GATE CS 2025 Set 2 | Question 44

MSQ · 2 marks

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?

  1. Option A: The system \(PX = Q\) can be solved by first solving \(LY = Q\) and then \(UX = Y\). (correct)
  2. Option B: If \(P\) is invertible, then both \(L\) and \(U\) are invertible. (correct)
  3. Option C: If \(P\) is singular, then at least one of the diagonal elements of \(U\) is zero. (correct)
  4. Option D: If \(P\) is symmetric, then both \(L\) and \(U\) are symmetric.

Answer

(A), (B), (C)

Explanation

  • A true. \(PX = Q\) is \(L(UX) = Q\). Put \(Y = UX\): solve \(LY = Q\) by forward substitution, then \(UX = Y\) by back substitution.
  • B true. \(L\) has a unit diagonal, so \(\det L = 1\) and \(\det P = \det L \cdot \det U = u_{11}u_{22}u_{33}\). If \(P\) is invertible this is non-zero, so \(U\) is invertible, and \(L\) always is.
  • C true. If \(P\) is singular, \(u_{11}u_{22}u_{33} = \det P = 0\), so some \(u_{ii} = 0\).
  • D false. A triangular matrix is symmetric only if it is diagonal. The symmetric \(P = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) has \(l_{21} = 1\) and \(u_{12} = 1\) with \(l_{12} = u_{21} = 0\), so neither factor is symmetric.

Answer: A, B, C