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: 
We start with polar coordinates of a point $P=(r,\theta)$ vs its Cartesian form $(x,y)$. You know $x=r\cos\theta, y=r\sin\theta$. Now rotate the point $P$ by an angle of $\alpha$ with respect to origin so its angle with respect to $x$ axis becomes $\theta+\alpha$. Its new polar coordinates is $P'=(r,\theta+\alpha)$ and its new Cartesian coordinates $(x',y')$, where $x'=r\cos(\theta+\alpha), y'=r\sin(\theta+\alpha)$. Use trig identities to expand these

$x'=r(\cos \theta \cos \alpha -\sin \theta \sin \alpha)=x\cos \alpha -y \sin\alpha$, and

$ y'=r(\sin\theta\cos\alpha+\cos\theta\sin\alpha)=y\cos\alpha+x\sin\alpha$. 

If you write these as a matrix equation you can finish your problem. 
$
\left[ \begin{array}{c} x' \\ y' \end{array} \right] = \begin{bmatrix} \cos \alpha & -\sin \alpha \\
 \sin \alpha & \cos \alpha \\ \end{bmatrix} \times \left[ \begin{array}{c} x \\ y \end{array} \right]
$

If you have several rotation $\alpha\_1, \cdots, \alpha\_n$ then the effect of multiplying all the related matrices is the same as adding the given angles. In particular if you have $n$ rotations by the same angle $\alpha$ the result will be same as one rotation of $n\alpha$. In short
$
 \begin{bmatrix} \cos \alpha & -\sin \alpha \\
 \sin \alpha & \cos \alpha \\ \end{bmatrix}^n= \begin{bmatrix} \cos n\alpha & -\sin n\alpha \\
 \sin n\alpha & \cos n\alpha \\ \end{bmatrix}$
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 $7$