Linear Algebra Solutions 22

# Linear Algebra Solutions 22 - 1 ‚ Hence A n = P •(5 12...

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Now 1 + b 2 + ( c - ab )( c + ab ) 1 + a 2 = 1 + b 2 + c 2 - ( ab ) 2 1 + a 2 = 1 + a 2 + b 2 + c 2 1 + a 2 6 = 0 . Hence B can be reduced to I 3 using four more row operations and conse- quently A is non–singular. 18. The proposition is clearly true when n = 1. So let n 1 and assume ( P - 1 AP ) n = P - 1 A n P . Then ( P - 1 AP ) n +1 = ( P - 1 AP ) n ( P - 1 AP ) = ( P - 1 A n P )( P - 1 AP ) = P - 1 A n ( PP - 1 ) AP = P - 1 A n IAP = P - 1 ( A n A ) P = P - 1 A n +1 P and the induction goes through. 19. Let A = 2 / 3 1 / 4 1 / 3 3 / 4 and P = 1 3 - 1 4 . Then P - 1 = 1 7 4 - 3 1 1 . We then verify that P - 1 AP = 5 / 12 0 0 1 . Then from the previous ques- tion, P - 1 A n P = ( P - 1 AP ) n = 5 / 12 0 0 1 n = (5 / 12) n 0 0 1 n = (5 / 12) n
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Unformatted text preview: 1 ‚ . Hence A n = P • (5 / 12) n 1 ‚ P-1 = • 1 3-1 4 ‚• (5 / 12) n 1 ‚ 1 7 • 4-3 1 1 ‚ = 1 7 • (5 / 12) n 3-(5 / 12) n 4 ‚• 4-3 1 1 ‚ = 1 7 • 4(5 / 12) n + 3 (-3)(5 / 12) n + 3-4(5 / 12) n + 4 3(5 / 12) n + 4 ‚ = 1 7 • 3 3 4 4 ‚ + 1 7 (5 / 12) n • 4-3-4 3 ‚ . 25...
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