NDA Mathematics Question (2026) — Solution

Step-by-step solution

Let a = p^2, b = q^2, and c = r^2. The given determinant can be written as: = vmatrix a+b & c & c \\ a & b+c & a \\ b & b & c+a vmatrix Applying the row operation R_1 R_1 - R_2 - R_3, we get: = vmatrix 0 & -2b & -2a \\ a & b+c & a \\ b & b & c+a vmatrix Expanding the determinant along the first row: = 0 - (-2b) vmatrix a & a \\ b & c+a vmatrix + (-2a) vmatrix a & b+c \\ b & b vmatrix = 2b(ac + a^2 - ab) - 2a(ab - b^2 - bc) = 2abc + 2a^2b - 2ab^2 - 2a^2b + 2ab^2 + 2abc = 4abc Substituting back a = p^2, b = q^2, and c = r^2: = 4(p^2)(q^2)(r^2) = 4(pqr)^2 Since p, q, r are the cube roots of unity (1, , ^2), their product is pqr = 1 ^2 = ^3 = 1. = 4(1)^2 = 4 Answer: 4