Which of the following statement('s) is/are TRUE?
(A) Skew symmetric matrix of even order is always symmetric
(B) Skew symmetric matrix of odd order is non-singular
(C) Skew symmetric matrix of odd order is singular
(D) Skew symmetric matrix is always square matrix
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (C) and (D) only
A matrix A is skew-symmetric when \(A^T = -A\). For this, \(A^T\) must have the same order as A, so A must be square.
If A is also symmetric, \(A^T = A\) as well:
\(A = -A\)
\(A = O\)
For a skew-symmetric A of order n:
\(\det A = \det A^T\)
\(\det A = \det(-A) = (-1)^n\det A\)
When n is odd, \(\det A = -\det A\), so \(\det A = 0\).
So a skew-symmetric matrix is always square, it is symmetric only when it is the zero matrix, and its determinant is 0 when its order is odd.
Checking the statements
(A) Skew symmetric matrix of even order is always symmetric — Incorrect. Only the zero matrix is both; the order-2 matrix with \(a_{12} = 1, a_{21} = -1\) is skew-symmetric but not symmetric.
(B) Skew symmetric matrix of odd order is non-singular — Incorrect. Its determinant is 0, so it is singular.
(C) Skew symmetric matrix of odd order is singular — Correct. Its determinant is 0.
(D) Skew symmetric matrix is always square matrix — Correct. \(A^T = -A\) needs \(A^T\) and A of the same order.