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,

Statement-1 is True, Statement-2 is False.

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.