GATE CS 2025 Set 2 Q.44: Linear Algebra

Consider a system of linear equations $PX = Q$ where $P \in \mathbb{R}^{3 \times 3}$ and $Q \in \mathbb{R}^{3 \times 1}$ . Suppose $P$ has an LU decomposition, $P = LU$ , where

L = \begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix} \text{ and } U = \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix}.

Which of the following statement(s) is/are TRUE?

The system $PX = Q$ can be solved by first solving $LY = Q$ and then $UX = Y$ .

If $P$ is invertible, then both $L$ and $U$ are invertible.

If $P$ is singular, then at least one of the diagonal elements of $U$ is zero.

If $P$ is symmetric, then both $L$ and $U$ are symmetric.

Official answer and solution

Official answer: A, B and C
From: Official GATE 2025 answer key

Solution

Use $P = LU$ and $\det(P) = \det(L)\det(U) = 1 \cdot (u_{11} u_{22} u_{33})$ to evaluate each statement.

(A) is TRUE: Substituting $P = LU$ into $PX = Q$ gives $L(UX) = Q$ . Defining $Y = UX$ , we first solve $LY = Q$ by forward substitution and then $UX = Y$ by back substitution.
(B) is TRUE: Since $\det(P) = \det(L)\det(U)$ , if $P$ is invertible ( $\det(P) \neq 0$ ), then both $\det(L) \neq 0$ and $\det(U) \neq 0$ , so both $L$ and $U$ are invertible.
(C) is TRUE: Because $L$ is unit lower-triangular, $\det(L) = 1$ , so $\det(P) = \det(U) = u_{11} u_{22} u_{33}$ . If $P$ is singular ( $\det(P) = 0$ ), then $u_{11} u_{22} u_{33} = 0$ , meaning at least one diagonal element $u_{ii}$ of $U$ is zero.
(D) is FALSE: A lower-triangular matrix $L$ and upper-triangular matrix $U$ are symmetric only if they are diagonal, which is false for any non-diagonal symmetric matrix $P$ .

Final answer

Options (A), (B), (C)

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2025 Computer Science and Information Technology - Session 2 (CS2)
Question: 44 of 65
Section: Computer Science and Information Technology

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