<s>[INST] If matrix $A$ in $\mathbb{R}^3 $ such that, $A^3 = I$, $\det A = 1$. Is there a such matrix which is not orthogonal, rotation and identity? I tried to use Cayley–Hamilton theorem to learn something about the matrix.
Using the theorem we have:
$p(A)=0 = - A^3 + \text{tr} A\cdot A^2 -\left(\begin{vmatrix}a_{11} && a_{12} \\ a_{21} && a_{22} \end{vmatrix} + \begin{vmatrix}a_{22} && a_{23} \\ a_{32} && a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} && a_{13} \\ a_{31} && a_{33} \end{vmatrix}\right) A + \det A$.
Since $-A^3 + \det A \cdot I = 0$, then $A = \frac{\left(\begin{vmatrix}a_{11} && a_{12} \\ a_{21} && a_{22} \end{vmatrix} + \begin{vmatrix}a_{22} && a_{23} \\ a_{32} && a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} && a_{13} \\ a_{31} && a_{33} \end{vmatrix}\right)}{\text{tr} A}I$. With this in mind I cannot obtain even rotation matrix to say nothing about something else. Any hints?
 [/INST] Here's a quick example: 
$$
A = \pmatrix{\cos(2\pi/3) & -10\sin(2 \pi /3) & 0\\
\frac 1{10}\sin(2 \pi /3) & \cos (2 \pi /3) & 0\\0 & 0 & 1}
$$
Is like the usual rotation, but with the off-diagonal entries slightly altered.  Show that $A^3 = I$, but $A$ is not orthogonal (or a rotation) since $A^TA \neq I$.

Another easy example:
$$
A = \pmatrix{\cos(2\pi/3) & \sin(2 \pi /3) & 2\\
\sin(2 \pi /3) & \cos (2 \pi /3) & 3\\0 & 0 & 1}
$$
the same argument applies.
</s>