FinalTestReviewSpring2011

# 001 301 031 2 so l e32 e31 e21 1 e211 e311 e321 100

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Unformatted text preview: − = 2 1 0 , E311 = 0 1 0 , and E321 = 0 1 0 . 001 301 031 2 − − − So, L = (E32 E31 E21 )−1 = E211 E311 E321 100 = 2 1 0 . 331 2 3 MATH 441.001 Instr. K. Ciesielski Spring 2011 NAME (print): TEST # 2 Review To be solved in class. Test will start with: “Solve the following exercises. Show your work. (No credit will be given for an answer with no supporting work shown.)” Ex. 1. Find a basis for a vector space V generated by the following vectors: 0 1 1 4 0 2 v . ￿1 = , ￿2 = , and ￿3 = v v 2 1 2 4 −2 5 ( pts) Solution: The space V is equal to the column space C (A) of A = [￿1￿2￿2 ]. Since vvv A reduces as follows: 11 0 1 1 0 −￿2 r 1 0 −2 0 2 0 0 1 4 ×1/2 1 2 2 A= → → =R 1 2 −￿1 0 −￿2 0 0 2 r 1 2 r 0 5 4 −2 −5￿1 r 0 −1 −2 +￿2 r 00 0 the pivot columns of A are the same as of R: columns #1 and #2. Therefore these columns of matrix A form the basis of V = C (A). Answer: {￿1 , ￿2 } is a basis of V . vv 11 0 2 0 2 −4 4 Ex. 2. Let A = . (a) For what...
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