Discussion Forum

Let $z=\frac{-1+\sqrt{3}i}{2}$ , where   $i=\sqrt{-1}$  , and  $r,s \epsilon (1,2,3)$. Let  $P=\begin{bmatrix}(-z)^{r} & z^{2s} \\z^{2s} & z^{r} \end{bmatrix}$  and  I be the identity matrix of  order2.Then, the total number of ordered pairs (r,s) for which   $p^{2}=-I$ is

Answer:

Explanation:

Here,

$z=\frac{-1+\sqrt{3}i}{2}$ =ω

$\because$     $P= \begin{bmatrix}(-\omega)^{r} & \omega^{2s} \\\omega^{2s} & \omega^{r}\end{bmatrix}$

   $P^{2}=\begin{bmatrix}(-\omega)^{r} & \omega^{2s} \\\omega^{2s} & \omega^{r} \end{bmatrix} \begin{bmatrix}(-\omega)^{r} & \omega^{2s} \\\omega^{2s} & \omega^{r} \end{bmatrix}$

         $=\begin{bmatrix}\omega^{2r}+\omega^{4s} & \omega^{r+2s}[(-1)^{r} +1\\\omega^{r+2s}[(-1)^{r}+1 & \omega^{4s}+\omega^{2r} \end{bmatrix}$

Given,           $P^{2}=-1$

$\therefore$       $\omega^{2r}+\omega^{4s}=-1$

  and    $\omega^{r+2s}[(-1)^{r}+1]=0$

Since,      $r\epsilon (1,2,3)$  and  $(-1)^{r}+1=0$

$\Rightarrow$          r={1,3}

Also,   $\omega^{2r}+\omega^{4r}=-1$

  If r=1  . then  $\omega^{2}+\omega^{4s}=-1$

 which is only possible , when s=1

As,    $\omega^{2}+\omega^{4}=-1$

$\therefore$    r=1, s=1

  Again, if r=3, then

$\omega^{6}+\omega^{4s}=-1$

$\Rightarrow$  $\omega^{4s}=-2$  [ never possible ]

   $\therefore$      $r\neq 3$

 $\Rightarrow$  (r,s) =(1,1) is the only solution.

 Hence , the total  number of ordered pairs is 1.