If $A$ is a symmetric matrix, then $A^3$ is a symmetric matrix.

$$\begin{aligned} \because\quad A^{\prime} & =A \\ \therefore\quad \left(A^3\right)^{\prime} & =A^3 \\ & =A^3\quad \left[\because\left(A^{\prime}\right)^n=\left(A^n\right)^{\prime}\right] \end{aligned}$$

If $A$ is a skew-symmetric matrix, then $A^2$ is a symmetric matrix.

$$\begin{aligned} \because\quad A^{\prime} & =-A \\ \therefore\quad \left(A^2\right)^{\prime} & =\left(A^{\prime}\right)^2 \\ & =(-A)^2 \\ & =A^2 \end{aligned} \quad\left[\because A^{\prime}=-A\right]$$

So, A$^2$ is a symmetric matrix.

(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.