JEE Advanced · Mathematics · 18. Matrices

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

Verified Solution

Answer & Solution

Correct Answer

(A) 1

Step-by-step Solution

Detailed explanation

z = ω

(

ω

is complex cube root of unity)

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

P^{2} = - I

\Rightarrow P^{2} = [\begin{matrix} 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{matrix}] = - I

= [\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}]

\Rightarrow {(- 1)}^{r} + 1 = 0 \Rightarrow r

is odd

\Rightarrow r = 1, 3

Also

ω^{2 r} + ω^{4 s} = - 1

∴

r \neq 3

r = 1 \Rightarrow ω^{2} + ω^{4 s} = - 1

\Rightarrow ω^{4 s} = ω = ω^{4}

\Rightarrow

s = 1

(r, s) = (1, 1)

Only 1 pair

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1	Column 2	Column 3
(I) $x^{2} + y^{2} = a^{2}$	(i) $m y = m^{2} x + a$	(P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II) $x^{2} + a^{2} y^{2} = a^{2}$	(ii) $y = m x + a \sqrt{m^{2} + 1}$	(Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III) $y^{2} = 4 a x$	(iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$	(R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV) $x^{2} - a^{2} y^{2} = a^{2}$	(iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$	(S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

The tangent to a suitable conic (Column 1) at

(\sqrt{3}, \frac{1}{2})

is found to be

\sqrt{3} x + 2 y = 4

, then which of the following options is the only Correct combination?

JEE Advanced 2017 Hard

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Let z=-1+ 3i2 , where i= -1 , and r, s∈1, 2, 3. Let P=-zrz2sz2szr and I be the identity matrix of order 2. Then the total number of ordered pairs (r,s) for which P2=-I is

Step-by-step Solution

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