<s>[INST] Find Matrix $A^{50}$? Find the matrix $A^{50}$ given
$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$
I was practicing some questions for my exam and I found questions of this form in a previous year's paper.
I don't know how to do such questions.
Please assist over this question.
Thank You
 [/INST] And yet another approach: Splitting $A$ into a diagonal matrix $D$ and a pertubation matrix $P$.
We look at the first powers of $P$:
$$
P^0 = 
\left(
\begin{matrix}
1 & 0\\
0 & 1 
\end{matrix}
\right)
\quad 
P^1 = 
\left(
\begin{matrix}
0 & -1 \\
0 & 0 
\end{matrix}
\right)
\quad
P^2 = 
\left(
\begin{matrix}
0 & 0 \\
0 & 0 
\end{matrix}
\right)
$$
and note that quadratic powers and higher of $P$ vanish.
Then
$$
DP = 2 P \quad PD = P
$$
Thus
\begin{align}
A^n 
&= (D + P)^n \\
&= (D^2 + DP + PD + P^2) (D+P)^{n-2} \\
&= (D^2 + (1+2^1)P) (D+P)^{n-2} \\
&= (D^3 + (1+2^1)PD + D^2P + 3P^2)(D+P)^{n-3} \\
&= (D^3 + (1+2^1)P + 2^2 P)(D+P)^{n-3} \\
&= (D^3 + (1+2^1+2^2) P)(D+P)^{n-3} \\
&= D^n + (1+2^1+\cdots+2^{n-1})P \\
&= D^n + (2^n-1)P
\end{align}
This gives
$$
A^n
=
\left(
\begin{matrix}
2^n  & 0 \\
0 & 1^n
\end{matrix}
\right)
+
\left(
\begin{matrix}
0 & 1-2^n \\
0 & 0 
\end{matrix}
\right)
=
\left(
\begin{matrix}
2^n & 1-2^n \\
0 & 1
\end{matrix}
\right)
$$
</s>