Matrices and Determinants 1 Question 21
Integer Type Question
21. Let $z=\frac{-1+\sqrt{3} i}{2}$, where $i=\sqrt{-1}$, and $r, s \in{1,2,3}$. Let $P=\begin{bmatrix}(-z)^{r} & z^{2 s} \\ z^{2 s} & z^{r}\end{bmatrix}$ and $I$ be the identity matrix of order 2. Then, the total number of ordered pairs $(r, s)$ for which $P^{2}=-I$ is
(2016 Adv.)
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Answer:
Correct Answer: 21. (1)
Solution:
- Here, $z=\frac{-1+i \sqrt{3}}{2}=\omega$
$ \begin{aligned} & \because \quad P=\begin{bmatrix} (-\omega)^{r} & \omega^{2 s} \\ \omega^{2 s} & \omega^{r} \end{bmatrix} \\ & P^{2}=\begin{bmatrix} (-\omega)^{r} & \omega^{2 s} & (-\omega)^{r} & \omega^{2 s} \\ \omega^{2 s} & \omega^{r} & \omega^{2 s} & \omega^{r} \end{bmatrix} \\ & =\begin{bmatrix} \omega^{2 r}+\omega^{4 s} & \omega^{r+2 s}\left[(-1)^{r}+1\right] \\ \omega^{r+2 s}\left[(-1)^{r}+1\right] & \omega^{4 s}+\omega^{2 r} \end{bmatrix} \end{aligned} $
Given, $P^{2}=-I$
$\therefore \quad \omega^{2 r}+\omega^{4 s}=-1$ and $\omega^{r+2 s}\left[(-1)^{r}+1\right]=0$
Since, $\quad r \in{1,2,3}$ and $(-1)^{r}+1=0$
$\Rightarrow \quad r={1,3}$
Also,
$\omega^{2 r}+\omega^{4 s}=-1$
If $\quad r=1$, then $\omega^{2}+\omega^{4 s}=-1$
which is only possible, when $s=1$.
As, $\quad \omega^{2}+\omega^{4}=-1$
$\therefore \quad r=1, s=1$
Again, if $r=3$, then
$ \begin{bmatrix} & & \omega^{6}+\omega^{4 s} & =-1 \\ & \therefore & \omega^{4 s} & =-2 \\ & r & \neq 3 \end{bmatrix} \quad \text { [never possible] } $
$\Rightarrow(r, s)=(1,1)$ is the only solution.
Hence, the total number of ordered pairs is 1 .
