Statement-1: Determinant of a skew-symmetric matrix of order 3 is zero.
Statement-2: For any matrix A, $Det (A) = Det (A^T)$ and $Det (-A)=-Det (A)$ where $Det (B)$ denotes the determinant of matrix B. Then,
Answer & explanation
Correct answer: option 3
Let A be a skew-symmetric matrix of order 3. Then,
$A^T =- A$
$⇒Det (A^T) = Det (-A)$
$⇒Det (A)=(-1)^3\, Det (A)$
$⇒Det (A)=-Det (A)$
$⇒2\, Det (A) = 0$
$⇒Det (A) = 0$
So, statement-1 is true.
For any square matrix of order n, we have
$Det (A^T) = Det (A)$ and $Det (-A)=(-1)^n\, Det (A)$
So, statement-2 is not true.