JEE Advanced 2026 — Paper 2 — MATH-II solution

JEE Advanced2026Paper 2MATH-II

Let $R$ denote the set of all real numbers and let $i = - 1$ . Consider the matrices

$S = [0 - 1 10]$ and $T = [1101]$ .

Let $a, b, c, d$ be real numbers such that

$S T = [ab c d]$ .

Let $H = {x + i y : x, y \in R and y > 0}$ .

Then which of the following statements is (are) TRUE?

(A) $\frac{b + ia}{d + i c} = i$

(B) If $ω = \frac{- 1 + i 3}{2}$ , then $\frac{aω + b}{c ω + d} = ω$

(D) If $z \in H$ , then $\frac{a z + b}{cz + d} \in H$

A
$\frac{b + ia}{d + i c} = i$
B
If $ω = \frac{- 1 + i 3}{2}$ , then $\frac{aω + b}{c ω + d} = ω$
C
If $m$ is an integer greater than $2$ such that $(S T)^{2} = (S T)^{m}$ , then $m$ is an integer multiple of $8$
D
If $z \in H$ , then $\frac{a z + b}{cz + d} \in H$

Solution

Compute $S T = [0 - 1 10] [1101] = [0 - 1 11]$ . So $a = 0$ , $b = - 1$ , $c = 1$ , $d = 1$ .

(A): $\frac{b + ia}{d + i c} = \frac{- 1}{1 + i} = \frac{- 1 ( 1 - i )}{2} = \frac{- 1 + i}{2} \neq = i$ . Incorrect.

(B): Since $ω$ is a primitive cube root of unity, $ω^{2} + ω + 1 = 0$ , so $ω + 1 = - ω^{2}$ . Then $\frac{aω + b}{c ω + d} = \frac{- 1}{ω + 1} = \frac{- 1}{- ω ^{2}} = \frac{1}{ω ^{2}} = ω$ (since $ω^{3} = 1$ ). Correct.

(C): Compute $(S T)^{2} = [- 1 - 1 10]$ , $(S T)^{3} = [- 10 0 - 1] = - I$ , so $(S T)^{6} = I$ . Therefore $(S T)^{m} = (S T)^{2} ⟺ m \equiv 2 (mod 6)$ . Such $m$ are $8, 14, 20, \dots$ ; $14$ is in the list but is not a multiple of $8$ . Incorrect.

(D): $\frac{a z + b}{cz + d} = \frac{- 1}{z + 1}$ . With $z = x + i y$ , $\frac{- 1}{x + 1 + i y} = \frac{- ( x + 1 ) + i y}{( x + 1 ) ^{2} + y ^{2}}$ . The imaginary part is $\frac{y}{( x + 1 ) ^{2} + y ^{2}} > 0$ when $y > 0$ , so the image lies in $H$ . Correct.

$∴$ Correct options: (B), (D).

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