CO227 Lecture Notes - Lecture 11: Invertible Matrix, Identity Matrix
30 May 2017
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Max z(x) = [-2 1 0 3 0 0 ]x s. t. We found a feasible solution because column 5 3 6 of a a gives the identity matrix. We were able to improve the solution since cb= Definition an lp in sef, max {ctx|ax=b, x>=0}, is in canonical form for the basis b if. Ab = i (for some permutation of b) Convert the previous example to canonical form for the basis b = {3,4,5}. Since b is a basis, a b is an invertible matrix. Recall: we found certificate of optimal by adding zero to the objective function by multiplying the equality constraints by some value y1,y2,y3 for equations 1,2,3. To satisfy canonical form, fro basis b = {3,4,5}: Note: although this lp is not in sef (objective function has a +3) we can subtract 3 from the objective function, solve the lp, then add it to the objective function afterwards.
