How to solve higher power of matrix?

Question

Given a matrix $A=\begin{pmatrix} 1 & 0 \\ {-1} & 1\end{pmatrix}$ We are asked to show $A^2 -2A +I = 0$ and hence find $A^{50}$. I've proved that $A^2 -2A +I = 0$ but couldn't figure out how to proceed for the next part.

Answer 1

First we try to find some powers of $A$:

$$A^2=2A-I$$

$$A^3=A^2.A=2A^2-A=4A-2I-A=3A-2I$$

$$A^4=A.A^3=3A^2-2A=4A-3I$$

So we guess that

$$A^n=nA-(n-1)I$$

we try to prove this using induction. Assume this is true for $n$. Therefore

$$A^{n+1}=A.A^n=nA^2-(n-1)A=2nA-nI-(n-1)A=(n+1)A-nI$$

so we could prove that $A^n=nA-(n-1)I$ then we can say:

$$A^{50}=50A-49I=\begin{pmatrix} 1 & 0\\ -50 &1 \ \end{pmatrix}$$