For a set of n points with distinct x-values, substitute the points into the polynomial p ( x) = a0 + a1 x + + an −1 x n −1 . This creates a system of linear equations in a0 , a1 , an −1. Solving the system gives values for the coefficients an , and the resulting polynomial fits the original points. (b) In a network, the total flow into a junction is equal to the total flow out of a junction. So, each junction determines an equation, and the set of equations for all the junctions in a network forms a linear system. In an electrical network, Kirchhoff’s Laws are used to determine additional equations for the system. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Review Exercises for Chapter 1 50 + 25 + T2 + T3 4 50 + 25 + T1 + T4 T2 = 4 25 + 0 + T1 + T4 T3 = 4 25 + 0 + T2 + T3 T4 = 4 21 36. T1 = 4T1 − ⇒ T2 − − T1 = 75 T3 − T1 + 4T2 − T4 = 75 + 4T3 − T4 = 25 − T2 − T3 + 4T4 = 25 Use Gauss-Jordan elimination to solve this system. ⎡ 4 −1 −1 0 ⎢ ⎢−1 4 0 −1 ⎢−1 0 4 −1 ⎢ ⎢⎣ 0 −1 −1 4 75⎤ ⎡1 ⎥ ⎢ 75⎥ 0 ⇒ ⎢ ⎥ ⎢ 25 0 ⎥ ⎢ ⎢⎣0 25⎥⎦ 0 0 0 31.25⎤ ⎥ 1 0 0 31.25⎥ 0 1 0 18.75⎥ ⎥ 0 0 1 18.75⎥⎦ So, T1 = 31.25°C, T2 = 31.25°C, T3 = 18.75°C, and T4 = 18.75°C. 38. 3x 2 − 7 x − 12 ( x + 4)( x − 4) 2 A B C + + x + 4 x − 4 ( x − 4)2 = 3x 2 − 7 x − 12 = A( x − 4) + B( x + 4)( x − 4) + C ( x + 4) 2 3x 2 − 7 x − 12 = Ax 2 − 8 Ax + 16 A + Bx 2 − 16 B + Cx + 4C 3x 2 − 7 x − 12 = ( A + B ) x 2 + ( −8 A + C ) x + 16 A − 16 B + 4C So, A + −8 A = B 3 + C = −7 16 A − 16 B + 4C = −12. Use Gauss-Jordan elimination to solve the system. 1 0 3⎤ ⎡ 1 ⎡ 1 0 0 1⎤ ⎢ ⎥ ⎢ ⎥ − 8 0 1 − 7 ⇒ ⎢ ⎥ ⎢0 1 0 2⎥ ⎢ 16 −16 4 −12⎥ ⎢0 0 1 1⎥ ⎣ ⎦ ⎣ ⎦ The solution is: A = 1, B = 2, and C = 1. So, 3x 2 − 7 x − 12 ( x + 4)( x − 4) 2 = 1 2 1 + + x + 4 x − 4 ( x − 4)2 40. Use Gauss-Jordan elimination to solve the system. ⎡0 2 2 −2⎤ ⎡ 1 0 0 25⎤ ⎢ ⎥ ⎢ ⎥ ⎢2 0 1 −1⎥ ⇒ ⎢0 1 0 50⎥ ⎢2 1 0 100⎥ ⎢0 0 1 −51⎥ ⎣ ⎦ ⎣ ⎦ So, x = 25, y = 50, and λ = −51. Review Exercises for Chapter 1 2. Because the equation cannot be written in the form a1 x + a2 y = b, it is not linear in the variables x and y. 4. Because the equation is in the form a1 x + a2 y = b, it is linear in the variables x and y. 6. Because the equation is in the form a1 x + a2 y = b, it is linear in the variables x and y. 8. Choosing x2 and x3 as the free variables and letting x2 = s and x3 = t , you have 3×1 + 2s − 4t = 0 3×1 = −2 s + 4t x1 = 13 ( −2 s + 4t ). © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 22 Chapter 1 Systems of Linear Equations 10. Row reduce the augmented matrix for this system. ⎡1 1 −1⎤ ⎡ 1 1 −1⎤ ⎡ 1 1 −1⎤ ⎡ 1 0 2⎤ ⎢ ⎥ ⇒ ⎢ ⎥ ⇒ ⎢ ⎥ ⇒ ⎢ ⎥ 3 2 0 0 1 3 0 1 3 − − ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣0 1 −3⎦ Converting back to a linear system, the solution is x = 2 and y = −3. 