Question

Let \(A\) be a \(5 \times 5\) skew-symmetric matrix, then the determinant of \(A\) is:

• A. 0
• B. 1
• C. -1
• D. 1/2

Hint:

Any skew-symmetric matrix of odd order (such as \(3 \times 3\), \(5 \times 5\), or \(7 \times 7\)) is singular, meaning its determinant is always 0.
If the order is even, the determinant is a perfect square and is generally non-zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A square matrix \(A\) of order \(n \times n\) is skew-symmetric if it is equal to the negative of its transpose:
\[ A^T = -A \]
The determinant of a skew-symmetric matrix depends on whether its order \(n\) is even or odd.

Step 2: Key Formula or Approach:
We use two fundamental properties of determinants:
1. The determinant of a matrix is equal to the determinant of its transpose:
\[ \det(A^T) = \det(A) \]
2. For any scalar \(k\) and an \(n \times n\) matrix \(A\):
\[ \det(k A) = k^n \det(A) \]

Step 3: Detailed Explanation:
Let \(A\) be a skew-symmetric matrix of order \(5 \times 5\).
By definition:
\[ A^T = -A \]
Taking the determinant on both sides of this equation:
\[ \det(A^T) = \det(-A) \]
Using the transpose property, we substitute \(\det(A^T) = \det(A)\):
\[ \det(A) = \det(-A) \]
Now, apply the scalar multiplication property with \(k = -1\) and \(n = 5\):
\[ \det(-A) = (-1)^5 \det(A) \]
Since the exponent 5 is an odd number, \((-1)^5 = -1\):
\[ \det(-A) = -\det(A) \]
Substitute this back into our equation:
\[ \det(A) = -\det(A) \]
Rearranging terms by adding \(\det(A)\) to both sides:
\[ \det(A) + \det(A) = 0 \]
\[ 2 \det(A) = 0 \implies \det(A) = 0 \]
Thus, the determinant of any skew-symmetric matrix of odd order is always exactly 0.

Step 4: Final Answer:
Therefore, the correct option is (A).