Showing $A^2 - 3A^{147} + 2I = \begin{pmatrix} 9 & -9\\ 3 & -3\end{pmatrix}$.

Question

Let $ A = $$ \begin{pmatrix} 9 & -9\\ 3 & -3\end{pmatrix} $$ $. Show $$A^2 - 3A^{147} + 2I = \begin{pmatrix} 9 & -9\\ 3 & -3\end{pmatrix}$$

Answer 1

Note that you can write $A$ as $A=3B$ where $$B=\begin{pmatrix} 3 & -3\\ 1 & -1\end{pmatrix}.$$

Now, you can compute easily that $B^2=2B$ and so by induction you'll have $B^n=2^{n-1}B$. Therefore $A^n=(3B)^n=3^n\times2^{n-1}B=6^{n-1}A$. I am sure from this point you can find what you wanted, can't you?

Answer 2

Once you compute the characteristic polynomial you get $\lambda^2-6\lambda=0$. Thus $A^2-6A=O$. Therefore $A^{n}=6^{n-1}A$. See if you can proceed from here.