129a_ans4 - 1 0 0 0 0 0 0 1 . No such C exists. Suppose...

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Spring 2010 Answer of HW4 2.1 #4 A - 5 I 3 = 4 - 1 3 - 8 2 - 6 - 4 1 3 , (5 I 3 ) A = 45 - 5 15 - 40 35 - 30 - 20 5 40 2.1 #9 k = 5 2.1 #12 B = " 2 - 2 1 - 1 # 2.1 #21 The columns of A are linearly dependent. 2.1 #27 u T v = - 2 a + 3 b - 4 c, v T u = - 2 a + 3 b - 4 c, uv T = - 2 a - 2 b - 2 c 3 a 3 b 3 c - 4 a - 4 b - 4 c , vu T = - 2 a 3 a - 4 a - 2 b 3 b - 4 b - 2 c 3 c - 4 c 2.2 #15 D = C - 1 B - 1 A - 1 2.2 #19 X = CB - A 2.2 #34 1 0 0 0 0 · · · 0 0 - 1 2 1 2 0 0 0 · · · 0 0 0 - 1 3 1 3 0 0 · · · 0 0 0 0 - 1 4 1 4 0 · · · 0 0 0 0 0 - 1 5 1 5 · · · 0 0 . . . . . . . . . . . . 0 0 0 0 0 · · · - 1 n 1 n 2.2 #35 Solve the linear system Ax = 0 0 1 to obtain the third column of A - 1 3 - 6 4 2.2 #38 D =
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Unformatted text preview: 1 0 0 0 0 0 0 1 . No such C exists. Suppose there is one, i.e. CA = I 4 . Then CA 1-1 1 = C " # = but I 4 1-1 1 = 1-1 1 contradiction!!!...
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This note was uploaded on 09/08/2010 for the course MATH 129A at San Jose State University .

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