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

$$A^3=I\iff A^2=A^{-1}$$ If the question is valid for all matrix then with a single example one can answer. Consider the matrix
$$A = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}$$
Calculation gives $$A^{-1}=\begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}\text { and } A^2=\begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}$$ We see then that $A^3=I$ and here the trace is obviously equal to $0$. Hence the answer is $0$.
