Mathematics - Matrices Question with Solution | TestHub

$\text{z}=\text{ω}$($\text{ω}$is complex cube root of unity)
$\text{P}=\left[\begin{array}{cc}{\left(-\text{ω}\right)}^{\text{r}}& {\text{ω}}^{2\text{s}}\\ {\text{ω}}^{2\text{s}}& {\text{ω}}^{\text{r}}\end{array}\right]$,${\text{P}}^2=-\text{I}$
$\Rightarrow {\text{P}}^2=\left[\begin{array}{cc}{\text{ω}}^{2\text{r}}+{\text{ω}}^{4\text{s}}& {\text{ω}}^{\text{r}+2\text{s}}\left({\left(-1\right)}^{\text{r}}+1\right)\\ {\text{ω}}^{\text{r}+2\text{s}}\left({\left(-1\right)}^{\text{r}}+1\right)& {\text{ω}}^{4\text{s}}+{\text{ω}}^{2\text{r}}\end{array}\right]=-\text{I}$$=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$
$\Rightarrow {\left(-1\right)}^{\text{r}}+1=0\Rightarrow r$is odd$\Rightarrow \text{r}=1,3$
Also${\text{ω}}^{2\text{r}}+{\text{ω}}^{4\text{s}}=-1$
$\therefore$$\text{r}\ne 3$
By$\text{r}=1\Rightarrow {\text{ω}}^2+{\text{ω}}^{4\text{s}}=-1$$\Rightarrow {\text{ω}}^{4\text{s}}=\text{ω}={\text{ω}}^4$
$\Rightarrow$$\text{s}=1$
$\left(\text{r},\text{s}\right)=\left(1,1\right)$
Only 1 pair