Question

The determinant of an odd order skew symmetric matrix is always:
A. Zero
B. One
C. Negative
D. Depends on the matrix

Answer 1

Hint: A matrix is skew- symmetric if and if it is the opposite of its transpose and the general properties of determinants is given as $det(A)=det(A^T)\text{ and }det(-A)=(-1{)}^{n}det(A)$ where $n$ is number of rows or columns of square matrix. Use these two properties to reach the answer.

Complete Step-by-Step solution:
Let A be a skew- symmetric matrix of $n\times n$ order, where $n$ is odd
We know that the determinant of A is always equal to the determinant of its transpose.
$det(A)=det(A^T)...................(1)$
However, since A is a skew-symmetric matrix where
$a_{ij}=-a_{ji}\text{ (i,j}$ are rows and column numbers).
Therefore, in case of skew-symmetric matrix
$$\begin{gathered} det(A^T)=(-1{)}^{n}det(A) \\ \because n\text{ is odd,}(-1)=-1 \\ \Rightarrow det(A^T)=-det(A)...............(2) \end{gathered}$$
Substituting the value of $det(A^T)$ in equation (1), we have
$$\begin{gathered} \Rightarrow det(A)=-det(A) \\ \Rightarrow 2det(A)=0 \\ \Rightarrow det(A)=0 \end{gathered}$$
Hence, the determinant of an odd skew- symmetric matrix is always zero and the correct option is A.

Note: In order to solve these types of questions, remember all the properties of the matrix. Some of the properties of skew symmetric matrix are - A scalar multiple of a skew symmetric matrix is skew- symmetric matrix. The elements on the diagonal of a skew-symmetric matrix are zero, and therefore its trace equals to zero. Trace of the matrix is the sum of its diagonal elements.