Another proof of complementary slackness consider the normal-form primal lp min primal: s. t. w = ctx ax> b x> 0 (objective function) (constraints #1 through #m) (sign restrictions #1 though #n) and its normal-form dual dual: max z = btp (objective function) s. t. alp 0 (sign restrictions #1 though #m) which, when placed in standard forms become min s. t. primal: w= ctx ax - ie = b x, e > 0 (objective function) (constraints #1 through #m) (sign restrictions #1 though #n) and max dual: s. t. z = btp atp + is = c p, s> 0 (objective function) (constraints #1 through #n) (sign restrictions #1 though #m) the purpose of this problem is for you to prove complementary slack- ness using a different approach than that taken in the notes and in class, and so using the above standard forms and without using the results of theorem #7, prove that at optimal solutions for both the primal and the dual, we must have xisi = 0 for all i = 1, 2, 3, n and pje; = 0 for all j = 1, 2, 3, m. hint: this should only take a little bit of matrix algebra.