Linear Algebra Solutions 2

# Linear Algebra Solutions 2 - 1 2-3 4 3-1 5 2 4 1 a 2-14 a 2...

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with x n arbitrary. Alternatively, writing the system in the form x 1 + ··· + x n = nx 1 x 1 + ··· + x n = nx 2 . . . x 1 + ··· + x n = nx n shows that any solution must satisfy nx 1 = nx 2 = ··· = nx n , so x 1 = x 2 = ··· = x n . Conversely if x 1 = x n ,...,x n - 1 = x n , we see that x 1 ,...,x n is a solution. 10. Let A = a b c d and assume that ad - bc 6 = 0. Case 1: a 6 = 0. a b c d R 1 1 a R 1 1 b a c d R 2 R 2 - cR 1 1 b a 0 ad - bc a R 2 a ad - bc R 2 1 b a 0 1 R 1 R 1 - b a R 2 1 0 0 1 . Case 2: a = 0. Then bc 6 = 0 and hence c 6 = 0. A = 0 b c d R 1 R 2 c d 0 b 1 d c 0 1 1 0 0 1 . So in both cases, A has reduced row–echelon form equal to 1 0 0 1 . 11. We simplify the augmented matrix of the system using row operations:
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Unformatted text preview: 1 2-3 4 3-1 5 2 4 1 a 2-14 a + 2 R 2 → R 2-3 R 1 R 3 → R 3-4 R 1 1 2-3 4-7 14-10-7 a 2-2 a-14 R 3 → R 3-R 2 R 2 →-1 7 R 2 R 1 → R 1-2 R 2 1 2-3 4 1-2 10 7 a 2-16 a-4 R 1 → R 1-2 R 2 1 1 8 7 1-2 10 7 a 2-16 a-4 . Denote the last matrix by B . 5...
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