Linear Algebra Solutions 71

# Linear Algebra Solutions 71 - -1 4-1 2-1 4 Î-1 2 f f f f 1...

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However, since we are assuming x (0) = 1 = y (0), we have x 1 (0) y 1 (0) = P - 1 x (0) y (0) = 1 - 4 0 - 1 - 4 - 2 ‚• 1 1 = 1 - 4 - 1 - 6 = 1 / 4 3 / 2 . Hence x 1 = e 6 t ( 3 2 t + 1 4 ) and y 1 = 3 2 e 6 t . Finally, solving for x and y , x y = - 2 1 4 0 ‚• x 1 y 1 = - 2 1 4 0 e 6 t ( 3 2 t + 1 4 ) 3 2 e 6 t = ( - 2) e 6 t ( 3 2 t + 1 4 ) + 3 2 e 6 t 4 e 6 t ( 3 2 t + 1 4 ) = e 6 t (1 - 3 t ) e 6 t (6 t + 1) . Hence x = e 6 t (1 - 3 t ) and y = e 6 t (6 t + 1). 9. Let A = 1 / 2 1 / 2 0 1 / 4 1 / 4 1 / 2 1 / 4 1 / 4 1 / 2 . (a) We ±rst determine the characteristic polynomial ch A ( λ ). ch A ( λ ) = det( λI 3 - A ) = f f f f f f λ - 1 / 2 - 1 / 2 0 - 1 / 4 λ - 1 / 4 - 1 / 2 - 1 / 4 - 1 / 4 λ - 1 / 2 f f f f f f = µ λ - 1 2 ¶f f f f λ
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Unformatted text preview: -1 / 4-1 / 2-1 / 4 Î»-1 / 2 f f f f + 1 2 f f f f-1 / 4-1 / 2-1 / 4 Î»-1 / 2 f f f f = Âµ Î»-1 2 Â¶â€°Âµ Î»-1 4 Â¶Âµ Î»-1 2 Â¶-1 8 Â¾ + 1 2 â€°-1 4 Âµ Î»-1 2 Â¶-1 8 Â¾ = Âµ Î»-1 2 Â¶Âµ Î» 2-3 Î» 4 Â¶-Î» 8 = Î» â€°Âµ Î»-1 2 Â¶Âµ Î»-3 4 Â¶-1 8 Â¾ 74...
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