Inverse matrices in an equation

Question

I know that $AX = b \iff X = A^{-1}b$, but how can I solve for X in this equation: $ACX = b$?

Is the answer $X = C^{-1}A^{-1}b$?

Answer 1

You are correct!

If $A$ and $C$ are both invertible matrices then there is a unique solution to the equation $AC\hat{x} = \hat{b}$ for which we can solve.

\begin{align*} AC\hat{x} &= \hat{b}\\ \implies C\hat{x} &= A^{-1}\hat{b}\\ \implies \hat{x} &= C^{-1}A^{-1}\hat{b}. \end{align*}

Or you could think of the matrix $AC$ as a single matrix (the result of computing the matrix multiplication). Since $A$ and $C$ are invertible, then $AC$ is invertible.

\begin{align*} (AC)\hat{x} &= \hat{b}\\ \implies \hat{x} &= (AC)^{-1}\hat{b}\\ \implies \hat{x} &= C^{-1}A^{-1}\hat{b}. \end{align*}

The credit really goes to Don Thousand for his comment that preceded my answer, I only chose to elaborate as an answer so the question can close.