If L and U are invertible, then (LU)²001-U001L001 Find A001 from the given LU factorization: = -2 0 0 -2 0 0 18 -2 6 0 -7 A = LU A = -1 -2 7 7 H 1 -9 -3 0 1 -7 ][ 1 0 0 -1
If L and U are invertible, then (LU)²001-U001L001 Find A001 from the given LU factorization: = -2 0 0 -2 0 0 18 -2 6 0 -7 A = LU A = -1 -2 7 7 H 1 -9 -3 0 1 -7 ][ 1 0 0 -1
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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![### Understanding Invertibility and LU Factorization
#### Problem Statement:
Given the matrices \( L \) and \( U \) are invertible, the equation \( (LU)^{-1} = U^{-1}L^{-1} \) holds true. The goal is to find \( A^{-1} \) from the given LU factorization.
Given Matrix:
\[ A = LU = \begin{bmatrix} -2 & 0 & 0 \\ 18 & -1 & -2 \\ 6 & 7 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -9 & 1 & 0 \\ -3 & -7 & 1 \end{bmatrix} \begin{bmatrix} -2 & 0 & 0 \\ 0 & -1 & -2 \\ 0 & 0 & -7 \end{bmatrix} \]
#### Task:
Find the inverse of matrix \( A \), denoted as \( A^{-1} \), using the given LU factorization.
#### Matrix Definitions:
- **L (Lower Triangular Matrix):**
\[
L = \begin{bmatrix}
1 & 0 & 0 \\
-9 & 1 & 0 \\
-3 & -7 & 1
\end{bmatrix}
\]
- **U (Upper Triangular Matrix):**
\[
U = \begin{bmatrix}
-2 & 0 & 0 \\
0 & -1 & -2 \\
0 & 0 & -7
\end{bmatrix}
\]
To find \( A^{-1} \):
\[
A^{-1} = (LU)^{-1} = U^{-1}L^{-1}
\]
#### Steps:
1. **Compute \( U^{-1} \):**
Find the inverse of the upper triangular matrix \( U \).
2. **Compute \( L^{-1} \):**
Find the inverse of the lower triangular matrix \( L \).
3. **Multiply \( U^{-1} \) and \( L^{-1} \):**
Multiply the results of \( U^{-1} \) and \( L^{-1} \) to obtain \( A^{-1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e366a85-c23b-450a-9bc7-d2ea0be52d3f%2Fb9ac8905-f546-46ff-ad3a-fcb5d8fa64e6%2Fbht5o08_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Invertibility and LU Factorization
#### Problem Statement:
Given the matrices \( L \) and \( U \) are invertible, the equation \( (LU)^{-1} = U^{-1}L^{-1} \) holds true. The goal is to find \( A^{-1} \) from the given LU factorization.
Given Matrix:
\[ A = LU = \begin{bmatrix} -2 & 0 & 0 \\ 18 & -1 & -2 \\ 6 & 7 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -9 & 1 & 0 \\ -3 & -7 & 1 \end{bmatrix} \begin{bmatrix} -2 & 0 & 0 \\ 0 & -1 & -2 \\ 0 & 0 & -7 \end{bmatrix} \]
#### Task:
Find the inverse of matrix \( A \), denoted as \( A^{-1} \), using the given LU factorization.
#### Matrix Definitions:
- **L (Lower Triangular Matrix):**
\[
L = \begin{bmatrix}
1 & 0 & 0 \\
-9 & 1 & 0 \\
-3 & -7 & 1
\end{bmatrix}
\]
- **U (Upper Triangular Matrix):**
\[
U = \begin{bmatrix}
-2 & 0 & 0 \\
0 & -1 & -2 \\
0 & 0 & -7
\end{bmatrix}
\]
To find \( A^{-1} \):
\[
A^{-1} = (LU)^{-1} = U^{-1}L^{-1}
\]
#### Steps:
1. **Compute \( U^{-1} \):**
Find the inverse of the upper triangular matrix \( U \).
2. **Compute \( L^{-1} \):**
Find the inverse of the lower triangular matrix \( L \).
3. **Multiply \( U^{-1} \) and \( L^{-1} \):**
Multiply the results of \( U^{-1} \) and \( L^{-1} \) to obtain \( A^{-1
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