Question: > 
> If $
>  \begin{bmatrix}
>  \cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\
>  \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7} \\
>  \end{bmatrix}^k
>  = 
>  \begin{bmatrix}
>  1 & 0 \\
>  0 & 1 \\
>  \end{bmatrix}
> $, then the least positive integral value of $k$ is?
> 
> 
> 

Actually, I got no idea how to solve this.

 I did trial and error method and got 7 as the answer.  

how do i solve this?  
 Can you please offer your assistance? Thank you

Answer: 
Look if your matrix is diagonizable. If it is, write it's diagonal form. Let say that $A$ is the matrix you describe in your post, then we will write $B=S^{-1}AS$ its diagonal form where $S$ is the matrix that allows you to change from the canonical base $E$ to a base $G$ where $B$ is diagonal.

Now assume that $A$ is diagonizable you have a diagonal matrix so the problem is to find the smallest integer $k$ for which $\lambda^k\_1=1$ and $\lambda^k\_2=1$ where $\lambda\_1$ and $\lambda\_2$ are the eigenvalues of $A$. The answer you will find will be the one you are looking for. Indeed, let's say that $m$ is the smallest value for which $B^m=I\_2$ then you have :

$$I\_2=B^m=\underbrace{(S^{-1}AS)…(S^{-1}AS)}\_{m \text{ times}}=S^{-1}A^mS \implies A^m=I\_2$$
I founded out that $$A^k=
 \begin{bmatrix}
 \cos \frac{k.2 \pi}{7} & -\sin \frac{k.2 \pi}{7} \\
 \sin \frac{k.2 \pi}{7} & \cos \frac{k.2 \pi}{7} \\
 \end{bmatrix}
$$

using appropriate trigonometric formulae. Now for $A=I$, $k$ should be equal to $7$