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Unformatted text preview: 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 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.responsible to check all the solutions and correct the errors if there is any....
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This note was uploaded on 06/18/2010 for the course COMPUTER S Math 225 taught by Professor Yosumhoca during the Spring '10 term at Bilkent University.
 Spring '10
 YosumHoca

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