find a matrix that satisfies $A^6= I$... 
How to solve this type of questions .....please explain....
I'm not getting how to start?
 A: The matrix $$\begin{pmatrix}\cos(\theta)&\sin(\theta)&0\\-\sin(\theta)&\cos(\theta)&0\\0&0&1\end{pmatrix}$$
represents a rotation on the $z-$axis (in the 3D space) of an angle $\theta \in [0,2\pi)$. Therefore your first matrix represents a rotation of $\frac{\pi}{4}$ on the $z-$axis, and applying $6$ times such a rotation you are actually applying a rotation of $6\cdot \frac{\pi}{4}=\frac{3\pi}{2}$, which is not the identity, because you are missing a quarter of a rotation to get a full one.
Similarly the second matrix represents a rotation on the $x-$axis of an angle of $\frac{\pi}{3}$. Applying $6$ times such a rotation you have a rotation of $6\cdot \frac{\pi}{3}=2\pi$, i.e: you are applying an entire rotation of $2\pi$ and therefore you will get the identity.
For the third one, the rotation is on the $y-$axis, and is of an angle of $\frac{\pi}{6}$. Applying it $6$ times leads to a rotation of $\pi$, which is not the identity.
Are you able to do the last case? The final answer is that only the second matrix satisfies $A^6=I$.
