If $A=\pmatrix{1 &0\\-1&1}$, show that $A^2-2A+I_2=0$. Hence find $A^{50}$ 
If $$A=\pmatrix{1 &0\\-1&1},$$ show that $$A^2-2A+I_2=0,$$ where $I_{2}$ is the $2x2$ Identity matrix. Hence find $A^{50}$.  

We have $$A^2-2A+I_2=A(A-2I_3)+I_=\pmatrix{1 &0\\-1&1}\pmatrix{-1 &0\\-1&-1}+I_2  
=-I_2+I_2=0.$$
How can I show the second part?

The characteristic polynomial of $A$ is
$$\chi_A(t) = \text{det}(A-t\Bbb{1})=t^2-2t+1$$
By the theorem of Caliley Hamilton, you get that $A^2-2A+\Bbb{1}=0$, where $\Bbb{1}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $0$ is the zero matrix.
Now we 

Claim $$A^n = \begin{pmatrix}1&0\\-n&1\end{pmatrix}$$

Proof
We procede by induction:
If $n=1$ we have $A = \begin{pmatrix}1&0\\-1&0\end{pmatrix}$, that is correct.
Assume that the claim is true for $n-1$. We want to prove that it holds also for $n$
\begin{align*}
A^n &= A^{n-1}\cdot A\\
&=\begin{pmatrix}1&0\\-(n-1)&1\end{pmatrix}\cdot \begin{pmatrix}1&0\\-1&1\end{pmatrix}\\&=\begin{pmatrix}1&0\\-n+1-1&1\end{pmatrix}=\begin{pmatrix}1&0\\-n&1\end{pmatrix}
\end{align*}
Which proves the claim. Now you should be able to evaluate $A^{50}$.
