<s>[INST] 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?
 [/INST] Since $(A-I_2)^2 = 0$, by binomial theorem, one has
$$\begin{align}A^{50}
&=  (I_2 + (A - I_2))^{50}\\
&= I_2 + \binom{50}{1} (A-I_2)^{1} + \binom{50}{2}(A-I_2)^2 + \binom{50}{3}(A-I_2)^3 +\cdots\\
&= I_2 + 50 (A-I_2) + 0 + 0 + \cdots\\
&= \begin{pmatrix}1 & 0\\-50 & 1\end{pmatrix}
\end{align}
$$
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