d
$\therefore A^{\prime}=A$
$B^{\prime}=B$
Now $(\mathrm{A}(\mathrm{BA}))^{\prime}=(\mathrm{BA})^{\prime} \mathrm{A}^{\prime}$
$=\left(A^{\prime} B^{\prime}\right) A^{\prime}=(A B) A=A(B A)$
Similarly $((\mathrm{AB}) \mathrm{A})^{\prime}=(\mathrm{AB}) \mathrm{A}$
So, $A(B A)$ and $(A B) A$ are symmetric matrices.
Again $(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\mathrm{BA}$
Now if $\mathrm{BA}=\mathrm{AB},$ then $\mathrm{AB}$ is symmetric matrix.