We suppose that $A$ is a symmetric matrix, then $A^{\prime}=A$       ……… $(1)$

Consider

$(B^{\prime} A B)^{\prime} =\left\{B^{\prime}(A B)\right\}^{\prime}$

$=(A B)^{\prime}(B)^{\prime}$                                    $[(A B)^{\prime}=B^{\prime} A^{\prime}]$

$=B^{\prime} A^{\prime}(B)$                                    $[(B^{\prime})^{\prime}=B]$

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

$=B^{\prime}(A B)$                        $[ $ Using $(1)]$

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

Thus, if $A$ is a symmetric matrix, then $B^{\prime}A B$ is a symmetric matrix.

Now, we suppose that $A$ is a skew – symmetric matrix.

Then, $A^{\prime}=A$

-Consider

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

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

$=-B^{\prime} A B$

$\therefore  $ $(B^{\prime} A B)^{\prime}=-B^{\prime} A B$

Thus, if $A$ is a skew – symmetric matrix then $B^{\prime} A B$ is a skew – symmetric matrix.

Hence, if $A$ is a asymmetric or skew – symmetric matrix, then $B^{\prime} A B$ is a symmetric or skew – ymmetric matrix accordingly.