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 $A=
        \begin{bmatrix}
        a & b  \\
        c & d
        \end{bmatrix}$?
 A: 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}
A: One can of course prove this directly by substituting and computing the entries of the $2 \times 2$ matrix on the left-hand side, but here's an outline for a solution that reduces the amount of actual computation one needs to do.
Hint


*

*Prove the claim for diagonal matrices.

*Prove the claim for matrices similar to diagonal matrices. (Recall that the trace and determinant of a matrix, which up to sign are the linear and constant terms of the polynomial in $A$, are invariant under similarity.)

*Use the fact that the set of diagonalizable matrices (those similar to diagonal matrices) is dense in the set of all $2 \times 2$ matrices.


Remark This is anyway a special case of a more general fact: That any matrix $A$ satisfies $p_A(A)$, where $p_A$ is the characteristic polynomial $p_A(t) := \det(t I - A)$ of $A$.
A: You can really just prove this by brute force, and its not that complicated.
$A=
        \begin{bmatrix}
        a & b  \\
        c & d
        \end{bmatrix}$
so 
$A^2=
        \begin{bmatrix}
        a^2-bc & ab-bd  \\
        ac-cd & bc-d^2
        \end{bmatrix}$
$(a+d)A=
        \begin{bmatrix}
        (a+d)a & (a+d)b  \\
        (a+d)c & (a+d)d
        \end{bmatrix}$
$(ad-bc)I=
        \begin{bmatrix}
        (ad-bc) & 0  \\
        0 & (ad-bc)
        \end{bmatrix}$
just add them together and you get 
$$A^2-(a+d)A+(ad-bc)I=O$$
A: Note that 
$$
A^2= \begin{bmatrix}
a^2+bc&ab+bd\\
ac+cd&cb+d^2
\end{bmatrix}
$$
and
$$
(tr A)A=(a+d)
 \begin{bmatrix}
a&b\\
c&d
\end{bmatrix}=
 \begin{bmatrix}
a^2+ad&ab+bd\\
ac+cd&ad+d^2
\end{bmatrix}
$$
so
$$
(tr A)A-A^2=
 \begin{bmatrix}
ad-bc&0\\
0&ad-bc
\end{bmatrix}
=
(det A) I
$$
Also note that 
$$
t^2-(tr A)t +det A=0
$$
is the characteristic equation of the $2\times 2$ matrix $A$ and, by Cayley-Hamilton theorem a matrix satisfies his characteristic equation.
