Question:
The determinant of a skew-symmetric matrix of odd order is
The determinant of a skew-symmetric matrix of odd order is
Show Hint
Skew-symmetric + Odd Order = Always Singular (Determinant is 0).
Updated On: Jul 3, 2026
- 1
- $x^2$
- -1
- 0
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The Correct Option is D
Solution and Explanation
Step 1: Concept
A matrix $A$ is skew-symmetric if $A^T = -A$.
Step 2: Meaning
Taking the determinant of both sides: $\det(A^T) = \det(-A)$. Since $\det(A^T) = \det(A)$ and $\det(-A) = (-1)^n \det(A)$ for an $n \times n$ matrix, we have $\det(A) = (-1)^n \det(A)$.
Step 3: Analysis
If $n$ is odd, then $(-1)^n = -1$, which leads to the equation $\det(A) = -\det(A)$.
Step 4: Conclusion
The only number that equals its negative is zero ($2\det(A) = 0 \Rightarrow \det(A) = 0$). Final Answer: (D)
A matrix $A$ is skew-symmetric if $A^T = -A$.
Step 2: Meaning
Taking the determinant of both sides: $\det(A^T) = \det(-A)$. Since $\det(A^T) = \det(A)$ and $\det(-A) = (-1)^n \det(A)$ for an $n \times n$ matrix, we have $\det(A) = (-1)^n \det(A)$.
Step 3: Analysis
If $n$ is odd, then $(-1)^n = -1$, which leads to the equation $\det(A) = -\det(A)$.
Step 4: Conclusion
The only number that equals its negative is zero ($2\det(A) = 0 \Rightarrow \det(A) = 0$). Final Answer: (D)
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