assign4_practice_soln

# assign4_practice_soln - Math 136 Practice Problems # 4 1....

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Math 136 Practice Problems # 4 1. For each of the following systems of linear equations: i) Write the augmented matrix. ii) Row-reduce the augmented matrix into row echelon form. iii) Find the general solution of the system or explain why the system is inconsistent. a) x 1 + x 2 = - 7 2 x 1 + 4 x 2 + x 3 = - 16 x 1 + 2 x 2 + x 3 = 9 Solution: i) 1 1 0 - 7 2 4 1 - 16 1 2 1 9 . ii) 1 1 0 - 7 2 4 1 - 16 1 2 1 9 R 2 - 2 R 1 R 3 - R 1 1 1 0 - 7 0 2 1 - 2 0 1 1 16 r 2 r 3 1 1 0 - 7 0 1 1 16 0 2 1 - 2 R 1 - R 2 R 3 - 2 R 2 1 0 - 1 - 23 0 1 1 16 0 0 - 1 - 34 ( - 1) R 3 1 0 - 1 - 23 0 1 1 16 0 0 1 34 R 1 + R 3 R 2 - R 3 1 0 0 11 0 1 0 - 18 0 0 1 34 iii) Writing the RREF into equation form we get x 1 = 11, x 2 = - 18, and x 3 = 34. Hence the general solution is ~x = x 1 x 2 x 3 = 11 - 18 34 . b) x 1 + x 2 + 2 x 3 + x 4 = 3 x 1 + 2 x 2 + 4 x 3 + 2 x 4 = 7 x 1 + x 4 = - 21 Solution: i) 1 1 2 1 3 1 2 4 1 7 1 0 0 1 - 21 . 1

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2 ii) 1 1 2 1 3 1 2 4 1 7 1 0 0 1 - 21 R 2 - R 1 R 3 - R 1 1 1 2 1 3 0 1 2 0 4 0 - 1 - 2 0 - 24 R 1 - R 2 R 3 + R 2 1 0 0 1 1 0 1 2 0 4 0 0 0 0 - 20 ( - 1 20 ) R 3 1 0 0 1 1 0 1 2 0 4 0 0 0 0 1 iii) Since the last row in RREF has the form [0 ··· 0 | 1] the system is inconsistent. c) x 2 + x 3 = 2 x 1 + x 2 + x 3 = 3 2 x 1 + 3 x 2 + 3 x 3 = 9 Solution: i) 0 1 1 2 1 1 1 3 2 3 3 9 . ii) 0 1 1 2 1 1 1 3 2 3 3 9 R 1 R 2 1 1 1 3 0 1 1 2 2 3 3 9 R 3 - 2 R 1 1 1 1 3 0 1 1 2 0 1 1 3 R 1 - R 2 R 3 - R 2 1 0 0 1 0 1 1 2 0 0 0 1 iii) Since the last row is ± 0 0 0 | 1 ² the system is inconsistent. d) 2 x 1 + x 2 - x 3 = 6 x 1 - 2 x 2 - 2 x 3 = 1 - x 1 + 12 x 2 + 8 x 3 = 7 Solution: i) 2 1 - 1 6 1 - 2 - 2 1 - 1 12 8 7 .
3 ii) 2 1 - 1 6 1 - 2 - 2 1 - 1 12 8 7 R 1 R 2 1 - 2 - 2 1 2 1 - 1 6 - 1 12 8 7 R 2 - 2 R 1 R 3 + R 1 1 - 2 - 2 1 0 5 3 4 0 10 6 8 R 1 + 2 5 R 2 R 3 - 2 R 2 1 0 - 4 / 5 13 / 5 0 5 3 4 0 0 0 0 (1 / 5) R 2 1 0 - 4 / 5 13 / 5 0 1 3 / 5 4 / 5 0 0 0 0 iii) Observe that x 3 is a free variable, so we let x 3 = t R . Writing the RREF in equation form we get x 1 - 4 5 x 3 = 13 5 , x 2 + 3 5 x 3 = 4 5 . Hence, we have x 1 = 13 5 + 4 5 t and x 2 = 4 5 - 3 5 t . Hence the general solution is x 1 x 2 x 3 = 13 5 + 4 5 t 4 5 - 3 5 t t = 13 / 5 4 / 5 0 + t 4 / 5 - 3 / 5 1 . e)

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## This note was uploaded on 07/13/2011 for the course MATH 136 taught by Professor All during the Winter '08 term at Waterloo.

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assign4_practice_soln - Math 136 Practice Problems # 4 1....

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