12. Rearrange the equations, form the augmented matrix, and row reduce. 7⎤ ⎡1 0 3⎤ ⎡ 1 −1 ⎡ 1 −1 3⎤ ⎡ 1 −1 3⎤ 3 ⎥. ⎢ ⎥ ⇒ ⎢ ⎥ ⇒ ⎢0 1 − 2 ⎥ ⇒ ⎢ 2 0 1 − 4 1 10 0 3 2 − − ⎢ ⎥ ⎢ ⎣ ⎦ ⎣ ⎦ 3⎦ ⎣ 3⎥ ⎣ ⎦ Converting back to a linear system, you obtain the solution x = 73 and y = − 23 . 14. Rearrange the equations, form the augmented matrix, and row reduce. ⎡ 4 1 0⎤ ⎡ 1 −1 0⎤ ⎡ 1 −1 0⎤ ⎡ 1 0 0⎤ ⎢ ⎥ ⇒ ⎢ ⎥ ⇒ ⎢ ⎥ ⇒ ⎢ ⎥. ⎣−1 1 0⎦ ⎣4 1 0⎦ ⎣0 5 0⎦ ⎣0 1 0⎦ Converting back to a linear system, the solution is x = y = 0. 16. Row reduce the augmented matrix for this system. 3⎤ 3⎤ ⎡ 1 3 ⎡1 3 ⎡40 30 24⎤ 4 5 ⇒ 5 ⎥ ⎢ 4 ⎥ ⎢ ⎥ ⇒ ⎢ ⎢⎣20 15 −14⎥⎦ ⎢⎣0 0 −26⎥⎦ ⎣20 15 −14⎦ Because the second row corresponds to the false statement 0 = −26, the system has no solution. 18. Use Gauss-Jordan elimination on the augmented matrix. ⎡1 ⎢3 ⎢⎣2 3⎤ ⎡ 1 0 −3⎤ ⎥ ⇒ ⎢ ⎥ 3 15⎥⎦ ⎣0 1 7⎦ 4 7 28. The matrix satisfies all three conditions in the definition of row-echelon form. Because each column that has a leading 1 (columns 2 and 3) has zeros elsewhere, the matrix is in reduced row-echelon form. 30. Use Gauss-Jordan elimination on the augmented matrix. 3 1 10⎤ 5⎤ ⎡2 ⎡1 0 0 ⎢ ⎥ ⎢ ⎥ ⎢2 −3 −3 22⎥ ⇒ ⎢0 1 0 2⎥ ⎢4 −2 3 −2⎥ ⎢0 0 1 −6⎥ ⎣ ⎦ ⎣ ⎦ So, the solution is: x = 5, y = 2, and z = −6. So, the solution is: x = −3, y = 7. 20. Multiplying both equations by 100 and forming the augmented matrix produces ⎡20 −10 7⎤ ⎢ ⎥. ⎣40 −50 −1⎦ Gauss-Jordan elimination yields the following. 7⎤ ⎡ 1 −1 7 ⎤ ⎡1 − 1 2 20 ⇒ 2 20 ⎢ ⎥ ⎢ ⎥ ⎣⎢40 −50 −1⎦⎥ ⎣⎢0 −30 −15⎦⎥ ⎡1 − 1 2 ⇒ ⎢ 1 ⎣⎢0 7⎤ ⎡1 20 ⇒ ⎥ ⎢ 1 ⎥ 2⎦ ⎣⎢0 32. Use the Gauss-Jordan elimination on the augmented matrix. ⎡1 0 2 3⎤ ⎡2 1 2 4⎤ 2 ⎢ ⎥ ⎢ ⎥ ⎢2 2 0 5⎥ ⇒ ⎢0 1 −2 1⎥ ⎢0 0 0 0⎥ ⎢2 −1 6 2⎥ ⎣ ⎦ ⎣ ⎦ Choosing z = t as the free variable, you can describe the solution as x = 32 − 2t , y = 1 + 2t , and z = t , where t is any real number. 0 1 3⎤ 5 ⎥ 1 ⎥ 2⎦ So, the solution is: x = 53 and y = 12 . 22. Because the matrix has 3 rows and 2 columns, it has size 3 × 2. 24. This matrix corresponds to the system x1 + 2 x2 + 3 x3 = 0 0 = 1. Because the second equation is impossible, the system has no solution. 26. The matrix satisfies all three conditions in the definition of row-echelon form. Because each column that has a leading 1 (columns 1 and 4) has zeros elsewhere, the matrix is in reduced row-echelon form. 34. Use Gauss-Jordan elimination on the augmented matrix. ⎡1 0 0 − 3⎤ ⎡2 0 6 −9⎤ 4 ⎢ ⎥ ⎢ ⎥ 0⎥ ⎢ 3 −2 11 −16⎥ ⇒ ⎢0 1 0 ⎢0 0 1 − 5 ⎥ ⎢ ⎥ ⎣ 3 −1 7 −11⎦ 4⎦ ⎣ So, the solution is: x = − 34 , y = 0, and z = − 54 . 