JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

We are given the matrices S = bmatrix 0 & -1 \\ 1 & 0 bmatrix and T = bmatrix 1 & 1 \\ 0 & 1 bmatrix . First, we compute the product ST: ST = bmatrix 0 & -1 \\ 1 & 0 bmatrix bmatrix 1 & 1 \\ 0 & 1 bmatrix = bmatrix 0 & -1 \\ 1 & 1 bmatrix Comparing this with ST = bmatrix a & b \\ c & d bmatrix , we get a = 0, b = -1, c = 1, and d = 1. Evaluating Option (A): b + ia d + ic = -1 + i(0) 1 + i(1) = -1 1 + i = -1(1 - i) (1 + i)(1 - i) = -1 + i 2 This is not equal to i, so statement (A) is FALSE. Evaluating Option (B): Given = -1 + i 3 2 , which is a complex cube root of unity, satisfying ^2 + + 1 = 0 and ^3 = 1. a + b c + d = 0 - 1 1 + 1 = -1 + 1 Since + 1 = - ^2, we have: -1 - ^2 = 1 ^2 = ^3 ^2 = Thus, statement (B) is TRUE. Evaluating Option (C): The characteristic equation of ST is ^2 - Tr (ST) + (ST) = 0, which gives ^2 - + 1 = 0. By the Cayley-Hamilton theorem, (ST)^2 - ST + I = 0. Multiplying by (ST + I), we get (ST)^3 + I = 0 (ST)^3 = -I. Squaring both sides gives (ST)^6 = I. We are given (ST)^2 = (ST)^m, which implies (ST)^ m-2 = I. This means m - 2 must be a multiple of 6, so m = 6k + 2 for some integer k. For k = 1, m = 8 (which is a multiple of 8). For k = 2, m = 14 (which is NOT a multiple of 8). Thus, statement (C) is FALSE. Evaluating Option (D): For z H, let z = x + iy where y > 0. az + b cz + d = -1 z + 1 = -1 (x + 1) + iy Multiplying the numerator and denominator by the conjugate (x + 1) - iy: -1 (x + 1) + iy (x + 1) - iy (x + 1) - iy = -(x + 1) + iy (x + 1)^2 + y^2 The imaginary part of this new complex number is y (x + 1)^2 + y^2 . Since y > 0 and (x + 1)^2 + y^2 > 0, the imaginary part is strictly positive. Therefore, az + b cz + d H, making statement (D) TRUE. Answer: If = -1 + i 3 2 , then a + b c + d = ; If z H, then az + b cz + d H