Cubic root of $3 \times 3$ identity matrix Is there a real $3 \times 3$ matrix $A$, such that: $A^3 = I_3$ and has at most one zero entry?  If so, how can I find it?

Since $A=(a_{ij})_{(3\times3)}$ satisfies $A^3-I=0$, so the characteristic equation of $A$ is $x^3-1=0$ 
Now $x^3-1=0\implies (x-1)(x^2+x+1)=0\implies\begin{vmatrix} 
1-x & 0 & 0 \\
0 & -x & 1 \\ 
0 & -1 & -1-x
\end{vmatrix}=0$
Construct a matrix $A=\begin{pmatrix} 
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & -1 
\end{pmatrix}$
Clearly $(x-1)(x^2+x+1)=0$ is  characteristic equation of $A$ and so it also satisfy $A^3-I=0$
