Let $A$ be the $2 \times 2$ matrix $\begin{pmatrix} \sin\frac{\pi}{18}&-\sin \frac{4\pi}{9} \\ \sin \frac{4\pi}{9}&\sin \frac {\pi}{18} \end{pmatrix}$. Then the smallest number $n \in \mathbb{N}$ such that $A^{n}=1$ is.
By solving Characteristic Equation $\lambda = sin(\pi/18) +\mathbf{i} sin(4\pi/9 )$
$\lambda = sin(\pi/18) -\mathbf{i} sin(4\pi/9 )$
By cayley hamilton theorem $\lambda ^n=1$ and solving for n(minimum) get n= 9 so Option B is correct