midterm 2 spring 2003 solutions

# Linear Algebra with Applications

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• davidvictor
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Math 21b Midterm II—Solutions Thursday, April 10, 2003 1. (12 points) True or False. No justification is necessary, simply circle T or F for each statement. T F (a) If A is an n × n invertible matrix, then det( A T A ) > 0. Solution. True. det( A T A ) = det( A T ) det( A ) = [det( A )] 2 > 0. T F (b) The subset W = { f C ( R ) : f ( x ) + f ( x ) = x 2 } is a subspace of the linear space of all infinitely differentiable functions C ( R ). Solution. False. f 0 is not in W . T F (c) If v R n and W is a subspace of R n , then v 2 = proj W v 2 + v - proj W v 2 . Solution. True. proj W v v - proj W v . T F (d) If det( A ) = det( B ), then two n × n matrices A and B cannot be similar. Solution. True. Consider the contrapositive. If A and B are similar matrices, then det( B ) = det( S - 1 AS ) = det( S - 1 ) det( A ) det( S ) = det( A ). T F (e) The vectors x +1, x - 1, and x 2 - 1 are linearly dependent in P 2 , the polynomials of degree less than or equal to 2. Solution. False. Let 0 = c 1 ( x + 1) + c 2 ( x - 1) + c 3 ( x 2 - 1) = ( c 1 - c 2 - c 3 ) + ( c 1 + c 2 ) x + c 3 x 2 . We must find a nontrivial solution for c 1 - c 2 - c 3 = 0 c 1 + c 2 = 0 c 3 = 0 . However, it is easy to see that c 1 = c 2 = c 3 = 0, and the vectors are linearly independent. 1

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T F (f) The determinant of A = 1 1000 2 3 4 5 6 1000 7 8 1000 9 8 7 6 5 4 3 2 1000 1 2 3 1000 4 is positive. Solution. False. The term 1000 5 is larger than the sum of all of the other terms, and this term is negative.
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