SATHEE: Matrices and Determinants 1 Question 21

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

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Answer:

Correct Answer: 21. (1)

Solution:

Here, $z = \frac{- 1 + i \sqrt{3}}{2} = ω$

$\begin{aligned} ∵ P = [\begin{array}{c} (- ω)^{r} & ω^{2 s} \\ ω^{2 s} & ω^{r} \end{array}] \\ P^{2} = [\begin{array}{c} (- ω)^{r} & ω^{2 s} & (- ω)^{r} & ω^{2 s} \\ ω^{2 s} & ω^{r} & ω^{2 s} & ω^{r} \end{array}] \\ = [\begin{array}{c} ω^{2 r} + ω^{4 s} & ω^{r + 2 s} [(- 1)^{r} + 1] \\ ω^{r + 2 s} [(- 1)^{r} + 1] & ω^{4 s} + ω^{2 r} \end{array}] \end{aligned}$

Given, $P^{2} = - I$

$∴ ω^{2 r} + ω^{4 s} = - 1$ and $ω^{r + 2 s} [(- 1)^{r} + 1] = 0$

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

$\Rightarrow r = 1, 3$

Also,

$ω^{2 r} + ω^{4 s} = - 1$

If $r = 1$ , then $ω^{2} + ω^{4 s} = - 1$

which is only possible, when $s = 1$ .

As, $ω^{2} + ω^{4} = - 1$

$∴ r = 1, s = 1$

Again, if $r = 3$ , then

$[\begin{matrix} ω^{6} + ω^{4 s} & = - 1 \\ ∴ & ω^{4 s} & = - 2 \\ r & \neq 3 \end{matrix}] [never possible]$

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

Hence, the total number of ordered pairs is 1 .

Matrices and Determinants 1 Question 21

21. 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

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Matrices and Determinants 1 Question 21

21. 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

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21. 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