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(C) Given that $A$ and $B$ are symmetric matrices,we have $A^{\prime} = A$ and $B^{\prime} = B$. For Statement $-1$: Consider $(A(BA))^{\prime} = (BA)

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Statement $-1$ is true,Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$.

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(C) Given that $A$ and $B$ are symmetric matrices,we have $A^{\prime} = A$ and $B^{\prime} = B$. For Statement $-1$: Consider $(A(BA))^{\prime} = (BA)^{\prime} A^{\prime} = (A^{\prime} B^{\prime}) A^{\prime} = (AB)A = A(BA)$. Since $(A(BA))^{\prime} = A(BA)$,$A(BA)$ is symmetric. Similarly,$((AB)A)^{\prime} = A^{\prime} (AB)^{\prime} = A(B^{\prime} A^{\prime}) = A(BA) = (AB)A$. Thus,$(AB)A$ is also symmetric. So,Statement $-1$ is true. For Statement $-2$: Consider $(AB)^{\prime} = B^{\prime} A^{\prime} = BA$. If $AB$ is symmetric,then $(AB)^{\prime} = AB$,which implies $BA = AB$. Conversely,if $BA = AB$,then $(AB)^{\prime} = BA = AB$,so $AB$ is symmetric. Thus,Statement $-2$ is true. However,Statement $-2$ describes the condition for $AB$ to be symmetric,whereas Statement $-1$ deals with the symmetry of $A(BA)$ and $(AB)A$ regardless of commutativity. Therefore,Statement $-2$ is not a correct explanation for Statement $-1$.

Let $A$ and $B$ be two symmetric matrices of order $3$.
Statement $-1$: $A(BA)$ and $(AB)A$ are symmetric matrices.
Statement $-2$: $AB$ is a symmetric matrix if the matrix multiplication of $A$ with $B$ is commutative.

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Let $A$ and $B$ be two symmetric matrices of order $3$. Statement $-1$: $A(BA)$ and $(AB)A$ are symmetric matrices. Statement $-2$: $AB$ is a symmetric matrix if the matrix multiplication of $A$ with $B$ is commutative.