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Linear Algebra Solutions 6

Linear Algebra Solutions 6 - is a solution of the...

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is a solution of the associated homogeneous system and x j = α j + y j , 1 j n , then reversing the argument shows that ( x 1 , . . . , x n ) is a solution of the system 1 . 16. We simplify the augmented matrix using row operations, working to- wards row–echelon form: 1 1 - 1 1 1 a 1 1 1 b 3 2 0 a 1 + a R 2 R 2 - aR 1 R 3 R 3 - 3 R 1 1 1 - 1 1 1 0 1 - a 1 + a 1 - a b - a 0 - 1 3 a - 3 a - 2 R 2 R 3 R 2 → - R 2 1 1 - 1 1 1 0 1 - 3 3 - a 2 - a 0 1 - a 1 + a 1 - a b - a R 3 R 3 + ( a - 1) R 2 1 1 - 1 1 1 0 1 - 3 3 - a 2 - a 0 0 4 - 2 a (1 - a )( a - 2) - a 2 + 2 a + b - 2 = B. Case 1: a 6 = 2. Then 4 - 2 a 6 = 0 and B 1 1 - 1 1 1 0 1 - 3 3 - a 2 - a 0 0 1 a - 1 2 - a 2 +2 a + b - 2 4 - 2 a . Hence we can solve for x, y and z in terms of the arbitrary variable
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