Question:

**Step 1: Verify Statement-1** Given that $A^T = A$ and $B^T = B$. Statement-1 asserts $A(BA)^T = A(BA)$. Let's evaluate the left-hand side (LHS): $$A(BA)^T = A(A^T B^T)$$ Since $A^T = A$ and $B^T = B$, we substitute these into the expression: $$A(A^T B^T) = A(AB) = A^2B$$ Now, let's evaluate the right-hand side (RHS): $$A(BA) = ABA$$ For Statement-1 to be true, we must have $A^2B = ABA$. This statement is true. **Step 2: Verify Statement-2** Statement-2 asserts $(AB)^T = B^T A^T = BA$ since $AB$ is commutative. First, let's verify the equality $(AB)^T = B^T A^T = BA$: Using the property of transpose of a product, $(AB)^T = B^T A^T$. Given $A^T = A$ and $B^T = B$, we substitute these into the expression: $$B^T A^T = BA$$ Thus, the equality $(AB)^T = BA$ is mathematically correct. Next, let's examine the reason provided: "since $AB$ is commutative". For matrices $A$ and $B$, $AB$ is commutative means $AB = BA$. However, for general symmetric matrices $A$ and $B$, it is not necessarily true that $AB = BA$. Consider the following counterexample: Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$. Both $A$ and $B$ are symmetric matrices. Calculate $AB$: $$AB = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 0 \cdot 1 & 1 \cdot 1 + 0 \cdot 1 \\ 0 \cdot 1 + 2 \cdot 1 & 0 \cdot 1 + 2 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$$ Calculate $BA$: $$BA = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 1 \cdot 0 & 1 \cdot 0 + 1 \cdot 2 \\ 1 \cdot 1 + 1 \cdot 0 & 1 \cdot 0 + 1 \cdot 2 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix}$$ Since $AB \neq BA$, the matrices $A$ and $B$ do not commute. Therefore, the claim that "$AB$ is commutative" is false. A statement that includes a false justification is considered false. Thus, Statement-2 is false.