The determinant of an odd order skew symmetric matrix is always:A. ZeroB. OneC. Negative D. Depends on the matrix

The determinant of an odd-order skew-symmetric matrix is always zero.

A skew-symmetric matrix is a square matrix (i.e., it has the same number of rows and columns) where the transpose of the matrix is equal to its negation. In mathematical terms, for a skew-symmetric matrix A, it satisfies the condition:

A^T = -A

Now, when you calculate the determinant of an odd-order skew-symmetric matrix, you'll find that it is always zero. Here's why:

For any square matrix A, the determinant of its transpose (det(A^T)) is equal to the determinant of the original matrix (det(A)). In the case of a skew-symmetric matrix, A^T = -A. Therefore:

det(A^T) = det(-A)

Now, the determinant of a scalar multiple of a matrix is equal to the scalar raised to the power of the matrix's order times the determinant of the original matrix. In this case, the scalar is -1, and the matrix is A:

det(-A) = (-1)^n * det(A)

where n is the order (number of rows or columns) of the matrix. Since we are dealing with an odd-order skew-symmetric matrix, n is an odd number. Thus:

(-1)^n = (-1)^odd = -1

So, the determinant of the odd-order skew-symmetric matrix is:

det(A) = (-1) * det(A)

Since the determinant of A is equal to its negative, it must be zero:

det(A) = 0

So, the correct answer is A. Zero.