0
$\begingroup$

I have these two equation systems. $$ \ \left\{ \begin{array} {A}xA\leq b \\ x \geq 0 \end{array} \right. \quad\text{and}\quad \ \left\{ \begin{array} {A}Ay\geq 0 \\ by < 0 \\ y \geq0 \end{array} \right. $$

How to prove that when first equation has solution the second one does not have and when the second one has the first one does not have. But not simultaneously.

Thanks for your time!

$\endgroup$
0
$\begingroup$

Hint:

Suppose on the contrary that both of them have solutions.

We have $(xA) \le b$, since $y \ge 0$,

We ahve $$(xA)y \le by$$

Use associative property of matrix product and the other constraints to find a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy