Answer
See solution
Work Step by Step
Transformation matrix is:
$\left[\begin{array}{ccc}1 / 2 & -\sqrt{3} / 2 & 3+4 \sqrt{3} \\ \sqrt{3} / 2 & 1 / 2 & 4-3 \sqrt{3} \\ 0 & 0 & 1\end{array}\right]$
Notice that this matrix has a form of $\left[\begin{array}{cc}A & p \\ 0^{T} & 1\end{array}\right],$ where $A=\left[\begin{array}{cc}1 / 2 & -\sqrt{3} / 2 \\ \sqrt{3} / 2 & 1 / 2\end{array}\right]$
and $p=\left[\begin{array}{l}3+4 \sqrt{3} \\ 4-3 \sqrt{3}\end{array}\right]$
We know that the starting matrix can be written as $\left[\begin{array}{cc}I & p \\ 0^{T} & 1\end{array}\right]\left[\begin{array}{cc}A & 0 \\ 0^{T} & 1\end{array}\right]$
This is a composition of a linear transformation in $\mathbb{R}^{2}$ and a translation for a vector $p$
Matrix $A$ is a rotation of $60^{\circ}=\frac{\pi}{3}$ about the origin in $\mathbb{R}^{2}$.
So, the transformation in exercise 7 is a composition of a rotation about the origin and a translation by vector $p=\left[\begin{array}{l}3+4 \sqrt{3} \\ 4-3 \sqrt{3}\end{array}\right]$
