<s>[INST] Show that a matrix $A=\pmatrix{a&b\\c&d}$ satisfies $A^2-(a+d)A+(ad-bc)I=O$ 
Let $A=
        \begin{bmatrix}
        a & b  \\
        c & d
        \end{bmatrix}
,a,b,c,d\in\mathbb{R}$ . Prove that every matrix $A$ satisfies the condition $$A^2-(a+d)A+(ad-bc)I=O .$$ Find $$        \begin{bmatrix}
        a & b  \\
        c & -a
        \end{bmatrix}^n .$$

For $a=1,b=2,c=3,d=4$ equality $A^2-(a+d)A+(ad-bc)I=O$ holds.
How to prove this equality for every $A=
        \begin{bmatrix}
        a & b  \\
        c & d
        \end{bmatrix}$?
 [/INST] The short answer is: The equality can be proofen by simply inserting A, performing a matrix multiplication and adding up the matrices. For the second part of your question, one may find a recursion formula.
The long answer including the calculation follows here:
Assuming $I=\begin{bmatrix}1& 0 \\ 0 &1\end{bmatrix}$ and $O = \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix}$, the equality can be shown by plugging  in $A$.
\begin{align}
A^2 &- (a+d)A&+&(ad-bc)I &=& O\\
\rightarrow \begin{bmatrix}a& b \\ c &d\end{bmatrix}\begin{bmatrix}a& b \\ c &d\end{bmatrix}& 
-\begin{bmatrix}(a+d)a& (a+d)b \\ (a+d)c &(a+d)d\end{bmatrix}&+&\begin{bmatrix}ad-bc& 0 \\ 0 &ad-bc\end{bmatrix}&=& \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix}
\end{align}
Now we still need to do the matrix multiplication
\begin{align}
\begin{bmatrix}a& b \\ c &d\end{bmatrix}\begin{bmatrix}a& b \\ c &d\end{bmatrix} = \begin{bmatrix}a^2+bc& ab+bd \\ ac+dc &bc+d^2\end{bmatrix}
\end{align}
Now, we may add the matrices component-wise:
\begin{align}
\begin{bmatrix}a^2+bc& ab+bd \\ ac+dc &bc+d^2\end{bmatrix}& 
-\begin{bmatrix}(a+d)a& (a+d)b \\ (a+d)c &(a+d)d\end{bmatrix}&+&\begin{bmatrix}ad-bc& 0 \\ 0 &ad-bc\end{bmatrix}&=& \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix}
\end{align}
\begin{align}
\rightarrow \begin{bmatrix}a^2+bc -(a+d)a+ad-bc& ab+bd -(a+d)b \\ ac+dc -(a+d)c&bc+d^2-(a+d)d+ad-bc\end{bmatrix}= \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix}
\end{align}
After rearranging the terms, we see that the quation holds.
\begin{align}
\rightarrow \begin{bmatrix}a^2-a^2 -ad+ad+bc-bc& ab-ab -ab-db \\ ac+dc -ac-cd&bc-bc+d^2-d^2+ad-ad\end{bmatrix}= \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix}
\end{align}
The second part is about how this helps evaluating
$ \begin{bmatrix}a& b \\ c &-a\end{bmatrix}^n=(\tilde A)^n$. Here we defined $\tilde A= \begin{bmatrix}a& b \\ c &-a\end{bmatrix}$.
Using our equation we just proofed with d=-a, we get:
\begin{align}
(\tilde A)^2-(a+(-a))\tilde A+(-a^2-bc)I=&O\\
\rightarrow (\tilde A)^2+(-a^2-bc)I=&O
\end{align}
Multiplying this equation with  $(\tilde A)^{n-2}$ gives the recursion formula
\begin{align}
(\tilde A)^n=(a^2+bc)(\tilde A)^{n-2}
\end{align}
This is useful since you may plug the right side back in yielding
\begin{align}
(\tilde A)^n=\begin{cases}(a^2+bc)^{n/2}I&\text{if  } n \text{ even} \\
(a^2+bc)^{(n-1)/2}\tilde A &\text{if  } n \text{ odd}
\end{cases}
\end{align}
</s>