pset3-soln

pset3-soln - 18.06 Problem Set 3 Due Wednesday, 27 February...

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18.06 Problem Set 3 Due Wednesday, 27 February 2008 at 4 pm in 2-106. Problem 1: Do problem 7 from section 2.7 (pg. 105) in the book. Solution (2+3+3+2 points) a) False. One example is when A = ± 1 2 3 4 ² . b) False. We check by taking the transpose: ( AB ) T = B T A T = BA So, any time we have symmetric matrices A and B which give different answers when we multiply them in different orders, the hypothesis will fail. One example is A = ± 0 1 1 0 ² and B = ± 1 2 2 3 ² , so that AB = ± 2 3 1 2 ² is not symmetric. c) True. If any matrix B is symmetric, then its inverse is also symmetric. So, if A - 1 were symmetric, then A would need to be symmetric as well. d) True. We check by taking the transpose: ( ABC ) T = C T B T A T = CBA Problem 2: Do problem 10 from section 3.1 (pg. 119). (Give explanations of just a sentence or two.) Solution (10 points) a) This is a subspace; it is the nullspace of the matrix A = ³ 1 - 1 0 ´ . b) This is not a subspace, as it does not contain the 0 vector. 1
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c) This is not a subspace; for example, the vectors v = 0 1 1 w = 1 0 1 both have b 1 b 2 b 3 = 0, but their sum does not. d) This is a subspace. Any time we take all linear combinations of a set of vectors, we get a subspace. e) This is a subspace; it is the nullspace of the matrix A = ± 1 1 1 ² . f) This is not a subspace. If we take the vector v = 1 2 3 and multiply it by the scalar - 1, it is no longer is in the set. Problem 3: Consider the system of equations 1 4 2 2 8 5 - 1 - 4 - 2 x 1 x 2 x 3 = b 1 b 2 b 3 a) For which right sides (find a condition on b 1 , b 2 , b 3 ) is this system solvable? b) Call the coefficient matrix A . Is the vector (2 , 5 , - 2) in the column space of A ? How about (1 , 2 , 3)? c) Suppose we add a fourth column (2 , 5 , - 2) to A . How does the column space change? What if we added the column (1 , 2 , 3) instead? Solution (5 points) a) We reduce the augmented matrix: 1 4 2 b 1 2 8 5 b 2 - 1 - 4 - 2 b 3 ± 1 4 2 b 1 0 0 1 b 2 - 2 b 1 - 1 - 4 - 2 b 3 ± 1 4 2 b 1 0 0 1 b 2 - 2 b 1 0 0 0 b 3 + b 1 2
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The conditions come from the 0 rows. In this case, there is only one 0 row, giving us the condition b 1 + b 3 = 0. b) The vector (2
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This note was uploaded on 08/10/2008 for the course MATH 250 taught by Professor Chanillo during the Spring '08 term at Rutgers.

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pset3-soln - 18.06 Problem Set 3 Due Wednesday, 27 February...

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