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Practice Problems 5 Solutions
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Practice Problems 5 Solutions

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  • 1. Practice Problems 5: Solutions  1 1. (a) 0 0  0 0 1 1 0 0 −1/2 1 1 0 1 0  0 1 0 −1 1 0  0 0 1 1 0  1 0 1 −1   1 0 1 5 2 = 0 1 1 0 0 1 0  0 1 0 −1 1 0  0 0 1 1 0  1 0 1 −1   1 0 2 1 5 2 8 = 0 1 1 3 0 1 4 0 0 2 0  2 6 2 =2 =6 = 2 ⇒ No solution    1 1 0 0 1 0 0 0 0 0 1 0 −1 1 0  1 0 0 1 −1 1 1 0 1   1 1 0 2 5 2 8 = 0 0 1 1 1 1 4 0 0 2 0  2 6 0 1 4 0  0 2 0 A has 2 pivots.  1 (b) 0 0  0 0 1 1 0 0 −1/2 1 1 x1 + x2 4x2 + 2x3 0x3  1 0 1 (c) 0 0 −1/2 x1 + x2 = 2 4x2 + 2x3 = 6 0x3 = 0 ⇒ Let x3 = 1, then x = (1, 1, 1) (d) Let x3 = t, then x = 1+t 3−t 2 , 2 ,t (e) No, A does not have an inverse; it does not have 3 pivots. 2. (a) Each matrix is n × n AB CAB (b) IB B =I = CI =C =C Or CA = I CAB = IB CI = B C=B (c) Yes, A is invertible, and A−1 = B = C 3. (I − A)2 = (I − A)(I − A) = I − 2A + A2 = I − 2A + A = I − A (I − A)5 = (I − A)2 (I − A)2 (I − A) = (I − A)2 (I − A) = (I − A)2 = (I − A)       1 6 9 4. (a) −3 2 + 2 4 =  2  3 2 −5 (b) No, the coefficient matrix only has 2 pivots.

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