Linear Algebra Solutions 1

Linear Algebra Solutions 1 - 1 1 0 0 Case 2: = 2. Then B =...

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Case 2: λ = 2. Then B = 1 - 1 0 0 and the solution is x = y , with y arbitrary. Case 3: λ = 4. Then B = 1 1 0 0 and the solution is x = - y , with y arbitrary. 8. 3 1 1 1 5 - 1 1 - 1 R 1 1 3 R 1 1 1 3 1 3 1 3 5 - 1 1 - 1 R 2 R 2 - 5 R 1 1 1 3 1 3 1 3 0 - 8 3 - 2 3 - 8 3 R 2 - 3 8 R 2 1 1 3 1 3 1 3 0 1 1 4 1 R 1 R 1 - 1 3 R 2 1 0 1 4 0 0 1 1 4 1 . Hence the solution of the associated homogeneous system is x 1 = - 1 4 x 3 , x 2 = - 1 4 x 3 - x 4 , with x 3 and x 4 arbitrary. 9. A = 1 - n 1 ··· 1 1 1 - n ··· 1 . . . . . . ··· . . . 1 1 ··· 1 - n R 1 R 1 - R n R 2 R 2 - R n . . . R n - 1 R n - 1 - R n - n 0 ··· n 0 - n ··· n . . . . . . ··· . . . 1 1 ··· 1 - n 1 0 ··· - 1 0 1 ··· - 1 . . . . . . ··· . . . 1 1 ··· 1 - n
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