Question

Let $z=\frac{-1+\sqrt{3}i}{2}$, where $i=\sqrt{-1}$ and $r,s\in \{1,2,3\}$. Let $P=\left[\begin{array}{cc}{(-z)}^r& z^{2s}\\ z^{2s}& z^r\end{array}\right]$ and $I$ be the identity matrix of order $2$. Then the total number of ordered pairs $(r,s)$ for which $P^2=-I$ is

Answer 1

$z=\omega$ (where $\omega$ is cube root of unity )
$P=\left[\begin{array}{cc}(-\omega {)}^r& {\omega }^{2s}\\ {\omega }^{2s}& {\omega }^r\end{array}\right]$
$P^2=\left[\begin{array}{cc}(-\omega {)}^r& {\omega }^{2s}\\ {\omega }^{2s}& {\omega }^r\end{array}\right]\left[\begin{array}{cc}(-\omega {)}^r& {\omega }^{2s}\\ {\omega }^{2s}& {\omega }^r\end{array}\right]=\left[\begin{array}{cc}(-\omega {)}^{2r}+{\omega }^{4s}& (-1{)}^r{\omega }^{2s+r}+{\omega }^{r+2s}\\ (-1{)}^r{\omega }^{2s+r}+{\omega }^{r+2s}& {\omega }^{2r}+{\omega }^{4s}\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$
$\Rightarrow {\omega }^{2r}+{\omega }^{4s}=-1$,

$\left(\right(-1{)}^r+1){\omega }^{2s+r}=0;r,s\in \{1,2,3\}$
$\Rightarrow$ second equation represent $r=1,3$
Case - 1: $r=1$
${\omega }^{4s}=-1-{\omega }^2=\omega \Rightarrow s=1$
Case - 2: $r=3$
${\omega }^{4s}=-1-1=-2\Rightarrow$ No value of s is possible
$\Rightarrow$ Total number of ordered pairs $(r,s)=1$