<s>[INST] Find $A^x$ where A is a $2\times 2$ matrix So, I need to calculate $A^x$ where 
$A=\pmatrix{4&-3\\3&-2}$ 
First of all,  $A^x = e^{x\ln A}$, so I get that 
$\ln A = \pmatrix{2&0\\-1&1}$
Then I use that 
$f(A)= T^{-1}f(J)T$
Here $J$ is the Jordan Canonical form. I got that 
$J=\pmatrix{1&1\\0&1}$
and 
$T=\pmatrix{3&1\\2&0}$, $T^{-1}=\pmatrix{0&\frac{1}{2}\\1&\frac{-3}{2}}$  
What should I do next? Any help would be greatly appreciated.
 [/INST] The result is this one
\begin{equation}
A^x=\left(\begin{array}{c}
\begin{array}{ccccc}
3x+1 & -3x\\
3x & -3x+1\\
\end{array}\end{array}\right)
\end{equation}
Now I have to exit, but as soon as I'm back I will explain it if needed...
Edit>>:
There are many ways to reach the same answer. The more plain one is that:
\begin{equation}
A^x=\left(\begin{array}{c}
\begin{array}{ccccc}
ax+b & cx+d\\
ex+f & gx+h\\
\end{array}\end{array}\right)
\end{equation}
since $A^0=I$ we have $d,f=0$ and $b,h=1$, and since $A^1=A$ we have $c=-3$ and $e=3$.
Then imposing
$$
a+b=4, g+h=-2, 
$$
we obtain $a=3$ and $g=-3$.
So for example
\begin{equation}
A^{10}=\left(\begin{array}{c}
\begin{array}{ccccc}
31 & -30\\
30 & -29\\
\end{array}\end{array}\right)
\end{equation}
</s>