Find $A^x$ where A is a $2\times 2$ matrix So, I need to calculate $A^x$ where 

$A=\pmatrix{4&amp;-3\\3&amp;-2}$ 

First of all,  $A^x = e^{x\ln A}$, so I get that 

$\ln A = \pmatrix{2&amp;0\\-1&amp;1}$

Then I use that 

$f(A)= T^{-1}f(J)T$

Here $J$ is the Jordan Canonical form. I got that 

$J=\pmatrix{1&amp;1\\0&amp;1}$

and 

$T=\pmatrix{3&amp;1\\2&amp;0}$, $T^{-1}=\pmatrix{0&amp;\frac{1}{2}\\1&amp;\frac{-3}{2}}$  

What should I do next? Any help would be greatly appreciated.
The result is this one
\begin{equation}
A^x=\left(\begin{array}{c}
\begin{array}{ccccc}
3x+1 &amp; -3x\\
3x &amp; -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 &amp; cx+d\\
ex+f &amp; 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 &amp; -30\\
30 &amp; -29\\
\end{array}\end{array}\right)
\end{equation}
Hint: using the Mercator series  $$\ln(1+x) = x - \frac{x^2}2 + \frac{x^3}3 + \ldots$$

You can write $$\ln(J)=\ln(I+X) = X$$ where $X=\pmatrix{0&amp;1\\0&amp;0}$ so $X^2$ and higher terms are zero.