<s>[INST] 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?
 [/INST] 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!
</s>