How to use powers on matrices In the questions compute $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6$ and $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99}$, how would you solve these?
 A: Since first part is answered by Doug M , for the second bit we can approach by the method of induction.
We consider this matrix ,$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{n}$
Lets check for n=2,
$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$.
Lets check for n=3,
$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$.
I think we got a pattern!
So, our hypthesis is $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{n} = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ .To prove our hypothesis we use first principle of mathematical induction.
Let us assume that this form is true for $n = k$ that is multiplying $k$ times which gives us $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k} =\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$
Now if we prove it for $n=k+1$ then it's true for all $n \geq 1$ 
So, consider $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$
Now $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ from our assumption, so $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & (k+1) \\ 0 & 1 \end{pmatrix}$ from our first case.
Hence this holds for any $n \geq 1$,so as a particular case of yours,for $n=99$,this case also holds that is $\begin{pmatrix} 1 & 1 \\ 0 & 1 
\end{pmatrix}^{99} = \begin{pmatrix} 1 & 99 \\ 0 & 1 \end{pmatrix}$.
Hope this helps!
A: $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}= 2 \begin{pmatrix} \cos \frac {\pi}{6} & -\sin\frac {\pi}{6}  \\ \sin\frac {\pi}{6} & \cos \frac {\pi}{6} \end{pmatrix}$
$\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6= 2^6 \begin{pmatrix} \cos \frac {\pi}{6} & -\sin\frac {\pi}{6}  \\ \sin\frac {\pi}{6} & \cos \frac {\pi}{6} \end{pmatrix}^6$
It is worth the exercise to see what happens when you multiply matrices that can be put into this form.
A: The complex numbers are isomorphic to the set of matrices of the form
$$
a+bi \sim
\begin{pmatrix} a & -b \\ b & a \end{pmatrix}
=a I + b J,
\text{ where }
J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},
J^2=-I
$$
Your first matrix corresponds to $z=\sqrt3 +i$.
Now $\dfrac{iz}{2}=\dfrac{-1+\sqrt3}{2}$ is a third root of unity and so $\left(\dfrac{iz}{2}\right)^6=1$.
Therefore $z^6=-64$ and the corresponding matrix is $-64I$.
The second matrix can be written $A=I+N$ where $N^2=0$.
Therefore, $A^k=I+kN$ by the binomial theorem.
