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: 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.
A: Clearly, anihilating polynomial for $A$ is $t^3-1$. So, the possible minimal polynomials are
$m_1(t)=t-1$
$m_2(t)=t^2+t+1$
$m_3(t)=(t-1)(t^2+t+1)$
Discard $m_1(t)$ as $A\neq I$. Discard $m_2(t)$ as $t^2+t+1=0$ gives two complex eigenvalues but where is the third one?
Now,  $m_3(t)=0$ gives eigenvalues as $1$,$\frac{-1\pm i√3}{2}$. So, tr($A$)$=$sum of eigenvalues$=0$.
A: Hint Since $A$ is $3 \times 3$, $A$ has exactly 1 or 3 real eigenvalues.
If $A$ has three real eigenvalue, then $\lambda=1$ is the only eignevalue of $A$.
Prove that in this case $A=I$ [Hint Jordan form and $A^3=I$], which is not possible.
Therefore, one of the eigenvalues is real, and the other two are complex. Therefore the eignevalues are ....
A: $$A^3=I\iff A^2=A^{-1}$$ 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 $0$.
