Skew-symmetric Determinant

A skew-symmetric matrix satisfies A^T = -A, and for odd order this forces its determinant to vanish, an elegant JEE Advanced result. Taking determinants, det(A^T) = det(-A). The left side equals det(A), and the right side equals (-1)^n det(A) for an n by n matrix. For odd n = 3, (-1)^3 = -1, so det(A) = -det(A), which gives 2 det(A) = 0, hence det(A) = 0. Thus every 3 by 3 skew-symmetric matrix is singular with determinant zero, independent of its specific entries. Option 1 and option -1 contradict the forced zero value. Option depends on the entries is false because the conclusion holds universally for odd order. Hence the determinant is 0. Plausibility check: a skew-symmetric matrix of odd order always has a zero eigenvalue, so it must be singular, and the product of eigenvalues being zero confirms a vanishing determinant regardless of entries. The vanishing of every odd-order skew-symmetric determinant is a clean parity result with no analogue for even orders, where the determinant is instead a perfect square known as the Pfaffian squared. Recognizing that A transpose equal to minus A forces a zero eigenvalue in odd dimensions gives an immediate, computation-free route to the conclusion that the matrix must be singular.