Find smallest $n \in \mathbb{N}$ s.t $A^n=I$ Let $A$ be $2 \times 2$ matrix: 
$$
\left(  
\begin{matrix} 
\sin\frac{\pi}{18} \\ \sin\frac{4\pi}{9} 
\end{matrix} 
\begin{matrix} 
-\sin\frac{4\pi}{9} \\ \sin\frac{\pi}{18} 
\end{matrix} \right) $$
Then find smallest  $n \in \mathbb{N}$ s.t $A^n=I$.
Looking at this question I seems I am missing some concept of of linear algebra without which it cannot be solved. How should one approach this problem?

Computing directly gives that the eigenvalues of $A$ are $\lambda_{\pm} := \exp \left(\pm \frac{4 \pi i}{9}\right)$, so there is a some matrix $P$ such that
$$A := P^{-1} \begin{pmatrix} \lambda_+ & 0 \\ 0 & \lambda_-\end{pmatrix} P.$$
So, for any integer $n$,
$$A^n = (P^{-1} A P)^n = P^{-1} \begin{pmatrix} \lambda_+ & 0 \\ 0 & \lambda_-\end{pmatrix}^n P = P^{-1} \begin{pmatrix} \lambda_+^n & 0 \\ 0 & \lambda_-^n \end{pmatrix} P.$$ Now, $\lambda_{\pm}^n = \exp\left(\pm \frac{4 \pi i n}{9}\right)$, so the smallest $n$ for which $\lambda_{\pm}^n = 1$ is $n = 9$, and for this $n$, we thus have $A^n = P^{-1} I^n P = I$. The only matrix similar to $I$ is $I$ itself, so no smaller positive $n$ satisfies the equation.
