How to find out whether a matrix to the $4$th power is the identity matrix? Last semester, I was following a Linear Algebra course and in the exam of that course, the following question was asked:
Of the following 5 matrices A, how many of them satisfy $A^4 = I$ ?
$$\begin{bmatrix}1&0\\0&-1\end{bmatrix}, \begin{bmatrix}\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\end{bmatrix}, \begin{bmatrix}\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}&-\frac{\sqrt2}{2}\end{bmatrix}, \begin{bmatrix}\frac{1}{2}&-\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&\frac{1}{2}\end{bmatrix}, \begin{bmatrix}0&-1\\1&0\end{bmatrix} $$
I answered this question by doing all the calculations to the power of 4, and as you would expect, this took way too much time but I eventually figured out the correct answer which is 2  of them, which are the following matrices:
$$\begin{bmatrix}1&0\\0&-1\end{bmatrix} and \begin{bmatrix}0&-1\\1&0\end{bmatrix} $$
I'm taking the exam again and was wondering if there's a different and easier way to solve such a question?

Here's how I would have approached the problem.
(1) Since this matrix is diagonal, we can calculate 
$$
\begin{bmatrix}1&0\\0&-1\end{bmatrix}^4=\begin{bmatrix}1^4&0\\0&(-1)^4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}
$$
so this matrix satisfies $A^4=I$.
(2) For this matrix, I notice that $\sqrt{2}/2$ appears a lot, so my brain goes to the fact that $\sin(\pi/4)=\cos(\pi/4)=\sqrt{2}/2$. Thinking of this matrix as corresponding to a linear transformation in the plane, I took a look at the form for a rotation matrix:
$$
\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix}.
$$
This tells me that our matrix corresponds to a rotation by $-\pi/4$, so $A^4$ corresponds to composing this rotation by itself four times, resulting in rotating by $-\pi$. Since, in general, rotating a vector by $-\pi$ doesn't result in the same vector, it follows that $A^4\neq I$.
(3) Here I noticed that the bottom row is a scalar multiple of the top row, so the matrix cannot be invertible. You can justify this by realizing this means the rows are not linearly independent, or simply by computing the determinant and seeing that it's zero. Since the matrix is not invertible, it is impossible that $A^4=I$. We can see this in a few ways: first, the properties of determinants tell me that $A^4=I$ implies that $\det(A)^4=\det(I)=1$ and so $\det(A)\neq{0}$, which we know isn't true for our $A$ since it's not invertible; second, $A^4=I$ can be written as $A(A^3)=I$, and this would imply that $A$ is invertible with $A^{-1}=A^3$, but $A$ isn't invertible.
(4) Like with (2), I see $1/2$ and $\sqrt{3}/2$, so I think of rotation matrices and see that this matrix corresponds to a rotation by $\pi/3$. Again, since for a vector $v\in\mathbb{R}^2$, $A^4v$ is an application of $A$ to the vector $v$ four times, it follows that $A^4$ corresponds to a rotation by $4\pi/3$. Thus, $A^4v\neq v$ for a general $v$.  
(5) Correcting my answer thanks to the comments below, if one did not wish to do the matrix multiplication, one could again realize that the matrix is a rotation matrix corresponding to $\theta=\pi/2.$ As applying this rotation four times to any vector results in a rotation by $2\pi$, or, equivalently, no rotation at all, we have that $A^4=I$.
