If A be a skew symmetric matrix of odd order, then `|A|` is equal to

Options

A. 0
B. 1
C. `-1`
D. None

Answer

To solve the problem, we need to show that the determinant of a skew-symmetric matrix of odd order is equal to zero. Here’s a step-by-step solution:

### Step 1: Define a Skew-Symmetric Matrix
A skew-symmetric matrix \( A \) is defined such that \( A^T = -A \), where \( A^T \) is the transpose of \( A \).

### Step 2: Properties of Determinants
Recall that the determinant of a matrix has the following property:
\[ |A^T| = |A| \]
This means the determinant of the transpose of a matrix is equal to the determinant of the matrix itself.

### Step 3: Apply the Skew-Symmetric Property
For a skew-symmetric matrix \( A \):
\[ |A^T| = |-A| \]
Using the property of determinants, we have:
\[ |-A| = (-1)^n |A| \]
where \( n \) is the order of the matrix.

### Step 4: Consider the Order of the Matrix
Since \( A \) is a skew-symmetric matrix of odd order, we have \( n \) as an odd number. Therefore:
\[ (-1)^n = -1 \]
This implies:
\[ |-A| = -|A| \]

### Step 5: Set Up the Equation
From the previous steps, we have:
\[ |A| = -|A| \]
Adding \( |A| \) to both sides gives:
\[ |A| + |A| = 0 \implies 2|A| = 0 \]
Thus:
\[ |A| = 0 \]

### Conclusion
Therefore, the determinant of a skew-symmetric matrix of odd order is:
\[ |A| = 0 \]