Question

hey pls answer this qn ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:z = \dfrac{ - 1 + \sqrt{3} i}{2}$

We know,

$\boxed{\rm :\longmapsto\: \omega \: = \dfrac{ - 1 + \sqrt{3} i}{2} }$

So,

$\bf\implies \:z = \omega$

Further given that,

$\rm :\longmapsto\:P = \bigg[ \begin{matrix} {( - z)}^{r} & {z}^{2s} \\ {z}^{2s} & {z}^{r} \end{matrix} \bigg]$

which can be rewritten as

$\rm :\longmapsto\:P = \bigg[ \begin{matrix} {( -\omega )}^{r} & {\omega}^{2s} \\ {\omega}^{2s} & {\omega}^{r} \end{matrix} \bigg]$

Consider,

$\rm :\longmapsto\: {P}^{2} = \bigg[ \begin{matrix} {( -\omega )}^{r} & {\omega}^{2s} \\ {\omega}^{2s} & {\omega}^{r} \end{matrix} \bigg]\bigg[ \begin{matrix} {( -\omega )}^{r} & {\omega}^{2s} \\ {\omega}^{2s} & {\omega}^{r} \end{matrix} \bigg]$

$\rm \: = \: \bigg[ \begin{matrix} {( \omega )}^{2r} + {\omega}^{4s} & {( - \omega)}^{r} {\omega}^{2s} + {\omega}^{2s} {\omega}^{r}\\ {( - \omega)}^{r} {\omega}^{2s} + {\omega}^{2s} {\omega}^{r} & {( \omega )}^{2r} + {\omega}^{4s} \end{matrix} \bigg]$

$\rm \: = \: \bigg[ \begin{matrix} {( \omega )}^{2r} + {\omega}^{4s} & ({( - \omega)}^{r} + {\omega}^{r}) {\omega}^{2s}\\ ({( - \omega)}^{r} + {\omega}^{r} ){\omega}^{2s} & {( \omega )}^{2r} + {\omega}^{4s} \end{matrix} \bigg]$

Now, it is given that,

$\red{\rm :\longmapsto\: {P}^{2} = - \: I}$

Given that, I is an identity matrix of order 2 × 2

So,

$\rm :\longmapsto\:I = \bigg[ \begin{matrix}1&0\\ 0&1 \end{matrix} \bigg]$

So, on substituting the values, we get

$\rm \: \bigg[ \begin{matrix} {( \omega )}^{2r} + {\omega}^{4s} & ({( - \omega)}^{r} + {\omega}^{r}) {\omega}^{2s}\\ ({( - \omega)}^{r} + {\omega}^{r} ){\omega}^{2s} & {( \omega )}^{2r} + {\omega}^{4s} \end{matrix} \bigg] = - \bigg[ \begin{matrix}1&0\\ 0&1 \end{matrix} \bigg]$

$\rm \: \bigg[ \begin{matrix} {( \omega )}^{2r} + {\omega}^{4s} & ({( - \omega)}^{r} + {\omega}^{r}) {\omega}^{2s}\\ ({( - \omega)}^{r} + {\omega}^{r} ){\omega}^{2s} & {( \omega )}^{2r} + {\omega}^{4s} \end{matrix} \bigg] = \bigg[ \begin{matrix} - 1&0\\ 0& - 1 \end{matrix} \bigg]$

Since, it is given that r and s can assume the values from 1, 2, 3.

So, on comparing, we get

$\rm :\longmapsto\:{( \omega )}^{2r} + {\omega}^{4s} = - 1 - - - (1)$

and

$\rm :\longmapsto\:({( - \omega)}^{r} + {\omega}^{r}) {\omega}^{2s} = 0$

$\bf\implies \:r = 1 \: \: \: or \: \: \: r = 3$

It holds when r = 1 or r = 3

Case :- 1

When r = 1

Substituting the value of r in equation (1), we get

$\rm :\longmapsto\:{( \omega )}^{2} + {\omega}^{4s} = - 1$

$\rm :\longmapsto\:+ {\omega}^{4s} = - 1 - {\omega}^{2}$

$\rm :\longmapsto\:+ {\omega}^{4s} = - (1 + {\omega}^{2} )$

$\rm :\longmapsto\ {\omega}^{4s} = - ( - {\omega})$

$\rm :\longmapsto\ {\omega}^{4s} = {\omega}$

can be rewritten as

$\rm :\longmapsto\ {\omega}^{3s} \times {\omega}^{s} = {\omega}$

$\rm :\longmapsto\ 1 \times {\omega}^{s} = {\omega}$

$\rm :\longmapsto\ {\omega}^{s} = {\omega}$

$\bf\implies \:s = 1$

So, one order pair is (1, 1).

Case :- 2

When r = 3

Put the value of r = 3 in equation (1), we get

$\rm :\longmapsto\:{( \omega )}^{6} + {\omega}^{4s} = - 1$

$\rm :\longmapsto\:1 + {\omega}^{4s} = - 1$

$\rm :\longmapsto\:{\omega}^{4s} = - 1 - 1$

$\rm :\longmapsto\:{\omega}^{3s} \times {\omega}^{s} = - 2$

$\rm :\longmapsto\:{\omega}^{s} = - 2$

which is not possible.

So, only one order pair ( 1, 1 ) exist.

Formula used :-

$\red{\boxed{ \rm{ {\omega}^{3} = 1}}}$

$\red{\boxed{ \rm{1 + \omega + {\omega}^{2} = 0}}}$