### Instruction: Compute $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6.$.### Response: To solve this problem, I will try to simplify the matrix by noticing some patterns.
First, I notice that the matrix has the form $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$, where $a = \sqrt{3}$ and $b = 1$.
This reminds me of complex numbers, where $a + bi$ can be written in polar form as $r(\cos \theta + i \sin \theta)$, where $r$ is the modulus and $\theta$ is the argument of the complex number.
In this case, I can see that $a + bi = \sqrt{3} + i$, which has modulus $r = 2$ and argument $\theta = \pi/6$.
So, I can think of the matrix as a linear transformation that corresponds to multiplying by the complex number $2(\cos \pi/6 + i \sin \pi/6)$.
In other words, the matrix rotates and scales vectors in the plane by a factor of 2 and an angle of $\pi/6$.
Therefore, raising the matrix to the sixth power means applying this transformation six times, which is equivalent to multiplying by the complex number $2^6(\cos \pi + i \sin \pi)$.
This complex number is equal to $-64$, since $\cos \pi = -1$ and $\sin \pi = 0$.
So, the matrix raised to the sixth power is equivalent to multiplying by -64, which means the matrix is $-64I$, where $I$ is the identity matrix.
In conclusion, $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6 = -64 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.