JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The cofactor matrix of A is calculated as follows: C_ 11 = 15, C_ 12 = -10, C_ 13 = 8 C_ 21 = 3, C_ 22 = 5 , C_ 23 = -4 C_ 31 = -6, C_ 32 = 4, C_ 33 = 3 - 2 The adjoint of A is the transpose of the cofactor matrix: adj (A) = bmatrix 15 & 3 & -6 \\ -10 & 5 & 4 \\ 8 & -4 & 3 - 2 bmatrix The matrix B is given by: B = bmatrix 1 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 4 & -2 bmatrix + bmatrix 15 & 3 & -6 \\ -10 & 5 & 4 \\ 8 & -4 & 3 - 2 bmatrix = bmatrix 16 & 3 & -6 \\ -10 & 0 & 4 \\ 8 & 0 & - 2 bmatrix Expanding the determinant of B along the second column: (B) = -3 vmatrix -10 & 4 \\ 8 & - 2 vmatrix = -3(-10( - 2) - 32) = -3(-10 - 12) = 30 + 36 Given (B) = 66, we get: 30 + 36 = 66 30 = 30 = 1 The determinant of A is: (A) = (15 - 0) - 1(10 - 0) + 2(8 - 0) = 15 + 6 Substituting = 1: (A) = 15(1) + 6 = 21 Using the property ( adj (A)) = ( (A))^ n-1 for a 3 3 matrix: ( adj (A)) = ( (A))^2 = (21)^2 = 441 Answer: 441