36. Use Gauss-Jordan elimination on the augmented matrix. ⎡2 5 −19 34⎤ ⎡ 1 0 3 2⎤ ⎢ ⎥ ⇒ ⎢ ⎥ 3 8 31 54 − ⎣ ⎦ ⎣0 1 −5 6⎦ Choosing x3 = t as the free variable, you can describe the solution as x1 = 2 − 3t , x2 = 6 + 5t , and x3 = t , where t is any real number. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Review Exercises for Chapter 1 38. Use Gauss-Jordan elimination on the augmented matrix. ⎡1 ⎢ ⎢0 ⎢0 ⎢ ⎢2 ⎢ ⎣2 0 14⎤ ⎡1 ⎥ ⎢ 0 3⎥ ⎢0 ⎥ 0 3 8 6 16 ⇒ ⎢0 ⎥ ⎢ ⎢0 4 0 0 −2 0⎥ ⎥ ⎢ 0 −1 0 0 0⎦ ⎣0 5 3 0 4 2 5 48. Use Gaussian elimination on the augmented matrix. 2⎤ ⎥ 0⎥ 0 1 0 0 4⎥ ⎥ 0 0 1 0 −1⎥ ⎥ 0 0 0 1 2⎦ ⎡1 ⎡ 1 −1 2 0⎤ ⎢ ⎢ ⎥ ⎢−1 1 −1 0⎥ ⇒ ⎢0 ⎢0 ⎢ 1 k 1 0⎥⎦ ⎣ ⎣ 0 0 0 0 1 0 0 0 ⎡1 ⎢ ⇒ ⎢0 ⎢0 ⎣ So, the solution is: x1 = 2, x2 = 0, x3 = 4, x4 = −1, and x5 = 2. 5 0 0⎤ ⎥ 0 1 0⎥ . 0 0 1⎥ ⎥ 0 0 0⎥⎦ 0 ( k + 1) −1 (k + 1) 0 2 0⎤ ⎥ −1 0⎥ 1 0⎥⎦ ⎡2 −1 1 a⎤ ⎢ ⎥ ⎢ 1 1 2 b⎥ ⎢0 3 3 c⎥ ⎣ ⎦ 42. Using a graphing utility, the augmented matrix reduces to ⎡ 1 0 0 1.5 0⎤ ⎢ ⎥ ⎢0 1 0 0.5 0⎥. ⎢0 0 1 0.5 0⎥ ⎣ ⎦ Choosing w = t as the free variable, you can describe the solution as x = −1.5t , y = −0.5t , z = −0.5t , w = t , where t is any real number. Use Gaussian elimination to reduce the matrix to row-echelon form. 1 ⎡ ⎢1 − 2 ⎢ 1 ⎢1 ⎢ 0 3 ⎣ 1 2 2 3 1 ⎡ a⎤ ⎢1 − 2 ⎢ 2⎥ ⎥ ⇒ ⎢ 3 b⎥ ⎢0 2 ⎥ ⎢ c⎦ 3 ⎣⎢0 44. Use Gauss-Jordan elimination on the augmented matrix. 3 2 − 52 0⎤ ⎥ 0⎦⎥ Letting x3 = t be the free variable, you have x1 = − 32 t , x2 = 52 t , and x3 = t , where t is any real number. 46. Use Gauss-Jordan elimination on the augmented matrix. ⎡ 1 0 37 ⎡1 3 5 0⎤ 2 ⎢ ⎥ ⇒ ⎢ 1 9 1 4 0 0 1 − ⎥⎦ ⎣⎢ 2 ⎣⎢ 2 0⎤ ⎥ 1 0⎥ −1 0⎥⎦ 2 50. Form the augmented matrix for the system. The system is inconsistent, so there is no solution. ⎡1 0 ⎡2 4 −7 0⎤ ⎢ ⎥ ⇒ ⎢ 1 − 3 9 0 ⎢⎣0 1 ⎣ ⎦ −1 So, there will be exactly one solution (the trivial solution x = y = z = 0) if and only if k ≠ −1. 40. Using a graphing utility, the augmented matrix reduces to ⎡1 ⎢ ⎢0 ⎢0 ⎢ ⎢⎣0 23 0⎤ ⎥ 0⎦⎥ Choosing x3 = t as the free variable, you can describe t , x2 = 92 t , and x3 = t , where the solution as x1 = − 37 2 − 1 2 3 2 3 a ⎤ 2 ⎥ ⎥ 2b − a ⎥ 2 ⎥ ⎥ c ⎦⎥ 1 ⎡ ⎢1 − 2 ⎢ ⇒ ⎢ 1 ⎢0 ⎢ 3 ⎢⎣0 1 2 1 ⎡ ⎢1 − 2 ⎢ ⇒ ⎢ 1 ⎢0 ⎢ 0 ⎣⎢0 a ⎤ ⎥ 2 ⎥ 2b − a ⎥ 1 ⎥ 3 ⎥ 0 c − 2b + a⎦⎥ 1 3 a ⎤ 2 ⎥ ⎥ 2b − a ⎥ 3 ⎥ ⎥ c ⎥⎦ 1 2 (a) If c − 2b + a ≠ 0, then the system has no solution. (b) The system cannot have one solution. (c) If c − 2b + a = 0, then the system has infinitely many solutions. t is any real number. 