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e208fk - Y instead (which got about 12 points of partial...

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MAS 3105 Sept 19, 2008 Quiz 2 Prof. S. Hudson 1) [20pt] Find the determinants of these matrices. If possible, find quick ways to do them (without using the definition) and explain briefly. A = 1 2 3 0 5 6 0 0 7 B = 0 0 7 0 5 6 1 2 3 C = 1 2 1 2 3 2 1 2 1 2) [20pt] Solve for X given that AX + B = X and A = ± 0 1 0 - 1 ² B = ± - 2 - 2 6 8 ² 3) [20pt] Choose one of these HW proofs; you can answer on the back: A) If A is row equivalent to both B and to C , then B is row equivalent to C . B) Suppose these 3 matrices are all nxn, and AB = C . Prove: if B is singular, then C is, too. Bonus: (about 5 points) Almost the same as part 3B, But prove: if A is singular, then C is, too. Hint: transpose. Remarks and Answers : I was short of time, and based this Quiz on one from Spring 2007. The average was about 47/60. The scale for this quiz is: A-’s start at 54, B-’s at 48, C-’s at 42, etc). 1) det A = 35, det B = -35, det C = 0. 2) From ( A - I ) X = - B , multiply by on the LEFT to get X = - ( A - I ) - 1 B = see below. If you multiplied on the right, and made no other mistakes, you should get the matrix
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Unformatted text preview: Y instead (which got about 12 points of partial credit). X = 1 2 3 4 Y = -2-2 6 7 Id strongly suggest checking your answer - which should be much easier than solving the problem. 3) These were HW problems from Ch 1. I already have answers or hints posted on this site, but if you need more help, you can always see me. Bonus: Suppose A is singular and C = AB . Then A T is singular, because det A T = det A = 0 [or, we could justify this step without determinants, using Exercise 17 of Ch 1.3). 1 So, B T A T is singular (by problem 3b above, since the A T is on the right side). So, the transpose of this is singular, namely AB . Done. Note: Transposes have many purposes, and this proof shows one of them - to swap the left and right factors in a product. 2...
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e208fk - Y instead (which got about 12 points of partial...

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