HW12_solution - choose c 22 = 1 c 11 c 31 = 0 ⇒ c 31 = 0...

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EE441 HW#12 Solution Spring 2007 Instructor : Dr. Jonckheere TA: Ben Raskob Due Date 4/26/2007 6.2.28 There are a few ways to do this problem. For instance, the LU factorization will also work. I am going to solve it by writing out the multiplication as A = CC T 9 0 0 0 1 2 0 2 8 = c 11 0 0 c 21 c 22 0 c 31 c 32 c 33 c 11 c 21 c 31 0 c 22 c 32 0 0 c 33 Then, starting with the top-left element and moving right we can solve for each element in C one at a time. c 2 11 = 9 , choose c 11 = 3 . c 11 c 21 = 0 , c 21 = 0 . c 2 21 + c 2 22 = 1 ,
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Unformatted text preview: choose c 22 = 1 . c 11 c 31 = 0 , ⇒ c 31 = 0 . c 21 c 31 + c 22 c 32 = 2 , ⇒ c 32 = 2 . c 2 31 + c 2 32 + c 2 33 = 8 , choose c 33 = 2 . This gives us C = 3 0 0 0 1 0 0 2 2 1 Notice that this answer is not unique, because of the squares we have freedom in negative signs. By similar calculation, for the next matrix we can arrive at C = 1 0 1 1 1 1 √ 5 2...
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