Computing matrices to a power of $6$ Compute
$\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6.$
How would I solve this question. I found out that it's square would be $\begin{pmatrix} 2 & -2\sqrt{3} \\ 2\sqrt{3} & 2 \end{pmatrix}.$ What next? Is there any easier way? 

Set
$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag{1}$
we seek $A^6$, where
$A = \begin{bmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{bmatrix} = \sqrt{3} I + J; \tag{2}$
we note that we may write $A$ as
$A = 2(\dfrac{\sqrt{3}}{2} I + \dfrac{1}{2}J); \tag{3}$
noting further that
$\dfrac{\sqrt{3}}{2} = \cos \dfrac{\pi}{6}, \tag{4}$
$\dfrac{1}{2} = \sin \dfrac{\pi}{6}, \tag{5}$
we see we may write
$A = 2(\cos \dfrac{\pi}{6} I + \sin \dfrac{\pi}{6} J); \tag{6}$
next, we observe that
$J^2 = - I; \tag{7}$
from (7) we may deduce that matrices of the general form $(\cos \theta) I + (\sin \theta) J$ satisfy a variant of de Moivre's formula;
$((\cos \theta) I + (\sin \theta) J)^n = (\cos n\theta) I + (\sin n\theta) J; \tag{8}$
(8) is, like its scalar equivalent, easy to prove by induction; since the proof is virtually identical to that for the scalar version, we won't repeat it here, but see
https://en.m.wikipedia.org/wiki/De_Moivre%27s_formula.  With the aid of (7), we compute
$A^6 = 2^6(\cos \dfrac{\pi}{6} I +  \sin \dfrac{\pi}{6} J)^6 = 64(\cos \pi I + \sin \pi J) = -64I. \tag{9}$
