Ans-265FE-F2003

# Ans-265FE-F2003 - (b B Â 1 = P Â 1 A Â 1 P since P Â 1 A Â...

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MA 265 FINAL EXAM Fall 2003 ANSWERS Part I 1. (a) a = b .( b ) ab 6 =0. (c) ab =0 , c =0. 2. (a) t = 7. (b) t =1. 3. v 1 , v 2 , v 4 . 4. (a) ( ¡ 2 ; 0 ; 1). (b) w =(3 = 5 ; 2 ; 14 = 5), u =( ¡ 2 = 5 ; 0 ; 1 = 5). 5. (a) ¸ =2 ; 3 ; 4. (b) A is diagonalizable because an n £ n matrix with n distinct eigenvalues is diagonalizable. 6. (a) P = · 13 15 ¸ .( b ) A 12 = I . 7. (a) x ( t )= c 1 e 2 t 2 4 1 0 0 3 5 + c 2 e 2 t 2 4 0 1 1 3 5 + c 3 e t 2 4 1 0 ¡ 1 3 5 . (b) x ( t )= e 2 t 2 4 2 2 2 3 5 + e t 2 4 ¡ 1 0 1 3 5 . 8. (a) det( B )=det( P ¡ 1 AP )=det( P ¡ 1 )det( A )det( P )= det( P ) ¡ 1 det( A )det( P )=det( A ). So if det( A ) 6 = 0, det( B ) 6 =0.
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Unformatted text preview: (b) B Â¡ 1 = P Â¡ 1 A Â¡ 1 P , since ( P Â¡ 1 A Â¡ 1 P ) B = ( P Â¡ 1 A Â¡ 1 P )( P Â¡ 1 AP ) = I . Part II 1. C. 2. A. 3. E. 4. D. 5. E. 6. B. 7. C. 8. (The question contains a mistake. There is no correct answer.) 9. C. 10. E. 11. D. 12. B....
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## This document was uploaded on 01/21/2012.

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