Assume $AX = C$. How to determine which entry of $BX - D$ is non-negative?

Let $$A,B$$ be $$n \times n$$ matrices and $$C,D$$ be $$n \times 1$$ matrices. Moreover, all entries of $$A,B,C,D$$ are non-negative. Assume that there is a unique matrix $$X$$ that solves $$AX = C$$.

My goal is not to calculate $$X$$, but to determine which entry of $$BX - D$$ is non-negative. In solving my problem, I need to repeat this procedure several times.

I would like to ask if there is an efficient method (or references) to do so. Thank you so much!

• Make your life easier and call vectors "vectors" rather than "$n\times 1$ matrices" :-) – Wolfgang Bangerth May 29 '20 at 21:57
• Hi @WolfgangBangerth, I've just posted the refinement of this question here. Because this is an essential part of my thesis, I hope that you have a look at it and lend me some suggestion ^_^ – LE Anh Dung May 29 '20 at 22:00
• I have no idea about what can or can not be shown in this area. – Wolfgang Bangerth May 30 '20 at 15:07
• Thank you for your response @WolfgangBangerth ^^ – LE Anh Dung May 30 '20 at 15:08