Show that the matrix $B^{\prime}AB$ is symmetric or skew-symmetric according as $A$ is symmetric or skew-symmetric.

(A) We suppose that $A$ is a symmetric matrix,then $A^{\prime} = A$ ......... $(1)$
Consider
$(B^{\prime}AB)^{\prime} = \{B^{\prime}(AB)\}^{\prime}$
$= (AB)^{\prime}(B^{\prime})^{\prime}$ $[(AB)^{\prime} = B^{\prime}A^{\prime}]$
$= B^{\prime}A^{\prime}(B)$ $[(B^{\prime})^{\prime} = B]$
$= B^{\prime}(A^{\prime}B)$
$= B^{\prime}(AB)$ $[$Using $(1)]$
$\therefore (B^{\prime}AB)^{\prime} = B^{\prime}AB$
Thus,if $A$ is a symmetric matrix,then $B^{\prime}AB$ is a symmetric matrix.
Now,we suppose that $A$ is a skew-symmetric matrix,then $A^{\prime} = -A$ ......... $(2)$
Consider
$(B^{\prime}AB)^{\prime} = \{B^{\prime}(AB)\}^{\prime} = (AB)^{\prime}(B^{\prime})^{\prime}$
$= (B^{\prime}A^{\prime})B$
$= B^{\prime}(-A)B$ $[$Using $(2)]$
$= -B^{\prime}AB$
$\therefore (B^{\prime}AB)^{\prime} = -B^{\prime}AB$
Thus,if $A$ is a skew-symmetric matrix,then $B^{\prime}AB$ is a skew-symmetric matrix.
Hence,the matrix $B^{\prime}AB$ is symmetric or skew-symmetric according as $A$ is symmetric or skew-symmetric.