hw1 - IEOR 162 Linear Programming Fall 2004 Homework 1...

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Unformatted text preview: IEOR 162 Linear Programming Fall 2004 Homework 1 Solutions 2.Review.1. 1 1 02 0 1 13 1 2 15 1 1 02 0 1 13 0 1 13 1 0 - 1- 1 0 1 0 1 1 3 1 3 1 0 - 1- 1 0 1 0 0 1 3 0 0 We now find that x3 = k, x1 = -1 + k, x2 = 3 - k 2.Review.11. 1 0 21 0 0 0 1 00 1 0 0 1 10 0 1 1 0 21 0 0 0 1 00 1 0 0 0 1 0 -1 1 1 0 01 2 -2 0 1 -1 0 1 00 1 0 0 1 0 -1 1 0 2 Thus 1 = 2 -2 0 1 0 1 0 0 1 1 0 1 0 -1 2.Review.14 1 0 0 1 0 0 1 0 0 0 1 0 0 0 0 rank 2<3, so the original set of vectors is 0 1 0 -1 -1 0 0 1 0 0 -1 0 The last matrix has linearly dependent. 2.Review.20 20a. Since rank A = m and the system has m variables, the Gauss-Jordan Method will yield a diagonal matrix with positive elements (i.e., N will be emplty). The unique solution to Ax = 0 will be x1 = x2 =...= xm = 0. 20b. If rank A < m, the Gauss-Jordan Method will yield at least one row of 0's on the bottom (with the right hand side for each of these equations still being 0). Thus N will be non-empty and we will be in Case 3 (an infinite number of solutions). ...
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hw1 - IEOR 162 Linear Programming Fall 2004 Homework 1...

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