Let$A$ be a $3\times3$ real symmetric matrix such that $A^6=I$ . Then $A^2=I$ Let$A$ be a $3\times3$ real symmetric matrix such $ A^6=I$ . Then $A^2=I$.
How can I prove this statement is true or false?
As it is given $A$ is symmetric so $A=A^T$. Also $ A^6=I$.
But the main problem is that I can't operate $A^{-1}$ on both sides whether it is invertible or not. Can any help me what should I do?

$A^6=I$ gives $x^6-1$ $=(x^2-1)(x^2-x+1)(x^2+x+1)$ is an annihilating polynomial of $A$. Let $m(x)$ be the minimal polynomial of $A$ then $m(x)$ divides $x^6-1$.
Since all eigenvalues of a real symmetric matrix are real and $x^2\pm x+1$ does not contain any real root, $m(x)$ cannot contain any factor of $x^2-x+1$ or $x^2+x+1$. Thus $m(x)$ divides $x^2-1$, which gives $x^2-1$ is an annihilating polynomial of $A$.