52. Find all possible first rows, where a and b are nonzero real numbers. [0 0 0], [0 0 1], [0 1 0], [0 1 a], [1 0 0], [1 a 0], [1 a b], [1 0 a] For each of these, examine the possible second rows. ⎡0 0 0⎤ ⎡0 0 1⎤ ⎡0 1 0⎤ ⎡0 1 0⎤ ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎣0 0 0⎦ ⎣0 0 0⎦ ⎣0 0 0⎦ ⎣0 0 1⎦ ⎡0 1 a⎤ ⎡1 0 0⎤ ⎡1 0 0⎤ ⎡1 0 0⎤ ⎡1 0 0⎤ ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎣0 0 0⎦ ⎣0 0 0⎦ ⎣0 1 0⎦ ⎣0 0 1⎦ ⎣0 1 a⎦ ⎡1 a 0⎤ ⎡1 a 0⎤ ⎡1 a b⎤ ⎡1 0 a⎤ ⎡1 0 a⎤ ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥ ⎣0 0 0⎦ ⎣0 0 1⎦ ⎣0 0 0⎦ ⎣0 0 0⎦ ⎣0 1 0⎦ © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 24 Chapter 1 Systems of Linear Equations 54. Use Gaussian elimination on the augmented matrix. ⎡(λ + 2) ⎢ ⎢ −2 ⎢ 1 ⎣ −2 (λ − 1) 2 ⎡1 3 0⎤ 2 λ 0⎤ 2 ⎡1 ⎢ ⎥ ⎢ ⎥ 6 0⎥ ⇒ ⎢0 λ + 3 6 + 2λ 0⎥ ⇒ ⎢0 λ + 3 ⎢ 2 ⎥ ⎢ ⎥ λ 0⎦ ⎢⎣0 0 ⎣0 −2λ − 6 −λ − 2λ + 3 0⎦ λ 6 + 2λ (λ − 2λ − 15) 2 0⎤ ⎥ 0⎥ ⎥ 0⎥⎦ So, you need λ 2 − 2λ − 15 = (λ − 5)(λ + 3) = 0, which implies λ = 5 or λ = −3. 56. (a) True. A homogeneous system of linear equations is always consistent, because there is always a trivial solution, i.e., when all variables are equal to zero. See Theorem 1.1 on page 21. (b) False. Consider, for example, the following system (with three variables and two equations). 62. (a) Because there are four points, choose a third-degree polynomial, p( x) = a0 + a1x + a2 x 2 + a3 x3 . By substituting the values at each point into this equation, you obtain the system a0 − a1 + a2 − a3 = −1 x+ y − z = 2 −2 x − 2 y + 2 z = 1. a0 + a1 + a2 + a3 = 1 It is easy to see that this system has no solution. a0 + 2a1 + 4a2 + 8a3 = 4. Use Gauss-Jordan elimination on the augmented matrix. 58. From the following chart, you obtain a system of equations. ⎡1 1 −1 −1⎤ ⎢ ⎥ 0 0⎥ ⎢0 ⇒ ⎢ ⎥ 0 1 1 1 ⎢ ⎥ ⎢ 4 8 4⎦⎥ ⎣0 A B C Mixture X 1 5 2 5 2 5 Mixture Y 0 0 1 ⎡1 −1 ⎢ ⎢1 0 ⎢1 1 ⎢ ⎣⎢1 2 Mixture Z 1 3 1 3 1 3 So, p( x) = 23 x + 13 x3. Desired Mixture 6 27 8 27 13 27 2 x + y + 1 z = 13 5 3 27 12 2 1 5 ⇒ y = 27 To obtain the desired mixture, use 10 gallons of spray X, 5 gallons of spray Y, and 12 gallons of spray Z. ( x + 1) ( x − 1) 2 = A B C + + x + 1 x − 1 ( x + 1)2 3x 2 + 3x − 2 = A( x + 1)( x − 1) + B( x + 1) + C ( x − 1) 2 3x 2 + 3x − 2 = Ax 2 − A + Bx 2 + 2 Bx + B + Cx − C 3x 2 + 3x − 2 = ( A + B ) x 2 + ( 2 B + C ) x − A + B − C So, A + = 3 2B + C = 3 B −A + (2, 4) 3 ⎬ x = 27 , z = 27 10 0 0 0 0⎤ ⎥ 1 0 0 23 ⎥ 0 1 0 0⎥ ⎥ 0 0 1 13 ⎥⎦ y (b) 2x + 1z = 8 5 3 27 ⎪ ⎭ 3x 2 + 3x − 2 0 4 1x + 1z = 6 ⎫ 5 3 27 ⎪ 60. = 0 a0 (− 1, − 1) (1, 1) x (0, 0) 2 3 64. Substituting the points, (1, 0), (2, 0), (3, 0), and (4, 0) into the polynomial p( x) yields the system a0 + a1 + a2 + a3 = 0 a0 + 2a1 + 4a2 + 8a3 = 0 a0 + 3a1 + 9a2 + 27 a3 = 0 a0 + 4a1 + 16a2 + 64a3 = 0. Gaussian elimination shows that the only solution is a0 = a1 = a2 = a3 = 0. B − C = −2. Use Gauss-Jordan elimination to solve the system. 