If $A$ and $B$ are symmetric matrices of the same order,then show that $AB$ is symmetric if and only if $A$ and $B$ commute,that is $AB = BA$.

(N/A) Since $A$ and $B$ are symmetric matrices,we have $A^{\prime} = A$ and $B^{\prime} = B$.
Part $1$: Assume $AB$ is symmetric.
Then $(AB)^{\prime} = AB$.
Using the property of transpose $(AB)^{\prime} = B^{\prime}A^{\prime}$,we get $B^{\prime}A^{\prime} = AB$.
Since $A^{\prime} = A$ and $B^{\prime} = B$,this implies $BA = AB$.
Part $2$: Conversely,assume $AB = BA$.
We need to show that $AB$ is symmetric,i.e.,$(AB)^{\prime} = AB$.
$(AB)^{\prime} = B^{\prime}A^{\prime}$.
Since $A$ and $B$ are symmetric,$B^{\prime}A^{\prime} = BA$.
Given $BA = AB$,we have $(AB)^{\prime} = AB$.
Thus,$AB$ is symmetric.