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assign3_soln

# assign3_soln - Math 136 Assignment 3 Solutions 1 For each...

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Math 136 Assignment 3 Solutions 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 3 - 2 r 2 1 1 0 - 7 0 1 1 16 0 0 - 1 - 34 ( - 1) r 3 1 1 0 - 7 0 1 1 16 0 0 1 34 iii) By back-substitution we get x 3 = 34, x 2 = 16 - 34 = - 18 and x 1 = - 7 - ( - 18) = 11. Hence the general solution is ( x 1 , x 2 , x 3 ) = (11 , - 18 , 34). 1

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2 b) x 1 + x 2 + 2 x 3 + x 4 = 3 x 1 + 2 x 2 + 4 x 3 + 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 . 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 3 + r 2 1 1 2 1 3 0 1 2 0 4 0 0 0 0 - 20 ( - 1 20 ) r 3 1 1 2 1 3 0 1 2 0 4 0 0 0 0 1 iii) Since the last row in REF has the form [0 · · · 0 | 1] the system is inconsistent. c) (1 + i ) z 1 + 2 iz 2 = 1 (1 + i ) z 2 + z 3 = 1 2 - 1 2 i z 1 - z 3 = 0 Solution: i) 1 + i 2 i 0 1 0 1 + i 1 1 2 - 1 2 i 1 0 - 1 a .
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