<s>[INST] Given matrices $A$ and $B$, solve for $Z$ in the following equation: $A^{-4}BZ^{3}A^{5}B^{2}A=-3A^{-6}$ $
A=
\begin{pmatrix}
2 & -4 & 3\\
4 & -1 & 2\\
-1 & 1 & -1\\
\end{pmatrix}
$
$
B=
\begin{pmatrix}
3 & -4 & 5\\
1 & -2 & 3\\
-2 & 3 & -4\\
\end{pmatrix}
$
Solve Z in the following equation: $A^{-4}BZ^{3}A^{5}B^{2}A=-3A^{-6}$.
I tried this way:
$A^{-4}BZ^{3}A^{5}B^{2}A=-3A^{-6}$
$\Leftrightarrow BZ^{3}A^{5}B^{2}A=-3A^{-2}$
$\Leftrightarrow Z^{3}A^{5}B^{2}A=-3B^{-1}A^{-2}$
$\Leftrightarrow Z^{3}A^{5}B^{2}=-3B^{-1}A^{-3}$
$\Leftrightarrow Z^{3}A^{5}=-3B^{-1}A^{-3}B^{-2}$
$\Leftrightarrow Z^{3}=-3B^{-1}A^{-3}B^{-2}A^{-5}$
But in  the final equation, $B^{-1}$ does not exist so how can I solve this?
 [/INST] I think this problem wasn't meant to be solved. Whoever set this question probably wanted you to point out that the equation is insolvable, because the LHS is singular (as $B$ is singular) but the RHS is not.
</s>