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homework set 22-Solutions - BILKENT UNIVERSITY Department...

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B ˙ ILKENT UNIVERSITY Department of Mathematics MATH 225, LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS, Solution of Homework set 1 # 22 U. Mu˘gan July 4, 2008 Homework problems from the 2 nd Edition, SECTION 6.3 15 (15) ) 2 The C.E. p ( λ ) = λ 2 - 3 λ + 2 = ( λ - 2)( λ - 1) 2 = 0 . Therefore, the eigenvalues are λ 1 = 1 , λ 2 = 2 . The eigenvectors for λ 1 = 1: 3 v 11 - 3 v 12 = 0 2 v 11 - 2 v 12 = 0 (1) ~v 1 = 1 1 For λ 2 = 2: 2 v 21 - 3 v 22 = 0 2 v 21 - 3 v 22 = 0 (2) ~v 2 = 3 2 Therefore, P = 1 3 1 2 , D = 1 0 0 2 , P - 1 = - 2 3 1 - 1 . A 5 = 1 3 1 2 ‚ • 1 0 0 32 ‚ • - 2 3 1 - 1 = 94 - 93 62 - 61 . 7 (7) ) The C.E. p ( λ ) = - ( λ - 1)( λ - 2) 2 = 0 . Therefore, the eigenvalues are λ 1 = 1 , λ 2 = 2 , λ 3 = 2 . The eigenvectors for λ 1 = 1: 3 v 12 = 0 v 12 = 0 v 13 = 0 1 I made every effort to avoid the calculation errors and/or typos while I prepared the solution set. You are responsible to check all the solutions and correct the errors if there is any. If you find any errors and/or misprints, please notify me.
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