Let A be a skew-symmetric matrix of even order, then `absA`

Options

A. is a square
B. is not a square
C. is always zero
D. none of these

Answer

To solve the problem, we need to determine the determinant of a skew-symmetric matrix \( A \) of even order. Here’s a step-by-step solution:

### Step 1: Understanding Skew-Symmetric Matrices
A matrix \( A \) is skew-symmetric if it satisfies the property:
\[ A^T = -A \]
This means that the transpose of the matrix is equal to the negative of the matrix itself.

### Step 2: Consider a Skew-Symmetric Matrix of Even Order
Let’s consider a skew-symmetric matrix of order 2 (which is even). A general form of a 2x2 skew-symmetric matrix can be written as:
\[ A = \begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix} \]
where \( a \) is any real number.

### Step 3: Calculate the Determinant of the Matrix
To find the determinant of \( A \), we use the formula for the determinant of a 2x2 matrix:
\[ \text{det}(A) = ad - bc \]
For our matrix \( A \):
\[ \text{det}(A) = (0)(0) - (a)(-a) = 0 + a^2 = a^2 \]

### Step 4: Generalize for Higher Even Orders
For a skew-symmetric matrix of higher even order (e.g., 4x4, 6x6, etc.), it can be shown that the determinant will also be a perfect square. This is because the determinant of any skew-symmetric matrix of even order can be expressed as the square of some value.

### Conclusion
Thus, we conclude that the determinant of a skew-symmetric matrix \( A \) of even order is always a perfect square.

\[ \text{det}(A) = k^2 \quad \text{for some real number } k \]