Find all the values of $\alpha \in \mathbb{R}$ for which the system $BA^2x = BAx$ has a unique solution. $$ A=\begin{pmatrix} \alpha-2 & 0 & 0 \\\ 2 & 0 & \alpha-2 \\\ 3 & 1 & 1 \end{pmatrix} $$
$$ B \in \mathbb{R}^{3x3} \ \text{a matrix such that} \ det(B)=-2$$
I found that the $det(A)=-(\alpha-2)^2$, then $\alpha$ must be different from 2 so that $det(A) \neq 0$. But I think that is not enough to show that the system has a unique solution.
Since $\det(B)\ne0$, $BA^2x=BAx\iff A^2x=Ax$. Besides, $A^2x=Ax\iff\left(A^2-A\right)x=0$. But$$A^2-A=\begin{bmatrix}\alpha ^2-5\alpha+6&0&0\\5\alpha-12&\alpha-2&0\\3\alpha-4&0&\alpha-2\end{bmatrix},$$whose determinant is $\left(\alpha ^2-5\alpha+6\right)(\alpha-2)^2$. Therefore,\begin{align}\det\left(A^2-A\right)=0&\iff\alpha ^2-5\alpha+6=0\vee\alpha-2=0\\&\iff\alpha=2\vee\alpha=3.\end{align}So, the system $BA^2x=BAx$ has a unique solution if and only if $\alpha\in\Bbb R\setminus\{2,3\}$.