How to construct a $2\times 2$ real matrix $A$ not equal to Identity such that $A^3=I$? How to construct a $2\times 2$ real matrix $A$ not equal to Identity such
that $A^3=I$?
There is a correspondence between the ring of complex numbers and the
ring  of  $2\times2$ matrices (0 matrix
is included!) i.e.,$$a+ib\leftrightarrow\begin{pmatrix}a&-b\\b&a\end{pmatrix}$$
Can I apply this result and construct such matrix?
 A: Find the three cube roots of $1$.  Let $a+bi$ be one of those.  They are solutions of $x^3=1$.  Hence $x^3-1=0$.
Since $1$ is one of the solutions, $x-1$ must be one of the factors, thus:
$$
x^3-1 = (x-1)(\cdots\cdots\cdots).
$$
Fill in the blanks by doing long division.  You should get
$$
(x-1)(x^2+x+1).
$$
So the equation is
$$
(x-1)(x^2+x+1) = 0.
$$
That implies
$$
x-1=0\quad\text{or}\quad x^2+x+1 = 0.
$$
Solve the quadratic equation.
A: Note that there are also $2\times2$ matrices which are not of the form indicated in the question which solve the problem that $A^3=I$. The general solution to your problem is ($\alpha,\beta \in \mathbb{R}, \beta\neq 0$)
$$A =\begin{pmatrix} \alpha & \beta\\
-\beta^{-1}(1+\alpha+\alpha^2) & -(1+\alpha)\end{pmatrix} .$$
The specific examples which correspond to complex numbers $a+ib$ need to fulfill
$$\alpha = -(1+\alpha) \quad \text{and} \quad\beta = \beta^{-1}(1+\alpha+\alpha^2)$$
with the solutions $\alpha=-1/2$, $\beta=\pm\sqrt{3}/2$ (corresponding to rotation by $\pm 2\pi/3$). 
A: The matrix $A=\begin{pmatrix} 0&1\\-c&-b\end{pmatrix}$ satisfies $A^2+bA+cI=0$.
So the matrix $\begin{pmatrix} 0&1\\-1&-1\end{pmatrix}$ satisfies $A^3-I=(A-I)(A^2+A+I)=0$
