Let `A=[a_(ij)]_(nxxn)`n is odd. Then det `((A-A^(T))^(2009))` is equal to

Answer

To solve the problem, we need to find the determinant of the matrix \( (A - A^T)^{2009} \), where \( A \) is an \( n \times n \) matrix and \( n \) is odd.

### Step-by-Step Solution:

1. **Understanding the Matrix \( A - A^T \)**:
- The matrix \( A^T \) is the transpose of \( A \).
- The difference \( A - A^T \) results in a skew-symmetric matrix. A skew-symmetric matrix \( B \) satisfies the property \( B = -B^T \).

2. **Properties of Skew-Symmetric Matrices**:
- A key property of skew-symmetric matrices is that their determinant is zero when the order of the matrix is odd. Since \( n \) is given to be odd, we can conclude that:
\[ \det(A - A^T) = 0 \]

3. **Calculating the Determinant of \( (A - A^T)^{2009} \)**:
- Using the property of determinants, we know that:
\[ \det(B^k) = (\det(B))^k \]
- Applying this to our case, we have:
\[ \det((A - A^T)^{2009}) = (\det(A - A^T))^{2009} \]
- Since we established that \( \det(A - A^T) = 0 \), we can substitute this into our equation:
\[ \det((A - A^T)^{2009}) = (0)^{2009} = 0 \]

4. **Conclusion**:
- Therefore, the final result is:
\[ \det((A - A^T)^{2009}) = 0 \]

### Final Answer:
\[ \det((A - A^T)^{2009}) = 0 \]