Chapter 8 Systems of Equations and Inequalities

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⎡ ⎢ 1 → ⎢0 ⎢ ⎣0 0 1 0 3 − 2 3 − 2 4 8 ⎤ ⎥ 2⎥ ⎥ 0⎦ ⎛R1 = r2 + r1 ⎞ ⎜ ⎟ ⎝ R3 =− 2r2 + r3⎠ ⎡ ⎢ 1 → ⎢0 ⎢ ⎣0 0 1 0 3 − 2 3 − 2 1 8 ⎤ ⎥ 2⎥ ⎥ 0⎦ 1 ( R3 = r 4 3) ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 0 1 0 0 0 1 8⎤ ⎥ 2⎥ 0⎥ ⎦ ⎛ 3 R1 = r 2 3 + r1 ⎞ ⎜ ⎟ ⎜ 3 R2 = r 2 3 + r ⎟ ⎝ 2⎠ The solution is x = 8, y = 2, z = 0 or (8, 2, 0). 48. ⎧2x+ y = −4 ⎪ ⎨ − 2y+ 4z = 0 ⎪ ⎩ 3x − 2z =−11 Write the augmented matrix: ⎡2 ⎢ ⎢0 ⎢ ⎣ 3 1 − 2 0 0 4 − 2 − 4⎤ ⎥ 0⎥ −11⎥ ⎦ ⎡ ⎢ 1 → ⎢0 ⎢ ⎢3 ⎣ 1 2 − 2 0 0 4 − 2 − 2 ⎤ ⎥ 0 ⎥ ⎥ −11⎥ ⎦ 1 R1 = r 2 1 ( ) ⎡ 1 ⎢ → ⎢0 ⎢ ⎣ ⎢0 1 2 − 2 3 − 2 0 4 − 2 − 2 ⎤ ⎥ 0 ⎥ ( R3 =− 3r1+ r3) ⎥ −5⎥ ⎦ ⎡ ⎢ 1 → ⎢0 ⎢ ⎢ ⎣ 0 1 2 1 3 − 2 0 − 2 − 2 − 2 ⎤ ⎥ 1 0 ⎥ ( R2 =− r 2 2) ⎥ −5⎥ ⎦ ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 0 1 0 1 − 2 −5 − 2⎤ 1 ⎥ ⎛R1 =− r 2 2 + r1⎞ 0 ⎥ ⎜ ⎟ ⎜ 3 R 5 3 = r 2 2 + r ⎟ − ⎥ ⎝ 3 ⎠ ⎦ ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 0 1 0 1 − 2 1 − 2⎤ ⎥ 1 0 ⎥ ( R3 = − r 5 3) 1⎥ ⎦ ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 0 1 0 0 0 1 −3⎤ ⎥ 2⎥ 1⎥ ⎦ ⎛R1 =− r3 + r1 ⎞ ⎜ ⎟ ⎝R2 = 2r3 + r2⎠ The solution is x =− 3, y = 2, z = 1 or ( − 3,2,1) . 799 Section 8.2: Systems of Linear Equations: Matrices © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 49. 50. ⎧ x− 2y+ 3z= 7 ⎪ ⎨ 2x+ y+ z = 4 ⎪ ⎩− 3x+ 2y− 2z = −10 Write the augmented matrix: ⎡ 1 ⎢ ⎢ 2 ⎢−3 ⎣ − 2 1 2 3 1 − 2 7⎤ ⎥ 4⎥ −10⎥ ⎦ ⎡1 ⎢ →⎢0 ⎢ ⎣ 0 − 2 5 − 4 3 −5 7 7⎤ ⎥ ⎛R2 =− 2r1+ r2⎞ −10 ⎥ ⎜ ⎟ 3 3 1 3 11⎥ ⎝R = r + r ⎠ ⎦ ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 − 2 1 − 4 3 −1 7 7⎤ ⎥ − 2⎥ 11⎥ ⎦ 1 ( R2 = r 5 2) ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 0 1 0 1 −1 3 3⎤ ⎥ − 2⎥ 3⎥ ⎦ ⎛R1 = 2r2 + r1 ⎞ ⎜ ⎟ ⎝R3 = 4r2 + r3⎠ ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 0 1 0 1 −1 1 3⎤ ⎥ − 2⎥ 1⎥ ⎦ 1 ( R3 = r 3 3) ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 0 1 0 0 0 1 2⎤ ⎥ −1⎥ 1⎥ ⎦ ⎛R1 =− r3 + r1⎞ ⎜ ⎟ ⎝R2 = r3 + r2 ⎠ The solution is x = 2, y =− 1, z = 1 or (2, − 1,1) . ⎧ 2x+ y− 3z= 0 ⎪ ⎨− 2x+ 2y+ z =−7 ⎪ ⎩ 3x−4y− 3z = 7 Write the augmented matrix: ⎡ 2 ⎢ ⎢ −2 ⎢⎣ 3 1 2 −4 −3 1 −3 0⎤ −7 ⎥ ⎥ 7⎥⎦ ⎡ 1 ⎢ → ⎢−2 ⎢ ⎣ 3 1 2 2 −4 3 − 2 1 −3 0⎤ ⎥ 1 − 7 ⎥ ( R1 = r 2 1) 7⎥ ⎦ ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 1 2 3 11 − 2 3 − 2 −2 3 2 0⎤ ⎥ ⎛R2 = 2r1+ r2 ⎞ −7 ⎥ ⎜ ⎟ R3 =− 3r1+ r3 7⎥ ⎝ ⎠ ⎦
