Find an LU factorization of the matrix A (with L unit lower triangular). 6 4 A= 12 5
Find an LU factorization of the matrix A (with L unit lower triangular). 6 4 A= 12 5
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![## Finding LU Factorization of a Matrix (with L unit lower triangular)
### Problem Statement
Find an LU factorization of the matrix \( A \) where \( L \) is a unit lower triangular matrix.
The matrix \( A \) is given by:
\[
A = \begin{pmatrix}
6 & 4 \\
12 & 5
\end{pmatrix}
\]
### Explanation
LU factorization of a matrix involves decomposing the matrix \( A \) into two matrices: \( L \) and \( U \).
- \( L \) is a lower triangular matrix with ones on the diagonal.
- \( U \) is an upper triangular matrix.
In this case, since \( L \) is a unit lower triangular matrix, it will have ones on the diagonal. The goal is to express \( A \) in the form \( A = LU \), where:
\[
L = \begin{pmatrix}
1 & 0 \\
l_{21} & 1
\end{pmatrix}
\]
\[
U = \begin{pmatrix}
u_{11} & u_{12} \\
0 & u_{22}
\end{pmatrix}
\]
Let's find the values of \( l_{21} \), \( u_{11} \), \( u_{12} \), and \( u_{22} \).
1. **Determine \( u_{11} \):**
From the first row of \( A \), we have:
\[
u_{11} = 6
\]
2. **Determine \( u_{12} \):**
Continuing from the first row of \( A \), we get:
\[
u_{12} = 4
\]
3. **Determine \( l_{21} \):**
From the second row of \( A \), since \( l_{21} \times u_{11} = 12 \), we have:
\[
l_{21} = \frac{12}{6} = 2
\]
4. **Determine \( u_{22} \):**
From the second row of \( A \), substituting known values:
\[
2 \times 4 + u_{22} = 5 \Rightarrow u_{22} = 5 -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13b3ce4d-7939-40c7-a8c2-01af91160cea%2Fbe6e7b24-478f-48df-a38e-829bf8422196%2Foj4vkit_processed.png&w=3840&q=75)
Transcribed Image Text:## Finding LU Factorization of a Matrix (with L unit lower triangular)
### Problem Statement
Find an LU factorization of the matrix \( A \) where \( L \) is a unit lower triangular matrix.
The matrix \( A \) is given by:
\[
A = \begin{pmatrix}
6 & 4 \\
12 & 5
\end{pmatrix}
\]
### Explanation
LU factorization of a matrix involves decomposing the matrix \( A \) into two matrices: \( L \) and \( U \).
- \( L \) is a lower triangular matrix with ones on the diagonal.
- \( U \) is an upper triangular matrix.
In this case, since \( L \) is a unit lower triangular matrix, it will have ones on the diagonal. The goal is to express \( A \) in the form \( A = LU \), where:
\[
L = \begin{pmatrix}
1 & 0 \\
l_{21} & 1
\end{pmatrix}
\]
\[
U = \begin{pmatrix}
u_{11} & u_{12} \\
0 & u_{22}
\end{pmatrix}
\]
Let's find the values of \( l_{21} \), \( u_{11} \), \( u_{12} \), and \( u_{22} \).
1. **Determine \( u_{11} \):**
From the first row of \( A \), we have:
\[
u_{11} = 6
\]
2. **Determine \( u_{12} \):**
Continuing from the first row of \( A \), we get:
\[
u_{12} = 4
\]
3. **Determine \( l_{21} \):**
From the second row of \( A \), since \( l_{21} \times u_{11} = 12 \), we have:
\[
l_{21} = \frac{12}{6} = 2
\]
4. **Determine \( u_{22} \):**
From the second row of \( A \), substituting known values:
\[
2 \times 4 + u_{22} = 5 \Rightarrow u_{22} = 5 -
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