Show that the matrix is symmetric or skew symmetric according as A is symmetric or skew symmetric.

We suppose that A is a symmetric matrix, then … (1)

Consider

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

$\begin{array}{ll}=(A B)^{\prime}\left(B^{\prime}\right)^{\prime} & {\left[(A B)^{\prime}=B^{\prime} A^{\prime}\right]} \\ =B^{\prime} A^{\prime}(B) & {\left[\left(B^{\prime}\right)^{\prime}=B\right]}\end{array}$

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

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

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

Thus, if A is a symmetric matrix, thenis a symmetric matrix.

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

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

Consider

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

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

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

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