Calculation Results
LU Decomposition: Definition and Formula
LU decomposition is a method that factors a square matrix \(A\) into two matrices:
\[ A = LU \]
Where:
- \(L\) is a lower triangular matrix (diagonal elements are 1 or non-zero depending on convention).
- \(U\) is an upper triangular matrix.
LU decomposition is widely used for solving systems of linear equations, computing determinants, and numerical analysis.
Formula
For a matrix: \[ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \] we find matrices \[ L = \begin{pmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{pmatrix}, \quad U = \begin{pmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{pmatrix}. \]
The goal is to perform Gaussian elimination on \(A\) and record the multipliers into \(L\).
Example 1
Decompose the matrix:
\[ A = \begin{pmatrix} 2 & 3 \\ 4 & 7 \end{pmatrix} \]Step 1: Compute \(L\) multipliers
\[ l_{21} = \frac{4}{2} = 2 \]Step 2: Form \(U\)
\[ U = \begin{pmatrix} 2 & 3 \\ 0 & 7 - 2\cdot 3 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix} \]Step 3: Form \(L\)
\[ L = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \]Final Answer:
\[ A = LU = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix} \]Example 2
Find the LU decomposition of:
\[ A = \begin{pmatrix} 1 & 2 & 4 \\ 3 & 8 & 14 \\ 2 & 6 & 13 \end{pmatrix} \]Step 1: Eliminate entries below pivot.
\[ l_{21} = 3,\quad l_{31} = 2 \]After elimination, compute next multiplier:
\[ l_{32} = 1 \]Final Result:
\[ L = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 2 & 1 & 1 \end{pmatrix}, \quad U = \begin{pmatrix} 1 & 2 & 4 \\ 0 & 2 & 2 \\ 0 & 0 & 1 \end{pmatrix} \]Thus:
\[ A = LU \]Common Mistakes (Important!)
- Forgetting that \(L\) must be lower triangular.
- Using wrong multipliers during elimination.
- Forgetting to update rows when computing entries of \(U\).
- Assuming all matrices have an LU decomposition—pivoting may be required.
Practice Problems
1. Compute the LU decomposition of:
\[ A = \begin{pmatrix} 1 & 3 \\ 2 & 7 \end{pmatrix} \]Show Answer
\[ L= \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}, \quad U= \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \]2. Find the LU decomposition of:
\[ A = \begin{pmatrix} 2 & 1 \\ 6 & 4 \end{pmatrix} \]Show Answer
\[ L= \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}, \quad U= \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix} \]3. Decompose the matrix:
\[ A = \begin{pmatrix} 4 & 2 & 0 \\ 4 & 4 & 2 \\ 2 & 2 & 3 \end{pmatrix} \]Show Answer
\[ L= \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0.5 & 0 & 1 \end{pmatrix}, \quad U= \begin{pmatrix} 4 & 2 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix} \]4. LU decomposition of:
\[ A = \begin{pmatrix} 3 & 1 & 1 \\ 6 & 3 & 4 \\ 3 & 1 & 5 \end{pmatrix} \]Show Answer
\[ L= \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}, \quad U= \begin{pmatrix} 3 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{pmatrix} \]LU decomposition is useful alongside matrix multiplication, eigenvalues, and matrix rank.
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