Question

If \( A \) is a skew-symmetric matrix of order \( 5 \times 5 \), what is the value of its determinant \( |A| \)?

• A. \( 0 \)
• B. \( 1 \)
• C. \( -1 \)
• D. \( 5 \)

Hint:

This is a pure property question that requires no manual arithmetic calculations. If you see the words skew-symmetric combined with an odd order number (like \( 1, 3, 5, \) or \( 7 \)), select 0 immediately to save time.

Answer 1

The Correct Option is A

Concept: A square matrix is classified as skew-symmetric if its transpose equals its negative (\( A^T = -A \)). We can find its scalar determinant value by combining the scaling properties of determinants with transpose laws.

Step 1: Apply determinant operations across the matrix definition.
Take the determinant of both sides of the structural definition equation: \[ |A^T| = |-A| \]

Step 2: Apply standard determinant property scaling rules.
1. The determinant of any matrix is equal to the determinant of its transpose: \( |A^T| = |A| \).
2. Factoring out a scalar multiplier of \( -1 \) from a matrix determinant requires raising it to the power of the matrix order \( n \): \( |-A| = (-1)^n |A| \).
Substitute these core properties back into our expression: \[ |A| = (-1)^n |A| \]

Step 3: Evaluate the expression for the given odd matrix order \( n = 5 \).
The problem states that the matrix has a spatial size order of \( 5 \). Since 5 is an odd integer, the factor \( (-1)^5 = -1 \): \[ |A| = -|A| \quad \Rightarrow \quad |A| + |A| = 0 \quad \Rightarrow \quad 2|A| = 0 \quad \Rightarrow \quad |A| = 0 \] This proves that the determinant of any odd-order skew-symmetric matrix is always zero.