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 $A B=B A$.

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

In applying one or more row operations while finding $A^{-1}$ by elementary row operations, we obtain all zeroes in one or more, then $A^{-1}$ does not exist.