Chapter 8: Systems of Equations and Inequalities ⎡1 ⎢ → ⎢0 ⎢ ⎣0 1 2 1 11 − 2 3 − 2 2 − 3 3 2 0⎤ ⎥ 7 − 3⎥ ⎥ 7⎦ 1 ( R2 = r 3 2) ⎡1 ⎢ → ⎢0 ⎢ ⎢⎣ 0 0 1 0 7 − 6 2 − 3 13 − 6 7⎤ 6 ⎥ 7 − 3⎥ 35⎥ − 6 ⎥⎦ ⎛ 1 R1 =− r 2 2 + r1⎞ ⎜ ⎟ ⎜ 11 R3 r 2 2 r ⎟ ⎝ = + 3 ⎠ ⎡1 ⎢ → ⎢0 ⎢ ⎢⎣ 0 0 1 0 7 − 6 2 − 3 1 7⎤ 6 ⎥ 7 − 3⎥ 35⎥ 13 ⎥⎦ 6 ( R3 =− r 13 3) ⎡1 ⎢ → ⎢0 ⎢ ⎢⎣ 0 0 1 0 0 0 1 56 ⎤ 13 ⎥ 7 −13⎥ 35⎥ 13 ⎥⎦ ⎛ 7 R1 = r 6 3 + r1 ⎞ ⎜ ⎟ ⎜ 2 R2 = r 3 3 + r ⎟ ⎝ 2⎠ 56 7 35 The solution is x = , y =− , z = or 13 13 13 ⎛56 7 35 ⎞ ⎜ , − , ⎟ ⎝13 13 13 ⎠ . 51. ⎧ 2x−2y− 2z= 2 ⎪ ⎨2x+ 3y+ z = 2 ⎪ ⎩ 3x+ 2y = 0 Write the augmented matrix: ⎡2 ⎢ ⎢2 ⎢3 ⎣ −2 3 2 −2 1 0 2⎤ ⎥ 2⎥ 0⎥ ⎦ ⎡1 ⎢ → ⎢2 ⎢3 ⎣ −1 3 2 −1 1 0 1⎤ ⎥ 1 2 ⎥ ( R1 = r 2 1) 0⎥ ⎦ ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 −1 5 5 −1 3 3 1⎤ ⎥ ⎛R2 =− 2r1+ r2⎞ 0 ⎥ ⎜ ⎟ R3 =− 3r1+ r3 −3⎥ ⎝ ⎠ ⎦ ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 −1 5 0 −1 3 0 1⎤ ⎥ 0 ⎥ ( R3 =− r2 + r3) −3⎥ ⎦ There is no solution. The system is inconsistent. 800 52. ⎧ 2x−3y− z = 0 ⎪ ⎨− x+ 2y+ z = 5 ⎪ ⎩ 3x−4y− z = 1 Write the augmented matrix: ⎡ 2 ⎢ ⎢−1 ⎢ 3 ⎣ −3 2 − 4 −1 1 −1 0⎤ ⎥ 5⎥ 1⎥ ⎦ ⎡1 ⎢ → ⎢2 ⎢3 ⎣ − 2 −3 − 4 −1 −1 −1 −5⎤ ⎥ 0⎥ 1⎥ ⎦ ⎛Interchange⎞ ⎜ r1 and r ⎟ ⎝ − 2 ⎠ ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 − 2 1 2 −1 1 2 −5⎤ ⎥ ⎛R2 =− 2r1+ r2⎞ 10 ⎥ ⎜ R3 =− 3r1+ r ⎟ 3 16⎥ ⎝ ⎠ ⎦ ⎡1 ⎢ → ⎢0 ⎢ ⎣ 0 − 2 1 0 −1 1 0 −5⎤ ⎥ 10 ⎥ ( R3 = − 2r2 + r3) − 4⎥ ⎦ There is no solution. The system is inconsistent. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 53. ⎧− x+ y+ z= −1 ⎪ ⎨ − x+ 2y− 3z =−4 ⎪ ⎩ 3x−2y− 7z = 0 Write the augmented matrix: ⎡−1 1 1 −1⎤ ⎢ ⎥ ⎢−1 2 −3 − 4⎥ ⎢ 3 − 2 −7 0⎥ ⎣ ⎦ ⎡ 1 −1 −1 1⎤ ⎢ ⎥ → ⎢− 1 2 −3 − 4⎥ ( R1 = −r1) ⎢ 3 − 2 −7 0⎥ ⎣ ⎦ ⎡1 −1 −1 1⎤ ⎢ ⎥ ⎛R2 = r1+ r2 ⎞ → ⎢0 1 − 4 −3 ⎥ ⎜ R3 =− 3r1+ r ⎟ ⎢ 3 0 1 − 4 3⎥ ⎝ ⎠ ⎣ − ⎦ ⎡10 −5 − 2⎤ ⎢ ⎥ ⎛R1 = r2 + r1 ⎞ → ⎢01−4−3 ⎥ ⎜ R3 =− r2 + r ⎟ ⎢ 3 0 0 0 0⎥ ⎝ ⎠ ⎣ ⎦ The matrix in the last step represents the system ⎧x− 5z =−2 ⎧x = 5z−2 ⎪ ⎪ ⎨ y− 4z =−3 or, equivalently, ⎨y = 4z−3 ⎪ ⎩ 0= 0 ⎪ ⎩0= 0 The solution is x = 5z− 2, y = 4z− 3, z is any real number or {( xyz , , ) | x= 5z− 2, y= 4z−3, z is any real number}.
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Chapter 8 Systems of Equations and Inequalities

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