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pset2-soln - 18.06 Problem Set 2 Due Wednesday 20 February...

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18.06 Problem Set 2 Due Wednesday, 20 February 2008 at 4 pm in 2-106. Problem 1: a) Do problem 5 from section 2.4 (pg. 65) in the book. b) Do problem 26 in section 2.4 (pg. 69). Solution (5+5 points) a) We have 1 b 0 1 n = 1 nb 0 1 and 2 2 0 0 n = 2 n 2 n 0 0 If you insist on a rigorous proof, you can use induction. b) 1 0 2 4 2 1 3 3 0 1 2 1 = 1 2 2 3 3 0 + 0 4 1 1 2 1 = 3 3 0 6 6 0 6 6 0 + 0 0 0 4 8 4 1 2 1 = 3 3 0 10 14 4 7 8 1 Problem 2: Do problem 24 from section 2.4 (pg. 68). Solution (5+5 points) In general, if we take an upper triangular matrix with 0s along the diagonal, some power of it will be 0. The matrices I picked are all instances of this principle. a) A = 0 1 0 0 will work. b) A = 0 1 0 0 0 1 0 0 0 . Do you see the pattern? 1
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Problem 3: Do problem 7 from section 2.5 (pg. 79). Solution (3+4+3 points) a) Suppose we had a solution x . The equation Ax = (1 , 0 , 0) amounts to the three equations row 1 · x = 1, row 2 · x = 0, row 3 · x = 0. But we know row 1 + row 2 = row 3. The three dot products should sum correctly, but they don’t: (row 1 + row 2) · x = 1 + 0 = 0 = row 3 · x Thus, there can’t be a solution x . This shows that A is not invertible. b) By the same reasoning as above, we must have b 1 + b 2 = b 3 for any allowable solution. It’s possible that even some of these won’t have solutions; we don’t have enough information about A to say for sure. For example, the 0 matrix satisfies the requirement to be A , but 0 x = b certainly won’t have any solutions unless all the b i are 0. We’ll learn a more precise way to discuss this when we talk about rank. c) After elimination row 3 will become all 0. How do we see this? We know from part a) that A is not invertible (if it were, every choice of b would have a a solution). So A can’t have three (non-zero) pivots. Since A has three rows but at most two pivots, at least one row must be all 0. (Remember that a pivot is the first non-zero entry in a row; a row without a pivot must not have any non-zero entries at all.) Because we always eliminate downwards, the row without a pivot will be the bottom one (it’s possible that the earlier rows also will be all 0).
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