3⎤ ⎡ 1 1 0 ⎡ 1 0 0 2⎤ ⎢ ⎥ ⎢ ⎥ ⇒ 0 2 1 3 ⎢ ⎥ ⎢0 1 0 1⎥ ⎢−1 1 −1 −2⎥ ⎢0 0 1 1⎥ ⎣ ⎦ ⎣ ⎦ The solution is: A = 2, B = 1, and C = 1. So, 3x 2 + 3x − 2 ( x + 1) ( x − 1) 2 = 2 1 1 + + . x + 1 x − 1 ( x + 1)2 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Review Exercises for Chapter 1 25 66. s = 12 at 2 + v0t + s0 (a) When t = 1: s = 134: 12 a(1) + v0 (1) + s0 = 134 ⇒ a + 2v0 + 2 s0 = 268 2 When t = 2: s = 86: 12 a( 2) + v0 ( 2) + s0 = 86 ⇒ 2a + 2v0 + s0 = 86 2 When t = 3: s = 6: 12 a(3) + v0 (3) + s0 = 6 ⇒ 9a + 6v0 + 2 s0 = 12 2 Use Gaussian elimination to solve the system. a + 2v0 + 2 s0 = 268 2a + 2v0 + s0 = 86 9a + 6v0 + 2 s0 = 12 a + 2v0 + 2s0 = 268 − 2v0 − 3s0 = −450 −12v0 − 16s0 = −2400 a + 2v0 + 2 s0 = − 2v0 − 3s0 = −450 3v0 + 4 s0 = 600 2v0 + 2 s0 = 268 a + − 2v0 − (−2)Eq.1 + Eq.2 (−9)Eq.1 + Eq.3 268 (− 14 )Eq.3 3s0 = −450 − s0 = −150 3Eq.2 + 2Eq.3 − s0 = −150 ⇒ s0 = 150 −2v0 − 3(150) = −450 ⇒ v0 = 0 a + 2(0) + 2(150) = 268 ⇒ a = −32 The position equation is s = 12 ( −32)t 2 + (0)t + 150, or s = −16t 2 + 150. 68. (a) First find the equations corresponding to each node in the network. input = output x1 + 200 = x2 + x4 x6 + 100 = x1 + x3 x2 + x3 = x5 + 300 x4 + x5 = (b) When x3 = 100 = r , x5 = 50 = s, and x6 = 50 = t , you have x1 = 100 − 100 + 50 = 50 x2 = 300 − 100 + 50 = 250 x4 = −50 + 50 = 0. x6 Rearranging this system and forming the augmented matrix, you have ⎡ 1 −1 ⎢ ⎢1 0 ⎢0 1 ⎢ ⎣⎢0 0 0 −1 0 1 0 −1 1 0 0 −200⎤ ⎥ 100⎥ . 0 −1 0 300⎥ ⎥ 1 1 −1 0⎥⎦ 0 The equivalent reduced row-echelon matrix is ⎡1 ⎢ ⎢0 ⎢0 ⎢ ⎢⎣0 0 1 0 0 −1 100⎤ ⎥ 0 300⎥ . 1 −1 0⎥ ⎥ 0 0 0⎥⎦ 1 1 0 −1 0 0 1 0 0 0 Choosing x3 = r , x5 = s, and x6 = t as the free variables, you obtain x1 = 100 − r + t x2 = 300 − r + s x4 = − s + t , where r , s, and t are any real numbers. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 26 Chapter 1 Systems of Linear Equations Project Solutions for Chapter 1 1 Graphing Linear Equations 3 ⎡1 ⎤ − 12 ⎡ 2 −1 3⎤ 2 1. ⎢ ⎥ ⎥ ⇒ ⎢ 1a 6 − 3a 0 + b 6 a b ⎢⎣ ⎥ ⎣ ⎦ 2 2 ⎦ (a) Unique solution if b + 12 a ≠ 0. For instance, a = b = 2. (b) Infinite number of solutions if b + 12 a = 6 − 32 a = 0 ⇒ a = 4 and b = −2. (c) No solution if b + 12 a = 0 and 6 − 32 a ≠ 0 ⇒ a ≠ 4 and b = − 12 a. For instance, a = 2, b = −1. y (d) 2x + 2y = 6 y 2x − y = 3 4 3 2 1 3 2 1 −4 x −2 −4 2 3 1 2 3 4 −2 −3 y = 3 (b) 2 x − 2x + 2 y = 6 x −2 x −2 −2 −3 (a) 2 x − y 2x − y = 3 −2 −3 2x − y = 3 2x − y = 6 y = 3 (c) 2 x − y = 3 4x − 2 y = 6 2x − y = 6 (The answers are not unique.) 