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

$z=\frac{-1+\sqrt{3}i}{2}=w,$ (cube root of unity)
$P=\left[\begin{array}{cc}(-w{)}^r& w^{2s}\\ w^{2s}& w^r\end{array}\right]P^2=-I\Rightarrow \left[\begin{array}{cc}(-w{)}^r& w^{2s}\\ w^{2s}& w^r\end{array}\right]\left[\begin{array}{cc}(-w{)}^r& w^{2s}\\ w^{2s}& w^r\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\Rightarrow \left[\begin{array}{cc}{w}^{2r}+w^{4s}& (-1{)}^r{w}^{r+2s}+w^{r+2s}\\ (-1{)}^r{w}^{r+2s}+w^{r+2s}& w^{2r}+w^{4s}\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\Rightarrow w^{2r}+w^{4s}=-1....\left(1\right)(-1{)}^r{w}^{r+2s}+w^{r+2s}=0....(2)\Rightarrow (-1{)}^r{w}^{r+2s}=-w^{r+2s}\Rightarrow (-1{)}^r=-1\Rightarrow r=1,3[\because r,s\in \{1,2,3\}]$
For, $r=1$
$w^2+w^{4s}=-1\Rightarrow w^{4s}=-1-w^2\Rightarrow w^{4s}=w\Rightarrow s=1$
For, $r=3$
$w^6+w^{4s}=-1\Rightarrow w^{4s}=-1-1=-2$
$\Rightarrow$ Which is not possible
$\therefore$ Only one solution i.e., $(1,1)$