let A be a $2\times 2$ matrix . Then the smallest number $n\in \mathbb N$ such that $A^n=I$ is

Question

let A be a $2\times 2$ 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 $A^n=I$ is

(a). 3

(b). 9

(c). 18

(d). 27

My work

I checked for the order for 3. But, I could know that I am wrong in this way. And I tried to find that "Is it nth roots of some complex numbers" but in that way also I could not achieve. How can I proceed?

If I give this matrix what is the answer $$\begin{pmatrix}1&0&0\\ 0& \sin \frac \pi {18} & -\sin \frac {4\pi} {9}\\ 0&\sin \frac {4\pi} {9}&\sin \frac \pi {18}\end{pmatrix}$$

Answer 1

Here's a hint (assuming you are indeed missing a minus sign): $\sin \frac{\pi}{18} = \cos \frac{4\pi}{9}$.

Answer 2

Using the identity

$$\sin x=\cos\left(x-\frac\pi2\right)$$ we see that the given matrix is a matrix of rotation of angle $\frac{4\pi }{9}$ hence the answer is $9$.