Let z = - 1 + 3 i 2 , where i = - 1 , and r , s { 1 , 2 , 3 } . Let P = [ ( - z ) r z 2 s z 2 s z r ] and I be the identity matrix of order 2. Then the total number of ordered pairs ( r , s ) for which P 2 = - I is

Question

Let $z = \frac{- 1 + \sqrt{3} i}{2},$ where $i = \sqrt{- 1}$ , and $r, s \in {1, 2, 3}$ . Let $P = [\begin{matrix} {(- z)}^{r} & z^{2 s} \\ z^{2 s} & z^{r} \end{matrix}]$ 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]

Smart Achievers · Accepted Answer

(1)

$z = \frac{- 1 + i \sqrt{3}}{2} \Rightarrow z^{3} = 1$ and $1 + z + z^{2} = 0$

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

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

For $P^{2} = - I$ , we should have

$z^{2 r} + z^{4 s} = - 1$ and $z^{2 s} ({(- z)}^{r} + z^{r}) = 0$

$\Rightarrow z^{2 r} + z^{4 s} + 1 = 0$ and ${(- z)}^{r} + z^{r} = 0$

$\Rightarrow r is odd and s = r but not a multiple of 3,$

which is possible when $s = r = 1$

$∴$ $only one pair is there.$

Solution

Q.
Let $z = \frac{- 1 + \sqrt{3} i}{2},$ where $i = \sqrt{- 1}$ , and $r, s \in {1, 2, 3}$ . Let $P = [\begin{matrix} {(- z)}^{r} & z^{2 s} \\ z^{2 s} & z^{r} \end{matrix}]$ 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]

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Solution

Q. Let z=-1+3i2, where i=-1, and r,s∈{1,2,3}. Let P=[(-z)rz2sz2szr] and I be the identity matrix of order 2. Then the total number of ordered pairs (r,s) for which P2=-I is [2016]

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Q.
Let $z = \frac{- 1 + \sqrt{3} i}{2},$ where $i = \sqrt{- 1}$ , and $r, s \in {1, 2, 3}$ . Let $P = [\begin{matrix} {(- z)}^{r} & z^{2 s} \\ z^{2 s} & z^{r} \end{matrix}]$ 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]