Let $A$ be an $n \times n$ matrix over the set of all real numbers $\mathbb{R}$. Let $B$ be a matrix obtained from $A$ by swapping two rows. Which of the following statements is/are TRUE?

Let $A$ be an $n \times n$ matrix and $B$ be obtained by swapping two rows of $A$.

• Option A: By elementary row operations, interchanging two rows of a matrix negates the determinant. Thus, $\det(B) = -\det(A)$. This statement is TRUE.

• Option B: A matrix is invertible if and only if its determinant is non-zero. Since $\det(B) = -\det(A)$, if $\det(A) \neq 0$, then $\det(B) \neq 0$. Therefore, if $A$ is invertible, $B$ is also invertible. This statement is TRUE.

• Option C: Swapping two rows of a symmetric matrix does not necessarily yield a symmetric matrix. For example, consider the symmetric matrix:

  $$A = \begin{pmatrix} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

  Swapping row 1 and row 3 gives:

  $$B = \begin{pmatrix} 0 & 0 & 1 \\ 2 & 1 & 0 \\ 1 & 2 & 0 \end{pmatrix}$$

  Here $B^T \neq B$, so $B$ is not symmetric. This statement is FALSE.

• Option D: Consider the matrix with trace $0$:

  $$A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$

  Here $\operatorname{trace}(A) = 2 - 1 - 1 = 0$. Swapping row 1 and row 3 gives:

  $$B = \begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ 2 & 0 & 0 \end{pmatrix}$$

  The trace of $B$ is $0 + (-1) + 0 = -1 \neq 0$. This statement is FALSE.

Hence, the correct statements are A and B.