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

Question

I posted this question on https://scicomp.stackexchange.com, but seems to receive no attention. As long as I get answer in one of them, I will inform in the other.

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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!

Answer 1

I post @Marc Dinh's comment to remove this question from unanswered list:

Since $AX=C$ has a unique solution, $A$ is necessarily invertible so that $X = A^{-1}C$.