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# H53 - Math 330 Homework 5.3(Pages 325,326(2 Let A = PDP 1...

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Unformatted text preview: Math 330 Homework 5.3 (Pages 325,326) (2) Let A = PDP- 1 and compute A 4 . P = bracketleftBigg 2- 3- 3 5 bracketrightBigg D = bracketleftBigg 1 0 1 2 bracketrightBigg SOLUTION: A 4 = ( PDP- 1 )( PDP- 1 )( PDP- 1 )( PDP- 1 ) = PD ( P- 1 P ) D ( P- 1 P ) D ( P- 1 P ) DP- 1 = PD 4 P- 1 Now using the formula for the inverse of a 2 × 2 matrix we have that P- 1 = 1 (2)(5)- (- 3)(- 3) bracketleftBigg 5 3 3 2 bracketrightBigg = bracketleftBigg 5 3 3 2 bracketrightBigg Therefore, A 4 = bracketleftBigg 2- 3- 3 5 bracketrightBigg bracketleftBigg 1 1 2 4 bracketrightBigg bracketleftBigg 5 3 3 2 bracketrightBigg = bracketleftBigg 2- 3- 3 5 bracketrightBigg bracketleftBigg 5 3 3 16 1 8 bracketrightBigg = bracketleftBigg 10- 9 16 6- 3 8 15 16- 15 5 8- 9 bracketrightBigg = bracketleftBigg 151 16 45 8- 225 16- 67 8 bracketrightBigg (4) Use the factorization A = PDP- 1 to compute A k for an arbitrary integer k . A = bracketleftBigg- 2 12- 1 5 bracketrightBigg = bracketleftBigg 3 4 1 1 bracketrightBigg bracketleftBigg 2 0 0 1 bracketrightBiggbracketleftBigg- 1 4 1- 3 bracketrightBigg SOLUTION: Since the two matrices at each end of the product are inverses, and 1 k = 1, we get that A k = bracketleftBigg...
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H53 - Math 330 Homework 5.3(Pages 325,326(2 Let A = PDP 1...

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