Question

Let $A$ be a $5 \times 5$ skew-symmetric matrix of real numbers, then which of the following is true

• A. $A$ is singular matrix
• B. $A$ is invertible
• C. All the eigen values are complex number
• D. $|A| \ne 0$

Hint:

Remember this general rule:
- Odd-order real skew-symmetric matrices are always singular (\( \det = 0 \)).
- Even-order real skew-symmetric matrices have non-negative, perfect square determinants (\( \det \ge 0 \)).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A real square matrix $A$ is skew-symmetric if it is equal to the negative of its transpose:
\[ A^T = -A \]
The properties of skew-symmetric matrices depend on whether their dimension (order) $n$ is even or odd.

Step 2: Detailed Explanation:
Let us analyze the determinant of a skew-symmetric matrix of odd order:
Let $A$ be an $n \times n$ skew-symmetric matrix where $n = 5$ (an odd integer).
Take the determinant of both sides of the defining relation \( A^T = -A \):
\[ \det(A^T) = \det(-A) \]
Using the properties of determinants:
1. The determinant of a transpose is equal to the determinant of the original matrix: \( \det(A^T) = \det(A) \).
2. For any $n \times n$ matrix, factoring out a scalar \( c \) yields: \( \det(c A) = c^n \det(A) \).
Applying these properties with \( c = -1 \):
\[ \det(A) = (-1)^n \det(A) \]
Since the order of our matrix is $n = 5$ (odd):
\[ \det(A) = (-1)^5 \det(A) = -\det(A) \]
Rearranging terms:
\[ \det(A) + \det(A) = 0 \]
\[ 2 \det(A) = 0 \implies \det(A) = 0 \]
Since the determinant of $A$ is zero, the matrix is singular and cannot be inverted.
This proves that any real skew-symmetric matrix of odd order is singular, matching Option (A).

Step 3: Final Answer:
$A$ is a singular matrix.