prelim1_fa06sol

# prelim1_fa06sol - Prelim 1 Short solutions (corrected)...

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Prelim 1 Short solutions (corrected) September 28, 2006 1. (a) x 1 x 2 x 3 x 4 x 5 x 6 = 1 0 2 0 0 3 + r - 3 1 0 0 0 0 + s 1 0 - 1 1 0 0 + t - 1 0 - 2 0 1 0 ; r, s, t R (b) p = 6 , q = 4 (c) rank( A ) = dim (image( T )) = 3 (d) Yes, because A~x = ~ b has a solution. (e) dim (ker( T )) = 3, basis is the last 3 vectors in (a). (f) Basis for image( T ) is { v 1 , v 3 , v 6 } ; not enough information to de- termine them explicitly. 2. (a) A - 1 = - 3 - 2 4 2 1 - 2 - 1 0 1 ; check AA - 1 = I 3 . (b) ~v = A - 1 1 1 1 = - 1 1 0 3. (a) For example, v 1 = 1 - 1 2 , v 2 = 1 1 0 , v 3 = 0 2 1 . (b) - 1 0 0 0 1 0 0 0 1 (c) 1 1 0 -

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## This note was uploaded on 10/01/2008 for the course MATH 2940 taught by Professor Hui during the Fall '05 term at Cornell University (Engineering School).

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prelim1_fa06sol - Prelim 1 Short solutions (corrected)...

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