1. **Problem statement:** Find the transformation matrix $M$ for (a) a reflection in the line $y=\frac{\sqrt{3}}{3}x$ followed by a clockwise rotation of $30^\circ$ about the origin. Then (b) find the angle $\theta$ such that $M$ corresponds to a single reflection in the line $y=\tan\theta \cdot x$. 2. **Reflection matrix formula:** For reflection about a line through the origin with angle $\alpha$ to the x-axis, the matrix is: $$R_\text{reflect}(\alpha) = \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{bmatrix}$$ 3. **Rotation matrix formula:** For a clockwise rotation by angle $\beta$, the matrix is: $$R_\text{rotate}(\beta) = \begin{bmatrix} \cos \beta & \sin \beta \\ -\sin \beta & \cos \beta \end{bmatrix}$$ 4. **Calculate reflection matrix:** Given line $y=\frac{\sqrt{3}}{3}x$, slope $m=\frac{\sqrt{3}}{3}$, so angle $\alpha=\arctan\left(\frac{\sqrt{3}}{3}\right)=30^\circ$. Calculate $2\alpha=60^\circ$. $$R_\text{reflect}(30^\circ) = \begin{bmatrix} \cos 60^\circ & \sin 60^\circ \\ \sin 60^\circ & -\cos 60^\circ \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix}$$ 5. **Calculate rotation matrix:** Clockwise rotation by $30^\circ$: $$R_\text{rotate}(30^\circ) = \begin{bmatrix} \cos 30^\circ & \sin 30^\circ \\ -\sin 30^\circ & \cos 30^\circ \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$$ 6. **Find combined transformation matrix $M$:** $$M = R_\text{rotate}(30^\circ) \times R_\text{reflect}(30^\circ)$$ Calculate: $$M = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \left(-\frac{1}{2}\right) \\ -\frac{1}{2} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} & -\frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \left(-\frac{1}{2}\right) \end{bmatrix}$$ Simplify each element: - Top-left: $\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$ - Top-right: $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ - Bottom-left: $-\frac{1}{4} + \frac{3}{4} = \frac{1}{2}$ - Bottom-right: $-\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$ So: $$M = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{\sqrt{3}}{2} \end{bmatrix}$$ 7. **Interpret $M$ as a reflection matrix:** Recall reflection matrix about line $y=\tan\theta x$ is: $$R_\text{reflect}(\theta) = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{bmatrix}$$ Compare with $M$: $$\cos 2\theta = \frac{\sqrt{3}}{2}, \quad \sin 2\theta = \frac{1}{2}$$ 8. **Find $2\theta$:** $$\cos 2\theta = \frac{\sqrt{3}}{2}, \quad \sin 2\theta = \frac{1}{2}$$ This corresponds to: $$2\theta = 30^\circ$$ 9. **Find $\theta$:** $$\theta = \frac{30^\circ}{2} = 15^\circ$$ **Final answers:** (a) $$M = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{\sqrt{3}}{2} \end{bmatrix}$$ (b) $$\theta = 15^\circ$$