If $A$ is a $3 \times 3$ real skew-symmetric matrix, then the determinant of $A$ is always equal to:
Options:
(A) 1
(B) -1
(C) 0
(D) Any positive real number
Answer Key:
C
Step-by-Step Solution & Explanation
For any skew-symmetric matrix $A$ of order $n$, $A^T = -A$.
Taking determinants on both sides:
$\det(A^T) = \det(-A) = (-1)^n \det(A)$
Since $\det(A^T) = \det(A)$ and here $n = 3$ is odd:
$\det(A) = -\det(A) \implies 2\det(A) = 0 \implies \det(A) = 0$
Therefore, the determinant of any odd-order skew-symmetric matrix is always $0$.