2. (a) x + y + z = 0 (b) x + y + z = 0 (c) x + y + z = 0 x + y + z = 0 y + z =1 x + y + z = 1 x− y − z = 0 z = 2 x− y − z = 0 (The answers are not unique.) There are other configurations, such as three mutually parallel planes or three planes that intersect pairwise in lines. 2 Underdetermined and Overdetermined Systems of Equations 1. Yes, x + y = 2 is a consistent underdetermined system. 2. Yes, x+ y = 2 2x + 2 y = 4 3x + 3 y = 6 is a consistent, overdetermined system. 3. Yes, x + y + z = 1 x + y + z = 2 is an inconsistent underdetermined system. 4. Yes, x + y =1 x + y = 2 5. In general, a linear system with more equations than variables would probably be inconsistent. Here is an intuitive reason: Each variable represents a degree of freedom, while each equation gives a condition that in general reduces number of degrees of freedom by one. If there are more equations (conditions) than variables (degrees of freedom), then there are too many conditions for the system to be consistent. So you expect such a system to be inconsistent in general. But, as Exercise 2 shows, this is not always true. 6. In general, a linear system with more variables than equations would probably be consistent. As in Exercise 5, the intuitive explanation is as follows. Each variable represents a degree of freedom, and each equation represents a condition that takes away one degree of freedom. If there are more variables than equations, in general, you would expect a solution. But, as Exercise 3 shows, this is not always true. x + y = 3 is an inconsistent underdetermined system. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. C H A P T E R Matrices 2 Section 2.1 Operations with Matrices ……………………………………………………………28 Section 2.2 Properties of Matrix Operations…………………………………………………..33 Section 2.3 The Inverse of a Matrix………………………………………………………………38 Section 2.4 Elementary Matrices…………………………………………………………………..43 Section 2.5 Applications of Matrix Operations ………………………………………………49 Review Exercises …………………………………………………………………………………………….55 Project Solutions……………………………………………………………………………………………..63 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. C H A P T E R Matrices 2 Section 2.1 Operations with Matrices 2. x = 13, y = 12 4. x + 2 = 2 x + 6 2 y = 18 −4 = x y = 9 2 x = −8 y + 2 = 11 x = −4 y = 9 ⎡ 1 2⎤ ⎡−3 −2⎤ ⎡ 1 − 3 2 − 2⎤ ⎡−2 0⎤ 6. (a) A + B = ⎢ ⎥ + ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎣2 1⎦ ⎣ 4 2⎦ ⎣2 + 4 1 + 2⎦ ⎣ 6 3⎦ ⎡ 1 2⎤ ⎡−3 −2⎤ ⎡ 1 + 3 2 + 2⎤ ⎡ 4 4⎤ (b) A − B = ⎢ ⎥ − ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎣2 1⎦ ⎣ 4 2⎦ ⎣2 − 4 1 − 2⎦ ⎣−2 −1⎦ ⎡ 2(1) 2( 2)⎤ ⎡ 