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,

If A is a skew-symmetric matrix of order 3, then prove that det A = 0 .

If A is a skew-symmetric matrix of order 3, then prove that det A=0.

Assertion: Determinant of a skew symmetric matrix of order 3 is zero. Reason: For any matix A, det(A^T)=det(A) and det(-S)=-det(S) (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true and R is not the correct explanation of A (C) A is true but R is false. (D) A is false but R is true.

If A is a skew symmetric matrix of order 3, then write the value of det A.

If A is skew symmetric matrix of order 3 then det A is (A)0 (B)-A (C)A (D)none of these

If A is a square matrix of order 3, then the true statement is (where I is unit matrix) (1)det(-A)=-det A(2) det A=0 (3) det(A+I)=1+det A(4)det(2A)=2det A

If A is a matrix of order 3, then det(kA) is

If det(adjA)=|A|^(2), then the order of matrix A is

Assertion (A): If A is skew symmetric matrix of order 3 then its determinant should be zero Reason (R): If A is square matrix then det(A)=det(A')=det(A')