Question

Which one of the following is correct
A. Skew-symmetric matrix of odd order is non-singular
B. Skew-symmetric matrix of odd order is singular
C. Skew-symmetric matrix of even order is always singular
D. None of these

Answer 1

Hint:
We have a skew-symmetric matrix, that is $A^{T} = -A$. Also, we know that $det(A) = det(A^{T})$. In the case of a skew symmetric matrix we have, $det(-A) = (-1)^{n}det(A)$. We use these concepts to solve this question.


Complete step-by-step answer:
If $A = [a_{ij}]_{n \times n}$ is a skew symmetric matrix then,
 $a_{ij} = -a_{ji}$ for all $i$ and $j$.
Also, $A^{T} = -A$.
We know that, $det(A) = det(A^{T})$
$\implies det(A^{T}) = det(-A)$
$\implies det(A) = det(-A)$
$\implies det(A) = (-1)^{n}det(A)$
Suppose, the skew-symmetric matrix is of odd order, then $(-1)^{n} = -1$.
Therefore, we have $det(A) = -det(A)$
$\implies det(A)+det(A) = 0$
$\implies 2det(A) = 0$
$\implies det(A) = 0$
This means that A is a singular matrix.

Hence, the correct answer is Option B.

Note: To solve these kinds of questions remember the properties of skew-symmetric matrices. Some of the properties are: the diagonal elements of skew-symmetric are zero, so its trace is zero. The scalar multiple of a skew-symmetric matrix is a skew-symmetric matrix.