Linear Algebra Solutions 10

Linear Algebra Solutions 10 - 1 + n-1 1 2 + + 1 n-1 2 + n 2...

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or X n = AX n - 1 , where X n = x n +1 x n and A = a b 1 0 . Then X n = A n X 0 if n 1. Hence by Question 3, x n +1 x n = (3 n - 1) 2 A + (3 - 3 n ) 2 I 2 ¾• x 1 x 0 = (3 n - 1) 2 4 - 3 1 0 + 3 - 3 n 2 0 0 3 - 3 n 2 ‚¾• x 1 x 0 = (3 n - 1)2 + 3 - 3 n 2 (3 n - 1)( - 3) 3 n - 1 2 3 - 3 n 2 x 1 x 0 Hence, equating the (2 , 1) elements gives x n = (3 n - 1) 2 x 1 + (3 - 3 n ) 2 x 0 if n 1 7. Note: λ 1 + λ 2 = a + d and λ 1 λ 2 = ad - bc . Then ( λ 1 + λ 2 ) k n - λ 1 λ 2 k n - 1 = ( λ 1 + λ 2 )( λ n - 1 1 + λ n - 2 1 λ 2 + ··· + λ 1 λ n - 2 2 + λ n - 1 2 ) - λ 1 λ 2 ( λ n - 2 1 + λ n - 3 1 λ 2 + ··· + λ 1 λ n - 3 2 + λ n - 2 2 ) = ( λ n 1 + λ n - 1 1 λ 2 + ··· + λ 1 λ n - 1 2 ) +( λ n - 1 1 λ 2 + ··· + λ 1 λ n - 1 2 + λ n 2 ) - ( λ n - 1 1 λ 2 + ··· + λ 1 λ n - 1 2 ) = λ n
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Unformatted text preview: 1 + n-1 1 2 + + 1 n-1 2 + n 2 = k n +1 If 1 = 2 , we see k n = n-1 1 + n-2 1 2 + + 1 n-2 2 + n-1 2 = n-1 1 + n-2 1 1 + + 1 n-2 1 + n-1 1 = n n-1 1 13...
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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