Question

Let $z=\frac{-1+\sqrt{3}i}{2}$, where $i=\sqrt{-1}$, and r, s $ϵ$ {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

Here, $z=\frac{-1+i\sqrt{3}}{2}=\omega$ [From cube root of unity]
$\because P=\left[\begin{array}{cc}(-\omega {)}^r& {\omega }^{2s}\\ {\omega }^{2s}& {\omega }^r\end{array}\right]$
$P^2=\left[\begin{array}{cc}(-\omega {)}^r& {\omega }^{2s}\\ {\omega }^{2s}& {\omega }^r\end{array}\right]\left[\begin{array}{cc}(-\omega {)}^r& {\omega }^{2s}\\ {\omega }^{2s}& {\omega }^r\end{array}\right]$
$=\left[\begin{array}{cc}{\omega }^{2r}+{\omega }^{4s}& {\omega }^{r+2s}\left[\right(-1{)}^r+1]\\ {\omega }^{r+2s}\left[\right(-1{)}^r+1]& {\omega }^{4s}+{\omega }^{2r}\end{array}\right]$
Given $P^2$ = -I
$\therefore {\omega }^{2r}+{\omega }^{4s}=-1and{\omega }^{r+2s}\left[\right(-1{)}^r+1]=0$
Since, $rϵ\{1,2,3\}$
and $(-1{)}^r$ + 1 = 0
$\Rightarrow$ r = {1, 3}
Also ${\omega }^{2r}+{\omega }^{4s}=-1$
r = 1, then ${\omega }^2+{\omega }^{4s}$ = -1
which is only possible, when s = 1.
As, ${\omega }^2+{\omega }^4={\omega }^2+\omega =-1[\because {\omega }^3=1and1+\omega +{\omega }^2=0]$
$\therefore$ r = 1, s = 1
Again, if r = 3, then
${\omega }^6+{\omega }^{4s}=-1$
$\Rightarrow {\omega }^{4s}=-2$ [never possible]
$\therefore r\ne 3$
$\Rightarrow$ (r, s) = (1, 1) is the only solution.
Hence, the total number of ordered pairs is 1.