1 2⎤ ⎡2 4⎤ (c) 2 A = 2 ⎢ ⎥ = ⎢ ⎥ = ⎢ 2 2 2 1 2 1 ( )⎥⎦ ⎣4 2⎥⎦ ⎢⎣ ( ) ⎣ ⎦ ⎡2 4⎤ ⎡−3 −2⎤ ⎡5 6⎤ (d) 2 A − B = ⎢ ⎥ − ⎢ ⎥ = ⎢ ⎥ ⎣4 2⎦ ⎣ 4 2⎦ ⎣0 0⎦ ⎡− 52 −1⎤ ⎡−3 −2⎤ 1 ⎡ 1 2⎤ ⎡−3 −2⎤ ⎡ 12 1⎤ = (e) B + 12 A = ⎢ ⎢ ⎥ ⎥ ⎥ + 2⎢ ⎥ = ⎢ ⎥ + ⎢ 5 1 ⎢⎣ 5 2 ⎦⎥ ⎣ 4 2⎦ ⎣2 1⎦ ⎣ 4 2⎦ ⎣⎢ 1 2 ⎦⎥ ⎡ 2 1 1⎤ ⎡ 2 −3 4⎤ ⎡ 2 + 2 1 − 3 1 + 4⎤ ⎡ 4 −2 5⎤ 8. (a) A + B = ⎢ ⎥ + ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎣−1 −1 4⎦ ⎣−3 1 −2⎦ ⎣−1 − 3 −1 + 1 4 − 2⎦ ⎣−4 0 2⎦ ⎡ 2 1 1⎤ ⎡ 2 −3 4⎤ ⎡ 2 − 2 1 + 3 1 − 4⎤ ⎡0 4 −3⎤ (b) A − B = ⎢ ⎥ − ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎣−1 −1 4⎦ ⎣−3 1 −2⎦ ⎣−1 + 3 −1 − 1 4 + 2⎦ ⎣2 −2 6⎦ ⎡ 2( 2) 2(1) 2(1)⎤ ⎡ 2 1 1⎤ ⎡ 4 2 2⎤ (c) 2 A = 2 ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎢⎣2( −1) 2( −1) 2( 4)⎥⎦ ⎣−1 −1 4⎦ ⎣−2 −2 8⎦ ⎡ 4 2 2⎤ ⎡ 2 −3 4⎤ ⎡2 5 −2⎤ (d) 2 A − B = ⎢ ⎥ − ⎢ ⎥ = ⎢ ⎥ ⎣−2 −2 8⎦ ⎣−3 1 −2⎦ ⎣ 1 −3 10⎦ 1 ⎡ 2 −3 4⎤ 1 ⎡ 2 1 1⎤ ⎡ 2 −3 4⎤ ⎡ 1 2 (e) B + 12 A = ⎢ ⎥ + 2⎢ ⎥ = ⎢ ⎥ + ⎢ 1 1 ⎣−3 1 −2⎦ ⎣−1 −1 4⎦ ⎣−3 1 −2⎦ ⎣⎢− 2 − 2 1⎤ 2 ⎡ 3 − 52 ⎥ = ⎢ 7 1 2⎦⎥ 2 ⎣⎢− 2 9⎤ 2 ⎥ 0⎦⎥ ⎡ 2 + 0 3 + 6 4 + 2⎤ ⎡2 3 4⎤ ⎡ 0 6 2⎤ ⎡2 9 6⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 10. (a) A + B = ⎢0 1 −1⎥ + ⎢ 4 1 0⎥ = ⎢ 0 + 4 1 + 1 −1 + 0⎥ = ⎢4 2 −1⎥ ⎢2 + (−1) 0 + 2 1 + 4⎥ ⎢2 0 1⎥ ⎢−1 2 4⎥ ⎢ 1 2 5⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ 2 − 0 3 − 6 4 − 2⎤ ⎡2 3 4⎤ ⎡ 0 6 2⎤ ⎡ 2 −3 2⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (b) A − B = ⎢0 1 −1⎥ − ⎢ 4 1 0⎥ = ⎢ 0 − 4 1 − 1 −1 − 0⎥ = ⎢−4 0 −1⎥ ⎢2 − ( −1) 0 − 2 1 − 4⎥ ⎢2 0 1⎥ ⎢−1 2 4⎥ ⎢ 3 −2 −3⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡2( 2) 2(3) 2( 4)⎤ ⎡2 3 4⎤ ⎡4 6 8⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (c) 2 A = 2 ⎢0 1 −1⎥ = ⎢2(0) 2(1) 2( −1)⎥ = ⎢0 2 −2⎥ ⎢2( 2) 2(0) ⎢2 0 1⎥ ⎢4 0 2⎥ 2(1)⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 28 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 2.1 Operations with Matrices 29 ⎡2 3 4⎤ ⎡ 0 6 2⎤ ⎡4 6 8⎤ ⎡ 0 6 2⎤ ⎡ 4 0 6⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 1 −2⎥ (d) 2 A − B = 2 ⎢0 1 −1⎥ − ⎢ 4 1 0⎥ = ⎢0 2 −2⎥ − ⎢ 4 1 0⎥ = ⎢−4 ⎢2 0 1⎥ ⎢−1 2 4⎥ ⎢4 0 2⎥ ⎢−1 2 4⎥ ⎢ 5 −2 −2⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ (e) ⎡1 2⎤ ⎡ 0 6 2⎤ ⎡2 3 4⎤ ⎡ 0 6 2⎤ ⎡ 1 32 ⎥ ⎢ ⎥ 1⎢ ⎥ ⎢ ⎥ ⎢ 1 1 1 0⎥ + 2 ⎢0 1 −1⎥ = ⎢ 4 1 0⎥ + ⎢0 2 − 2 ⎥ = ⎢4 ⎢0 1⎥ ⎢−1 2 4⎥ ⎢2 0 1⎥ ⎢−1 2 4⎥ ⎢ 1 0 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 2⎦ ⎣ ⎢ B + 12 A = ⎢ 4 12. (a) A + B is not possible. A and B have different sizes. (b) A − B is not possible. A and B have different sizes. ⎡ 3⎤ ⎡ 6⎤ ⎢ ⎥ ⎢ ⎥ (c) 2 A = 2 ⎢ 2⎥ = ⎢ 4⎥ ⎢−1⎥ ⎢−2⎥ ⎣ ⎦ ⎣ ⎦ 4⎤ ⎥ − 12 ⎥ 9⎥ 2 2⎦ 15 2 3 2 16. Simplifying the right side of the equation produces ⎡ w x⎤ ⎡−4 + 2 y 3 + 2w⎤ ⎢ ⎥ = ⎢ ⎥. ⎣ y x⎦ ⎣ 2 + 2 z −1 + 2 x⎦ By setting corresponding entries equal to each other, you obtain four equations. (d) 2A − B is not possible. A and B have different sizes. w = −4 + 2 y (e) B + 12 A is not possible. A and B have different y = sizes. 