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$$\begin{bmatrix}3 & -2\\-1 & -3\end{bmatrix}X + \begin{bmatrix}1 & 5\\0 & 7\end{bmatrix} = \begin{bmatrix}-5 & 8\\-1 & 3\end{bmatrix}X$$

Not sure how I am supposed to go at this ? Do I move the + to the other side first ? Can someone let me know how to tackle this, thanks.

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2 Answers

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Taking the $x$ to LHS, we get $$\begin{pmatrix}3+5&-2-8\\-1+1&-3-3\end{pmatrix}X=-\begin{pmatrix}1&5\\0&7\end{pmatrix}$$ $$\Rightarrow \begin{pmatrix}8&-10\\0&-6\end{pmatrix}X=-\begin{pmatrix}1&5\\0&7\end{pmatrix}$$

Now note that $\begin{pmatrix}8&-10\\0&-6\end{pmatrix}$ is invertible and let A be its inverse. (To find the inverse you may look here). So premultiplying with $A$, we get $$X=-A\begin{pmatrix}1&5\\0&7\end{pmatrix}$$

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$$AX+B=CX\implies X=(C-A)^{-1}\cdot B\quad (\text{provided $C-A$ is invertible})$$

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