Consider a system of linear equations \( P X = Q \) 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:
Which of the following statement(s) is/are TRUE?
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{In LU decomposition, the determinant of \( P \) is the product of the diagonal elements of \( U \), which helps determine invertibility.}
The system \( P X = Q \) can be solved by first solving \( L Y = Q \) and then \( U X = Y \).
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.
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The Correct Option isA, B, C
Solution and Explanation
- LU decomposition expresses \( P \) as the product of a lower triangular matrix \( L \) and an upper triangular matrix \( U \), allowing for easier solving of systems of linear equations.
- Option (A) is correct because solving \( P X = Q \) involves first solving \( L Y = Q \) to find \( Y \), and then solving \( U X = Y \) to find \( X \). This two-step process is a fundamental property of LU decomposition.
- If \( P \) is invertible, then neither \( L \) nor \( U \) has zero determinants, meaning both matrices are invertible (Option B). This ensures that the system has a unique solution.
- If \( P \) is singular, at least one of the diagonal elements of \( U \) must be zero, which implies that \( P \) does not have full rank, as stated in Option (C).
- However, if \( P \) is symmetric, \( L \) and \( U \) are not necessarily symmetric. This makes Option (D) incorrect, as the symmetry of \( P \) does not guarantee the symmetry of \( L \) and \( U \).