14. (a) c23 = 5a23 + 2b23 = 5( 2) + 2(11) = 32 (b) c32 = 5a32 + 2b32 = 5(1) + 2( 4) = 13 x = x = ⎧−2 y + w = −4 ⎪ ⎪x − 2w = 3 ⇒ ⎨ 2 + 2z ⎪y − 2z = 2 ⎪x = 1 −1 + 2 x ⎩ 3 + 2w The solution to this linear system is: x = 1, y = 32 , z = − 14 , and w = −1. ⎡2( 4) + ( − 2)( 2) 2(1) + ( − 2)( − 2)⎤ 1⎤ 6⎤ ⎡ 2 − 2⎤ ⎡4 ⎡4 18. (a) AB = ⎢ ⎥ = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ −1(1) + 4(− 2) ⎥⎦ 4⎦ ⎣2 − 2⎦ ⎢⎣−1( 4) + 4( 2) ⎣−1 ⎣4 − 9⎦ ⎡4( 2) + 1( −1) 4( − 2) + 1( 4) ⎤ 1⎤ ⎡ 2 − 2⎤ ⎡4 ⎡7 − 4⎤ (b) BA = ⎢ ⎥ = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ 4⎦ ⎣2 − 2⎦ ⎣−1 ⎣6 −12⎦ ⎣⎢2( 2) + ( − 2)( −1) 2( − 2) + ( − 2)( 4)⎦⎥ ⎡ 1 −1 7⎤ ⎡ 1 1 2⎤ ⎡1(1) + ( −1)( 2) + 7(1) 1(1) + ( −1)(1) + 7( −3) 1( 2) + (−1)(1) + 7( 2)⎤ ⎡6 −21 15⎤ ⎥ ⎢ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 20. (a) AB = ⎢2 −1 8⎥ ⎢2 1 1⎥ = ⎢2(1) + ( −1)( 2) + 8(1) 2(1) + (−1)(1) + 8( −3) 2( 2) + (−1)(1) + 8( 2)⎥ = ⎢8 −23 19⎥ ⎢3 1 −1⎥ ⎢ 1 −3 2⎥ ⎢ 3(1) + 1( 2) + ( −1)(1) 3(1) + 1(1) + ( −1)( −3) 3( 2) + 1(1) + ( −1)( 2)⎥ ⎢4 7 5⎥⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ 1(−1) + 1( −1)1 + 2(1) 1(7) + 1(8) + 2(−1)⎤ ⎡9 0 13⎤ ⎡ 1 1 2⎤ ⎡ 1 −1 7⎤ ⎡ 1(1) + 1( 2) + 2(3) ⎥ ⎢ ⎢ ⎥⎢ ⎥ ⎢ ⎥ (b) BA = ⎢2 1 1⎥ ⎢2 −1 8⎥ = ⎢ 2(1) + 1( 2) + 1(3) 2(−1) + 1( −1) + 1(1) 2(7) + 1(8) + 1(−1)⎥ = ⎢7 −2 21⎥ ⎢ 1 −3 2⎥ ⎢3 1 −1⎥ ⎢1(1) + ( −3)( 2) + 2(3) 1( −1) + ( −3)( −1) + 2(1) 1(7) + ( −3)(8) + 2(−1)⎥ ⎢ 1 4 −19⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎦ ⎣ 2 1⎤ ⎡ 1 2⎤ ⎡3(1) + 2( 2) + 1(1) 3( 2) + 2(−1) + 1( − 2) ⎤ 2⎤ ⎡ 3 ⎡ 8 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ − 3( 2) + 0( −1) + 4( − 2) 0 4⎥ ⎢2 −1⎥ = ⎢− 3(1) + 0( 2) + 4(1) 22. (a) AB = ⎢− 3 ⎥ = ⎢ 1 −14⎥ ⎢4(1) + ( − 2)( 2) + ( − 4)(1) 4( 2) + ( − 2)( −1) + (− 4)(− 2)⎥ ⎢ 4 − 2 − 4⎥ ⎢ 1 − 2⎥ ⎢− 4 18⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ (b) BA is not defined because B is 3 × 2 and A is 3 × 3. ⎡ −1( 2) −1(1) −1(3) −1( 2)⎤ ⎡ −1⎤ ⎡− 2 −1 − 3 − 2⎤ ⎢ ⎥ ⎢ ⎥ 2 2 2 1 2 3 2 2 2 () () ( )⎥ ⎢⎢ 4 2 6 4⎥⎥ ⎢ ( ) = 24. (a) AB = ⎢ ⎥ [2 1 3 2] = ⎢ ⎢− 2⎥ ⎢− 4 − 2 − 6 − 4⎥ − 2( 2) − 2(1) − 2(3) − 2( 2)⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 1( 2) 1(1) 1(3) 1( 2)⎥⎦ ⎢⎣ 1⎥⎦ 1 3 2⎦⎥ ⎣⎢ 2 ⎡ −1⎤ ⎢ ⎥ 2 (b) BA = [2 1 3 2] ⎢ ⎥ = ⎡⎣2( −1) + 1( 2) + 3(− 2) + 2(1)⎦⎤ = [− 4] ⎢− 2⎥ ⎢ ⎥ ⎢⎣ 1⎥⎦ © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.