• Solve failed: Matrix is singular or nearly singular - 1 If L and U are invertible, then (LU) 1 = U -L-!, Find A from the given LU factorization: ГО 6. [1 0 07 6 5 A = LU = 0 21 28 4 10 0. - 3 8. %3D 0 6 18 2 2 1 L 0 -8 A-1 =
• Solve failed: Matrix is singular or nearly singular - 1 If L and U are invertible, then (LU) 1 = U -L-!, Find A from the given LU factorization: ГО 6. [1 0 07 6 5 A = LU = 0 21 28 4 10 0. - 3 8. %3D 0 6 18 2 2 1 L 0 -8 A-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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![### Matrix Inversion Using LU Factorization
**Note:** Solve failed: Matrix is singular or nearly singular
When \( L \) and \( U \) are invertible matrices, then the inverse of the product of \( L \) and \( U \), denoted as \((LU)^{-1}\), can be computed by multiplying the inverses of \( U \) and \( L \) in reverse order. Specifically,
\[
(LU)^{-1} = U^{-1}L^{-1}
\]
To find \( A^{-1} \) from the given LU factorization:
Given:
\[
A = LU = \begin{bmatrix} 0 & 6 & 5 \\ 0 & 21 & 28 \\ 0 & 6 & 18 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 6 & 5 \\ 0 & -3 & 8 \\ 0 & 0 & -8 \end{bmatrix}
\]
We need to find \( A^{-1} \):
\[
A^{-1} = \begin{bmatrix} & & \\ & & \\ & & \end{bmatrix} \begin{bmatrix} & & \\ & & \\ & & \end{bmatrix} \begin{bmatrix} & & \\ & & \\ & & \end{bmatrix}
\]
Here, we focus on inverting the lower triangular matrix \( L \) and the upper triangular matrix \( U \). Then we multiply these inverses in reverse order to obtain \( A^{-1} \).
**Notes:**
- The first matrix \( L \) is a lower triangular matrix.
- The second matrix \( U \) is an upper triangular matrix.
- The solution involves computing \( L^{-1} \) and \( U^{-1} \) separately and then multiplying them as \( U^{-1}L^{-1} \).
The blank matrices on the left are placeholders for the step-by-step computations for \( A^{-1} \). This educational example helps in understanding the LU decomposition method and the calculation of the inverse of a matrix using this factorization technique, despite the given matrix being singular or nearly singular.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F088399be-9a43-44b3-a949-bec438b8f98b%2Fc1de157c-d1b9-427a-b96b-f720264bc653%2Fmfg0ee_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Matrix Inversion Using LU Factorization
**Note:** Solve failed: Matrix is singular or nearly singular
When \( L \) and \( U \) are invertible matrices, then the inverse of the product of \( L \) and \( U \), denoted as \((LU)^{-1}\), can be computed by multiplying the inverses of \( U \) and \( L \) in reverse order. Specifically,
\[
(LU)^{-1} = U^{-1}L^{-1}
\]
To find \( A^{-1} \) from the given LU factorization:
Given:
\[
A = LU = \begin{bmatrix} 0 & 6 & 5 \\ 0 & 21 & 28 \\ 0 & 6 & 18 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 6 & 5 \\ 0 & -3 & 8 \\ 0 & 0 & -8 \end{bmatrix}
\]
We need to find \( A^{-1} \):
\[
A^{-1} = \begin{bmatrix} & & \\ & & \\ & & \end{bmatrix} \begin{bmatrix} & & \\ & & \\ & & \end{bmatrix} \begin{bmatrix} & & \\ & & \\ & & \end{bmatrix}
\]
Here, we focus on inverting the lower triangular matrix \( L \) and the upper triangular matrix \( U \). Then we multiply these inverses in reverse order to obtain \( A^{-1} \).
**Notes:**
- The first matrix \( L \) is a lower triangular matrix.
- The second matrix \( U \) is an upper triangular matrix.
- The solution involves computing \( L^{-1} \) and \( U^{-1} \) separately and then multiplying them as \( U^{-1}L^{-1} \).
The blank matrices on the left are placeholders for the step-by-step computations for \( A^{-1} \). This educational example helps in understanding the LU decomposition method and the calculation of the inverse of a matrix using this factorization technique, despite the given matrix being singular or nearly singular.
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