sol_practicemid2_20F_fall04

# sol_practicemid2_20F_fall04 - 1 PRACTICE MIDTERM 2 MATH 20F...

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PRACTICE MIDTERM 2 MATH 20F, LECTURE C SOLUTIONS Problem 1: a ) 1 - 3 5 - 2 1 6 0 7 0 1 - 3 5 0 - 5 16 0 7 0 1 - 3 5 0 - 5 16 0 0 112 5 . There are 3 pivot columns, therefore Rank( A ) = 3. b ) By properties of determinants, det( AB - 1 ) = det( A )det( B - 1 ) = det( A ) 1 det( B ) det( A ) = 1 - 3 5 - 2 1 6 0 7 0 = 1 - 3 5 0 - 5 16 0 7 0 = 1 - 3 5 0 - 5 16 0 0 112 5 = 1 · ( - 5) · 112 5 = - 112 . det( B ) = ( - 3) · 1 4 · 2 = - 3 2 , det( AB - 1 ) = 224 3 . Problem 2: a ) H is not a subspace of R 4 because it does not contain the zero vector. b ) Note that 2 a - 3 b - 4 a + b b a = a 2 - 4 0 1 + b - 3 1 1 0 It follows that W = Col( A ), where A = 2 - 3 - 4 1 0 1 1 0 This shows that W is a subspace of R 4 . Problem 3:

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## This note was uploaded on 04/14/2008 for the course MATH 20F taught by Professor Buss during the Winter '03 term at UCSD.

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sol_practicemid2_20F_fall04 - 1 PRACTICE MIDTERM 2 MATH 20F...

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