(i) $(A B)^{\prime}=B^{\prime} A^{\prime}$

(ii) $(k A)^{\prime}=k A^{\prime}$

(iii) $[k(A-B)]^{\prime}=k\left(A^{\prime}-B^{\prime}\right)$

If $A$ is a skew-symmetric, then $k A$ is a skew-symmetric matrix (where, $k$ is any scalar).

$$\left[\because A^{\prime}=-A \Rightarrow(k A)^{\prime}=k(A)^{\prime}=-(k A)\right]$$

(i) $A B-B A$ is a skew-symmetric matrix.

Since,

$$\begin{array}{rlr} {[A B-B A]^{\prime}} & =(A B)^{\prime}-(B A)^{\prime} & \\ & =B^{\prime} A^{\prime}-A^{\prime} B^{\prime} & {\left[\because(A B)^{\prime}=B^{\prime} A^{\prime}\right]} \\ & =B A-A B & {\left[\because A^{\prime}=A \text { and } B^{\prime}=B\right]} \\ & =-[A B-B A] & \end{array}$$

So, $[A B-B A]$ is a skew-symmetric matrix.

(ii) $[B A-2 A B]$ is a neither symmetric nor skew-symmetric matrix.

$$\begin{aligned} \therefore \quad(B A-2 A B)^{\prime} & =(B A)^{\prime}-2(A B)^{\prime} \\ & =A^{\prime} B^{\prime}-2 B^{\prime} A^{\prime} \\ & =A B-2 B A \\ & =-(2 B A-A B) \end{aligned}$$

So, $[B A-2 A B]$ is neither symmetric nor skew-symmetric matrix.

If $A$ is a symmetric matrix, then $B^{\prime} A B$ is a symmetric metrix.

$$\begin{aligned} \because\quad{\left[B^{\prime} A B\right]^{\prime} } & =\left[B^{\prime}(A B)\right]^{\prime} \\ & =(A B)^{\prime}\left(B^{\prime}\right)^{\prime} \quad \left[\because(A B)^{\prime}=B^{\prime} A^{\prime}\right]\\ & =B^{\prime} A^{\prime} B\\ & =\left[B^{\prime} A^{\prime} B\right] \quad\left[\because A^{\prime}=A\right] \end{aligned}$$

So, $B^{\prime} A B$ is a symmetric matrix.

If $A$ and $B$ are symmetric matrices of same order, then $A B$ is symmetric if and only if $A B=B A$.

$$\begin{aligned} \therefore\quad & (A B)^{\prime} \\ = & B^{\prime} A^{\prime}=B A \\ = & A B \end{aligned} \quad[\because A B=B A]$$