253Solutions7

# Linear Algebra with Applications (3rd Edition)

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• davidvictor
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ma253 , Fall 2007 — Problem Set 7 Solutions 1. Problems from the textbook: a. Section 3 . 3 , problems * 18 , * 22 , * 26 , * 28 , * 30 , * 36 , * 38 . *18 Well, if el professor is going to ask you to do things involving nasty matrices, it’s nice when they come pre-row-reduced, eh? Reading off from the matrix, the image is R 3 , so a basis is the usual e 1 , e 2 and e 3 . The kernel is ker ( A ) = Span 1 0 - 5 1 0 , 2 1 0 0 0 . *22 Rats, this time we have to row-reduce. rref ( A ) = 1 0 - 6 0 1 5 0 0 0 , so a basis of the kernel is the single vector 6 - 5 1 and a basis of the image consists of the two vectors 2 4 7 and 4 5 9 . *26 For part (a), start by finding the kernel of C : it is the span of 0 1 - 1 . Now just find which of the others have this vector in the kernel: L is the only one. (Notice that this vector is in the kernel if and only if the last two columns are equal, so this is easy to check!). Now row-reduce L to make sure its kernel is also just the span of 0 1 - 1 ; it is.

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For (b), start by noting that the image of C is the span of 1 1 1 and 1 0 1 . It is not hard to see that this span is just the set of all vectors in R 3 such that the first and the third coordinates are equal. Now it’s easy to see that H and X have the same image as C . For (c), we can eliminate C , H , and X at once, by part (b). Now notice that T and Y have the same image: both images contain 1 0 0 , and the fact that 1 0 0 + 0 1 1 = 1 1 1 shows that they are the same. So L is the matrix whose image is different from that of all the others. *28 Since there are 4 vectors, they will be a basis if and only if they are independent. Now notice that if α 1 0 0 2 + β 0 1 0 3 + γ 0 0 1 4 + δ 2 3 4 k = 0 0 0 0 , we can read off from the first three lines that α = - , β = - , γ = - . Plugging those into the fourth line, we get 0 = - - - 16δ + = ( k - 29 ) δ.
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