<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] This is for the updated version of question where $A = \begin{bmatrix}2 & -1\\0 & 1\end{bmatrix}$. For the original version of $A$, the derivation is similar.
The characteristic polynomial for matrix $A$ is
$$\chi_A(\lambda) = \det\begin{bmatrix}\lambda-2 & 1\\0 & \lambda - 1\end{bmatrix}
= (\lambda - 2)(\lambda - 1) = \lambda^2 - 3\lambda + 2$$
By Cayley Hamilton theorem, we have
$$A^2 - 3A + 2I_2 = 0$$
If we divide the polynomial $x^{50}$ by $x^2 - 3x + 2$ with long division, 
we know there are polynomial $p(x)$ and coefficients $\alpha, \beta$ such that
$$x^{50} = p(x)(x^2 - 3x + 2) + \alpha x + \beta$$
To determine $\alpha, \beta$, substitute $x$ by $1$ and $2$ in above expression. We get
$$\begin{cases}
1 &= \alpha + \beta\\
2^{50} &= 2\alpha + \beta
\end{cases}
\quad\implies\quad
\begin{cases}
\alpha &= 2^{50} - 1\\
\beta  &= 2 - 2^{50}
\end{cases}$$
Form this, we get
$$\require{cancel}
\begin{align}
A^{50} &= p(A)\color{red}{\cancelto{0}{\color{gray}{(A^2 - 3A + 2I_2)}}} + \alpha A + \beta = \alpha A + \beta\\
&= (2^{50}-1)\begin{bmatrix}2 & -1\\0 & 1\end{bmatrix}
 +(2 - 2^{50})\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}
= \begin{bmatrix}2^{50} & 1 - 2^{50}\\0 & 1\end{bmatrix}
\end{align}
$$
</s>