Let $A = [a_{ij}]_{n \times n}$, $n$ is odd natural number. Then determinant of matrix $(A - A^T)^{2015}$ is _____.
Step-by-Step Solution
Key Concept: Recognize that $(A - A^T)$ is skew-symmetric since $(A - A^T)^T = A^T - A = -(A - A^T)$. For any skew-symmetric matrix $B$ of odd order $n$, the determinant satisfies $|B^T| = (-1)^n|B| = -|B|$ (since $n$ is odd), which forces $|B| = 0$.
For a skew-symmetric matrix $A$ of odd order $n$, we have $A^T = -A$, so $|A^T| = |A|$. Also $|A^T| = (-1)^n|A| = -|A|$ for odd $n$. Therefore $|A| = -|A|$, which implies $|A| = 0$. Thus $(A - A^T)^{2015} = 0$.
Correct Answer: 0