How can I prove the following matrix equality?

Question

Let $n\in\mathbb{N}$ be arbitrary. Why is the following equality true? $$ \left(\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix} + 3 \begin{pmatrix}4 & 6\\-2 & -3\end{pmatrix}\right)^n = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix} + (4^n-1)\begin{pmatrix}4 & 6\\-2 & -3\end{pmatrix} . $$

Also, it might help that:

$$\begin{pmatrix}4 & 6\\\ -2 & -3\end{pmatrix}^n=\begin{pmatrix}4 & 6\\\ -2 & -3\end{pmatrix}$$

Thanks in advance!

Answer 1

Just use binomial formula and the fact that $A^2 = A \Rightarrow A^k = A$

$$(I + 3A)^n = \sum_{k=0}^n \binom{n}{k}3^kA^k =I + A\sum_{k=1}^n \binom{n}{k}3^k = I + A ((1+3)^n-1) = I +(4^n-1)A$$

Note that you can apply the binomial formula since $IA = AI$.