Q4.
Styles
\[
\mathbf{M}=\left(\begin{array}{cc}
1 & -\sqrt{3} \\
\sqrt{3} & 1
\end{array}\right)
\]
(a) Show that $\mathbf{M}$ is non-singular.
(2)
The hexagon $R$ is transformed to the hexagon $S$ by the transformation represented by the matrix $M$.
Given that the area of hexagon $R$ is 5 square units,
(b) find the area of hexagon $S$.
(1)
The matrix $\mathbf{M}$ represents an enlargement, with centre $(0,0)$ and scale factor $k$, where $k> 0$, followed by a rotation anti-clockwise through an angle $\theta$ about $(0,0)$.
(c) Find the value of $k$.
(2)
(d) Find the value of $\theta$.
(2)
(Total for question $=7$ marks)
Answer
\(\cos\theta = \frac{1}{2}\), \(\theta = \cos^{-1}\left(\frac{1}{2}\right) = \boxed{60^\circ}\)
Step 1 ：\(\text{det}(\mathbf{M}) = \left|\begin{array}{cc} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{array}\right| = 1(1) - (-\sqrt{3})(\sqrt{3}) = 1 + 3 = 4 \neq 0\)
Step 2 ：\(\text{Area of hexagon S} = \text{Area of hexagon R} \times |\text{det}(\mathbf{M})| = 5 \times 4 = \boxed{20}\)
Step 3 ：\(k = \sqrt{\left(\sqrt{3}\right)^2 + 1^2} = \sqrt{3 + 1} = \boxed{2}\)
Step 4 ：\(\cos\theta = \frac{1}{2}\), \(\theta = \cos^{-1}\left(\frac{1}{2}\right) = \boxed{60^\circ}\)
