If $A$ is a real $n \times n$ matrix satisfying $A^3 = I$ then Trace of $A$ is always If $A$ is a real $3 \times 3$ matrix satisfying $A^3$ = I such that $ A \neq I $   .Then, Trace of A is always


*

*$0$

*$1$

*$-1$

*$3$


I proceed as follows: from given,
$\min(x)=x-1$ or $x^2+x+1$  or  $(x-1)(x^2+x+1)$
$\min(x)\ne x-1$ as $A\ne I$
so, $\min(x)=x^2+x+1$ or $(x-1)(x^2+x+1)$
now, how to proceed after this step
any help would be appreciated

Here are two matrices $A$ with $A^3 = A$:
$$
A = \begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}\qquad A = \begin{pmatrix}-1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1\end{pmatrix}
$$
Since these have diffferent traces, the question does not have a definite answer.
Edit: The matrix
$$
A = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}
$$
has the desired property (for the edited question) and has trace 0. Hence, if the question is well-posed, the answer must be 0.
