GATE CS 2024 Set 2 Q37 — Linear Algebra: Matrices, Determinants, System of Linear Equations

Let $A$ be an $n \times n$ matrix, and let $B$ be the matrix obtained by swapping two rows of $A$. This row operation can be represented as multiplying $A$ on the left by an elementary matrix $E$, i.e., $B = EA$.

Let's analyze each option:

• Option A: Swapping any two rows of a matrix changes the sign of its determinant.
  Mathematically, $\det(B) = \det(E) \cdot \det(A)$. Since the determinant of a row-swap elementary matrix $E$ is $-1$, we have:$$\det(B) = -\det(A)$$Thus, Option A is TRUE.
• Option B: A matrix is invertible if and only if its determinant is non-zero.
  Since $\det(B) = -\det(A)$, we have $\det(B) \neq 0$ if and only if $\det(A) \neq 0$. Therefore, if $A$ is invertible, $B$ must also be invertible.
  Thus, Option B is TRUE.
• Option C: If $A$ is symmetric, $B$ is not necessarily symmetric.
  For example, consider the symmetric matrix:$$A = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}$$Swapping row 1 and row 2 gives:$$B = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$$Since $B^T = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \neq B$, $B$ is not symmetric.
  Thus, Option C is FALSE.
• Option D: If the trace of $A$ is zero, the trace of $B$ is not necessarily zero.
  For example, consider:$$A = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}$$Here, $\text{Tr}(A) = 1 + (-1) = 0$. Swapping row 1 and row 2 gives:$$B = \begin{pmatrix} 3 & -1 \\ 1 & 2 \end{pmatrix}$$The trace of $B$ is $\text{Tr}(B) = 3 + 2 = 5 \neq 0$.
  Thus, Option D is FALSE.