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Linear Algebra Solutions 9

# Linear Algebra Solutions 9 - A(3-3 n 1 2 I 2 =(3 n 1-1 2...

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Section 2 . 4 2. Suppose B = a b c d e f and that AB = I 2 . Then - 1 0 1 0 1 0 a b c d e f = 1 0 0 1 = - a + e - b + f c + e d + f . Hence - a + e = 1 c + e = 0 , - b + f = 0 d + f = 1 ; e = a + 1 c = - e = - ( a + 1) , f = b d = 1 - f = 1 - b ; B = a b - a - 1 1 - b a + 1 b . Next, ( BA ) 2 B = ( BA )( BA ) B = B ( AB )( AB ) = BI 2 I 2 = BI 2 = B 4. Let p n denote the statement A n = (3 n - 1) 2 A + (3 - 3 n ) 2 I 2 . Then p 1 asserts that A = (3 - 1) 2 A + (3 - 3) 2 I 2 , which is true. So let n 1 and assume p n . Then from (1), A n +1 = A · A n = A n (3 n - 1) 2 A + (3 - 3 n ) 2 I 2 o = (3 n - 1) 2 A 2 + (3 - 3 n ) 2 A = (3 n - 1) 2 (4 A - 3 I 2 ) + (3 - 3 n ) 2 A = (3 n - 1)4+(3 - 3 n ) 2 A + (3 n - 1)( - 3) 2 I 2 = (4 · 3 n - 3 n ) - 1 2
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Unformatted text preview: A + (3-3 n +1 ) 2 I 2 = (3 n +1-1) 2 A + (3-3 n +1 ) 2 I 2 . Hence p n +1 is true and the induction proceeds. 5. The equation x n +1 = ax n + bx n-1 is seen to be equivalent to • x n +1 x n ‚ = • a b 1 ‚• x n x n-1 ‚ 12...
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