NDA Mathematics Question (2025) — Solution

Step-by-step solution

Given the determinant: vmatrix 2 & 3+i & -1 \\ 3-i & 0 & i-1 \\ -1 & -1-i & 1 vmatrix = A + iB Expanding the determinant along the first row, we get: D = 2 vmatrix 0 & i-1 \\ -1-i & 1 vmatrix - (3+i) vmatrix 3-i & i-1 \\ -1 & 1 vmatrix - 1 vmatrix 3-i & 0 \\ -1 & -1-i vmatrix Evaluating each 2 2 minor: First minor: vmatrix 0 & i-1 \\ -1-i & 1 vmatrix = (0)(1) - (i-1)(-1-i) = 0 - (-(i^2 - 1)) = 0 - (-(-1 - 1)) = -2 Second minor: vmatrix 3-i & i-1 \\ -1 & 1 vmatrix = (3-i)(1) - (i-1)(-1) = 3 - i + i - 1 = 2 Third minor: vmatrix 3-i & 0 \\ -1 & -1-i vmatrix = (3-i)(-1-i) - (0)(-1) = -3 - 3i + i + i^2 = -3 - 2i - 1 = -4 - 2i Substituting these values back into the expansion: D = 2(-2) - (3+i)(2) - 1(-4 - 2i) D = -4 - (6 + 2i) + (4 + 2i) D = -4 - 6 - 2i + 4 + 2i D = -6 Comparing this with A + iB, we get: A = -6 and B = 0 Therefore, A + B = -6 + 0 = -6. Answer: -6