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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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This note was uploaded on 09/19/2009 for the course MATH 330 taught by Professor Johnson,j during the Spring '08 term at Nevada.
 Spring '08
 Johnson,J
 Math, Linear Algebra, Algebra

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