Question:

Step 1: If $A$ is a non-singular matrix, then its inverse $A^{-1}$ exists. Given the matrix equation $AB = O$, multiply both sides by $A^{-1}$ from the left: $$ A^{-1}(AB) = A^{-1}O $$ $$ (A^{-1}A)B = O $$ $$ IB = O $$ $$ B = O $$ Thus, $B$ must be a null matrix. Step 2: For square matrices $A, B, C$ of order 2, with $\det(A) = 2$, $\det(B) = 3$, and $\det(C) = 4$. The property of determinants states that for a scalar $k$ and an $n \times n$ matrix $M$, $\det(kM) = k^n \det(M)$. Here, the order of the matrices is $n=2$. Also, $\det(XYZ) = \det(X)\det(Y)\det(Z)$. Therefore, $$ \det(3ABC) = 3^2 \det(A)\det(B)\det(C) = 9 \times 2 \times 3 \times 4 = 216. $$ Step 3: For a square matrix $A$ of order 3 with $\det(A) = \dfrac{1}{2}$. First, calculate the determinant of $A^{-1}$: $$ \det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{1/2} = 2. $$ Next, use the property that for an $n \times n$ matrix $M$, $\det(\text{adj}(M)) = (\det M)^{n-1}$. In this case, $M = A^{-1}$ and the order $n=3$. $$ \det(\text{adj}(A^{-1})) = (\det(A^{-1}))^{3-1} = (2)^2 = 4. $$ Step 4: A skew-symmetric matrix of odd order has a determinant of 0, which means it is singular. For example, any $3 \times 3$ skew-symmetric matrix is singular. However, a skew-symmetric matrix of even order can be non-singular. Consider the $2 \times 2$ skew-symmetric matrix: $$ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $$ Its determinant is $(0)(0) - (1)(-1) = 1$. Since the determinant is non-zero, this matrix is non-singular. Therefore, it is not true that every skew-symmetric matrix is singular.