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
$
  \det ({A^T}) = {( - 1)^n}\det (A) \\
  \because n{\text{ is odd,}}( - 1) = - 1 \\
   \Rightarrow \det ({A^T}) = - \det (A)...............(2) \\
$
Substituting the value of $\det ({A^T})$ in equation (1), we have
$
   \Rightarrow \det (A) = - \det (A) \\
   \Rightarrow 2\det (A) = 0 \\
   \Rightarrow \det (A) = 0 \\
$
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.