GATE CS 2025 Set 2 — Question 44

We analyze each statement:(A) Given $PX = Q$ and $P = LU$, we have $LUX = Q$. Let $UX = Y$. Then the equation becomes $LY = Q$. Since $L$ is lower triangular, we can solve $LY = Q$ for $Y$ using forward substitution. Once $Y$ is known, we solve $UX = Y$ for $X$ using backward substitution (since $U$ is upper triangular). This is the standard procedure for solving linear systems using LU decomposition. Thus, statement (A) is TRUE.(B) The determinant of a product is the product of determinants: $\det(P) = \det(L)\det(U)$. Since $L$ is a lower triangular matrix with all diagonal entries equal to 1, its determinant is the product of its diagonal entries: $\det(L) = 1 \times 1 \times 1 = 1$. Therefore, $\det(P) = 1 \cdot \det(U) = \det(U)$.
If $P$ is invertible, then $\det(P) \neq 0$, which implies $\det(U) \neq 0$. Thus, $U$ is invertible. Since $\det(L) = 1 \neq 0$, $L$ is always invertible. Hence, both $L$ and $U$ are invertible. Statement (B) is TRUE.(C) If $P$ is singular, then $\det(P) = 0$. From the relation $\det(P) = \det(U)$, it follows that $\det(U) = 0$. Since $U$ is an upper triangular matrix, its determinant is the product of its diagonal elements: $\det(U) = u_{11} u_{22} u_{33}$. For the product to be zero, at least one of the diagonal elements $u_{11}, u_{22}, u_{33}$ must be zero. Statement (C) is TRUE.(D) If $P$ is symmetric, $P = P^T$. However, $L$ is lower triangular and $U$ is upper triangular. A triangular matrix is symmetric if and only if it is a diagonal matrix. In general LU decomposition, $L$ and $U$ are not diagonal matrices, so they are not symmetric. For example, let $P = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$. Then $L = \begin{bmatrix} 1 & 0 \\ 0.5 & 1 \end{bmatrix}$ and $U = \begin{bmatrix} 2 & 1 \\ 0 & 1.5 \end{bmatrix}$. Neither $L$ nor $U$ is symmetric. Statement (D) is FALSE.