HW8_solution - EE441 HW#8 Solution Spring 2007 Instructor...

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Unformatted text preview: EE441 HW#8 Solution Spring 2007 Instructor : Dr. Jonckheere TA: Ben Raskob Due Date 3/29/2007 5.1.2 A= |A - I| = 1- -1 2 4- 1 -1 2 4 = 2 - 5 + 6 = ( - 3)( - 2). the eigenvalues are = 3, 2. Now find the eigenvectors: = 3 : [A - I]x1 = choose x1 = 1 -2 = 2 : [A - I]x2 = choose x2 = 1 -1 1 1 . Then S -1 = -2 -1 = -1 -1 . 2 1 -1 -1 2 2 x2 = 0 -2 -1 2 1 x1 = 0 Let S = [x1 x2 ] = Let D = 1 0 0 2 3 0 . Then A = SDS -1 and eAt = SeDt S -1 . 0 2 1 If we let c = S -1 u(0) = -1 -1 2 1 0 6 = -6 6 1 -2 then + 6e2t 1 -1 u(t) = c1 e1 t x1 + c2 e2 t x2 = -6e3t . 2 ...
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  • Spring '08
  • Neely
  • Eigenvalue, eigenvector and eigenspace, x1 x2, Dr. Jonckheere TA, Ben Raskob
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