Linear Algebra Solutions 70

# Linear Algebra Solutions 70 - P −1 AP = α1 0α . 8. The...

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Unformatted text preview: P −1 AP = α1 0α . 8. The system of diﬀerential equations is equivalent to the single matrix 4 −1 ˙ equation X = AX , where A = . 4 8 The characteristic polynomial of A is λ2 − 12λ + 36 = (λ − 6)2 , so we can use the previous question with α = 6. Let −2 −1 4 2 B = A − 6I2 = −2 = 4 P = [X1 |X2 ], we have Then BX2 = 1 . Also let X1 = BX2 . Then if 0 0 , if X2 = 0 P −1 AP = . 61 06 . Now make the change of variables X = P Y , where Y = ˙ Y = (P −1 AP )Y = 61 06 x1 y1 . Then Y, or equivalently x1 = 6x1 + y1 and y˙1 = 6y1 . ˙ Solving for y1 gives y1 = y1 (0)e6t . Consequently x1 = 6x1 + y1 (0)e6t . ˙ Multiplying both side of this equation by e−6t gives d −6t (e x1 ) = e−6t x1 − 6e−6t x1 = y1 (0) ˙ dt e−6t x1 = y1 (0)t + c, where c is a constant. Substituting t = 0 gives c = x1 (0). Hence e−6t x1 = y1 (0)t + x1 (0) and hence x1 = e6t (y1 (0)t + x1 (0)). 73 ...
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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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