FinalTestReviewSpring2011

2 find the matrix a for which a 10 pts solution a 1 0

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Unformatted text preview: Find the matrix A for which A (10 pts) Solution: A= 1 0 0 1 1 1 0 0 0 0 1 0 1 −1 0 −1 x y z t = x+y y+z z−t x−t Ex. 3. Using block multiplication, evaluate 1 3 1 3 (13 1 3 1 3 212 5 4 3 4 1 2 1 2 5 434 1 pts) Solution: 2 4 2 4 1 3 1 3 2 4 2 4 ￿ 5 1 5 1￿ 12 34 0 2 0 2 5 1 5 1 0 2 0 2 0 2 0 2 5 1 5 1 0 2 0 2 = ￿ ￿ 5 where A = and B = 1 ￿ ￿￿ 74 7 and AB + AB = + 19 8 19 14 8 14 8 38 16 38 16 14 8 14 8 38 16 38 16 AA AA ￿￿ ￿ BB BB ￿ = ￿ ￿ ￿ AB + AB AB + AB , AB + AB AB + AB ￿￿ ￿ 0 12 50 . Since AB = = 2 34 12 ￿ ￿ ￿ 4 14 8 = , the final answer is 8 38 16 ￿ 74 19 8 ￿ 11 0 Ex. 4. Using Gauss-Jordan elimination, find the inverse of the matrix A = 0 1 −1 21 0 (13 pts) Solution: 11 010 [A : I ] = 0 1 −1 0 1 21 000 10 0 −1 0 1 0 2 0 −1 → 0 1 0 0 −1 −2 1 1 0 0 → 1 1 → 0 0 1 0 0 0 1 0 1 0 10 1 −1 01 −1 0 −2 0 0 −1 0 1 0 2 0 −1 1 2 −1 −1 2 0 10 1 1 −1 0 0 → 0 1 −1 0 1 0 1 0 0 −1 −2 11 −1 0 1 0 −1 , so A−1 = 2 2 −1 −1 Ex. 5....
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This document was uploaded on 02/19/2014 for the course MATH 441 at WVU.

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