Observe the real matrices
$$ B = \begin{pmatrix} 4 & 2 \\ -6 & -3 \\ \end{pmatrix}, \ C = \begin{pmatrix} -3 & -2 \\ 6 & 4 \\ \end{pmatrix} $$ in $Mat_2(\mathbb{R})$. Let $\alpha\in\mathbb{R}$ be a real number and let: $$ A =\alpha \cdot B+C\in Mat_2(\mathbb{R})$$
Find the real values of $\alpha$ where A is invertible and prove by induction in $n$ that for all integers $n\ge 1$ it is true that $$ A^n = \alpha{^n}\cdot B+C$$
I know that A is invertible if $det(A) \ne 0 $, so in order to find the real values $\alpha$ would it be possible to add the two matrices $\alpha \cdot B$ and $C$ together and find the determinant?
I am unsure how to do the proof by induction part of the question.
Note that $B^2=B$, that $C^2=C$, and that $BC=\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right)$. Therefore,$$(\alpha B+C)^n=\alpha^nB^n+C^n=\alpha^nB+C.$$On the other hand, an easy computation show that $\det\left(\alpha B+C\right)=\alpha$.