Let $z=\frac{-1+\sqrt{3} i}{2}$, where $i=\sqrt{-1}$, and $r, s \in\{1,2,3\}$. Let $P=\left[\begin{a

Question

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

Professor PC Thomas Classes · Answer

(1) : Note that $z=\omega$\ $P \equiv\left[\begin{array}{cc}(-\omega)^{r} & \omega^{2 s} \ \omega^{2 s} & \omega^{r}\end{array}ight]$\ $P^{2}=\left[\begin{array}{cc}\omega^{2 r}+\omega^{4 s} & \omega^{r+2 s}\left\{(-1)^{r}+1ight\} \ \omega^{r+2 s}\left\{(-1)^{r}+1ight\} & \omega^{4 s}+\omega^{2 r}\end{array}ight]=-I$\ $\Rightarrow \quad(-1)^{r}+1=0 \quad 	herefore r=1$ or 3\ Again, $\omega^{2 r}+\omega^{4 s}=-1 . r$ and $s$ both can't be 3 .\ $	herefore \quad r=1$, then $s=1 	herefore(r, s)=(1,1)$ only.\