<s>[INST] Matrix inversion question with different dimensions 
Let $A=\left[\begin{matrix}0&0&0\\-1&0&0\\0&-1&0\end{matrix}\right]$ and $B=\left[\begin{matrix}4&1\\5&2\\6&3\end{matrix}\right]$ verify the matrix equation $X=AX+B$

*

*What are the dimensions of $X$?


*Find the matrix $X$. Is the matrix you found the unique solution of the equation?

Can you please discuss how to answer this type of question, I already know that $X$ has to be $3\times 2$ and this is clear through the equation given $X = AX + B$
Please discuss the second part of the question about how we could determine Matrix $X$
 [/INST] We know $X=\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]$ so we can say
\begin{align}X&=AX+B\\
\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]&=\left[\begin{matrix}0&0&0\\-1&0&0\\0&-1&0\end{matrix}\right]\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]+\left[\begin{matrix}4&1\\5&2\\6&3\end{matrix}\right]\\
\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]&=\left[\begin{matrix}0&0\\-a&-b\\-c&-d\end{matrix}\right]+\left[\begin{matrix}4&1\\5&2\\6&3\end{matrix}\right]\\
\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]&=\left[\begin{matrix}4&1\\5-a&2-b\\6-c&3-d\end{matrix}\right]
\end{align}
Therefore we have the following equalities
\begin{align}a&=4\\
b&=1\\
c&=5-a=1\\
d&=2-b=1\\
e&=6-c=5\\
f&=3-d=2\end{align}
and thus $$X=\left[\begin{matrix}4&1\\1&1\\5&2\end{matrix}\right]$$
</s>