I was thinking about the following problem:
Let A be a $3\times3$ real valued matrix such that $A^{3}=I$ but $A \neq I$ . Then trace of A must be
(a)0,
(b)1,
(c)-1,
(d)3.
My attempts: I take A to be $\begin{pmatrix} x &0 &0 \\ 0& x &0 \\ 0&0 & x \end{pmatrix}.$ Now we see $A^{3}=I$ gives $x^{3}=1$ which gives $x=1,w,w^{2}$ where $w$ being the cube root of unity. Thus we see that trace of A is $1+w+w^{2}=0$. So ,i think that (a) is the right choice.Am i going in the right direction? Please give your valuable opinion. Thanks in advance for your time.