It is given that

$z=\frac{-1}{2}+i \frac{\sqrt{3}}{2}=\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}=e^{i 2 \pi / 3}=\omega$

( $\omega$ is cube root of unity), where $r, s \in\{1,2,3\}$.

It is also given that $P=\left[\begin{array}{cc}(-z)^r & z^{2 s} \\ z^{2 s} & z^r\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$.

Since $P^2=-1$, we have

$P^2=\left[\begin{array}{cc}(-z)^r & z^{2 s} \\ z^{2 s} & z r\end{array}\right]\left[\begin{array}{ll}(z)^r & z^{2 s} \\ z^{2 s} & z^r\end{array}\right]=-\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$=\left[\begin{array}{cc}(-z)^{2 r}+z^{4 s} & (-z)^r \cdot z^{2 s}+z^r z^{2 s} \\ (-z)^r z^{2 s}+z^r \cdot z^{2 s} & z^{4 s}+(z)^{2 r}\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$

That is, we have

and

$(-z)^{2 r}+z^{4 s}=-1 \text { and } z^{4 s}+z^{2 r}=-1$

$\left((-z)^r+z^7 z^{2 s}=0 \text { and } z^{2 r}+z^{4 s}=-1\right.$

$\Rightarrow\left((-\omega)^r+(\omega)^r\right) \cdot \omega^{2 s}=0$

Now, $\omega^{2 s} \neq 0$; therefore,

$(-\omega)^r+(\omega)^r=0$

where $r$ is the odd number and hence $r=1,3$.

When $r=1(-\omega)^2+\omega^4 s=-1$

$\Rightarrow \omega^4 s=-1-\omega^2=+\omega$

Now, s can be $1$ (since $s \neq 3$ ).