Let $M=XAX^{-1}$ where $ X= \begin{bmatrix} 1 & 2 \\ 2 & 3 \\ \end{bmatrix}$, $A= \begin{bmatrix} 1 & 0 \\ 2 & 1 \\ \end{bmatrix}$
Find $M^{n},n\in\mathbb{N}$
Attempt:
Inductive method:
$M^{1}= \begin{bmatrix} -11 & 8\\ -18 & 13 \\ \end{bmatrix}$
$M^{2}= \begin{bmatrix} -23 & -64\\ 144 & -135 \\ \end{bmatrix}$
$M^{3}= \begin{bmatrix} 1405 & -376\\ 846 & 747 \\ \end{bmatrix}$
In order to find $M^{n}$, this method requires finding the closed form of sequences for all entries of a matrix $M$ (first sequence would be $-11,-23,1405,...$).
What is the simpler method to find $M^{n}$?
Hint: Notice that $$ M^2=(XAX^{-1})(XAX^{-1})=XAX^{-1}XAX^{-1}=XA^2X^{-1}. $$ You can similarly show that $M^n=XA^nX^{-1}$.
So, you only need to take powers of $A$. And, perhaps, that will be a bit easier to spot patterns for.