Linear Algebra Solutions 4

# Linear Algebra Solutions 4 - 1 1 3 2 4 4 4 3 3 R 2 → 3 R...

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where x 4 and x 5 are arbitrary elements of Z 2 . Hence there are four solutions: x 1 x 2 x 3 x 4 x 5 0 0 0 0 0 1 1 0 0 1 1 1 1 1 0 0 0 1 1 1 . 13. (a) We reduce the augmented matrix to reduced row–echelon form: 2 1 3 4 4 1 4 1 3 1 2 0 R 1 3 R 1 1 3 4 2 4 1 4 1 3 1 2 0 R 2 R 2 + R 1 R 3 R 3 + 2 R 1 1 3 4 2 0 4 3 3 0 2 0 4 R 2 4 R 2 1 3 4 2 0 1 2 2 0 2 0 4 R 1 R 1 + 2 R 2 R 3 R 3 + 3 R 2 1 0 3 1 0 1 2 2 0 0 1 0 R 1 R 1 + 2 R 3 R 2 R 2 + 3 R 3 1 0 0 1 0 1 0 2 0 0 1 0 . Consequently the system has the unique solution x = 1 , y = 2 , z = 0. (b) Again we reduce the augmented matrix to reduced row–echelon form: 2 1 3 4 4 1 4 1 1 1 0 3 R 1 R 3 1 1 0 3 4 1 4 1 2 1 3 4 R 2 R 2 + R 1 R 3 R 3 + 3 R 1
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Unformatted text preview: 1 1 3 2 4 4 4 3 3 R 2 → 3 R 2 1 1 3 1 2 2 4 3 3 R 1 → R 1 + 4 R 2 R 3 → R 3 + R 2 1 3 1 1 2 2 . We read oF the complete solution x = 1-3 z = 1 + 2 z y = 2-2 z = 2 + 3 z, where z is an arbitrary element of Z 5 . 7...
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