The GATE Grind

GATE 2025 CS (CS2) – Question 44

Engineering Mathematics · Linear Algebra · 2 marks · Multiple select

Consider a system of linear equations $PX = Q$ where $P \in \mathbb{R}^{3\times3}$ and $Q \in \mathbb{R}^{3\times1}$. Suppose $P$ has an LU decomposition, $P = LU$, where

$L = \begin{bmatrix}1 & 0 & 0\\l_{21} & 1 & 0\\l_{31} & l_{32} & 1\end{bmatrix}$ 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?

  1. The system $PX = Q$ can be solved by first solving $LY = Q$ and then $UX = Y$.
  2. If $P$ is invertible, then both $L$ and $U$ are invertible.
  3. If $P$ is singular, then at least one of the diagonal elements of $U$ is zero.
  4. If $P$ is symmetric, then both $L$ and $U$ are symmetric.

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Show answer and explanation

Correct answer: (A) The system $PX = Q$ can be solved by first solving $LY = Q$ and then $UX = Y$.; (B) If $P$ is invertible, then both $L$ and $U$ are invertible.; (C) If $P$ is singular, then at least one of the diagonal elements of $U$ is zero.

Explanation

Forward then back substitution solves $LUX = Q$. $\det P = \det L \cdot \det U = \prod u_{ii}$ since $\det L = 1$, which gives B and C. Symmetric $P$ gives $U = DL^T$, not a symmetric $U$ in general, so D is false.