Which of the following options specifies the default matrix of the linear image $T$ corresponding to counterclockwise rotation $\pi/6$ radians followed by mirroring in the $y$-axis?
Options:
$\dfrac12\begin{bmatrix}-\sqrt3&1\\1&\sqrt3\end{bmatrix}$
$\dfrac12\begin{bmatrix}-\sqrt3&-1\\-1&\sqrt3\end{bmatrix}$
$\dfrac12\begin{bmatrix}1&\sqrt3\\\sqrt3&-1\end{bmatrix}$
$\dfrac12\begin{bmatrix}1&-\sqrt3\\-\sqrt3&-1\end{bmatrix}$
Counter-clockwise rotation by angle $\theta$ in the $xy$-plane is represented by the rotation matrix $$ \left[ \begin{array}{cc} \cos{\theta} & -\sin{\theta}\\ \sin{\theta} & \cos{\theta} \end{array} \right] $$ In other words, if you have a point or vector, whose (end) point is represented by a column vector, multiplying by this matrix from the left rotates the point or vector around the origin.
The problem definition also mentions mirroring about the line $y=0$. This corresponds to just flipping the sign of the $x$-coordinate, so the representative matrix of this operation is $$ \left[ \begin{array}{cc} -1 & 0\\ 0 & 1 \end{array} \right] $$ Putting this together, and plugging in $\theta=\frac{\pi}{6}$ results in $$ \left[ \begin{array}{cc} -1 & 0\\ 0 & 1 \end{array} \right] \left[ \begin{array}{cc} \cos{\theta} & -\sin{\theta}\\ \sin{\theta} & \cos{\theta} \end{array} \right] = \left[ \begin{array}{cc} -\cos{\theta} & \sin{\theta}\\ \sin{\theta} & \cos{\theta} \end{array} \right] = \left[ \begin{array}{cc} -\frac{\sqrt{3}}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array} \right] $$ Therefore, the correct answer is the first option on the list.