### Matrix Inversion Using LU Factorization**Note:** Solve failed: Matrix is singular or nearly singularWhen $ 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.

Answered: Image /qna-images/answer/c1de157c-d1b9-427a-b96b-f720264bc653.jpg

• 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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Concept explainers

Matrices

‘Matrices’ is the plural form of the word matrix, and it is basically a spreadsheet in the form of a box. In mathematics, various functions can be carried out with matrices. Generally, a matrix comes in the shape of a square or rectangle. The elements are distributed horizontally in the form of rows, and vertically in the form of columns.

Question

### 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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