Section 2.5 -4 27 --- -4 5 2 and b = 0. Given that A has an LU factorization 6 -9 1. Let A = 2 1 0 -2 1 0 0 -1 1 دن use this LU factorization to solve Ax = b. 2 -4 2 0-3 6 0 0 1

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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**Section 2.5**

1. Let \( A = \begin{bmatrix} 2 & -4 & 2 \\ -4 & 5 & 2 \\ 6 & -9 & 1 \end{bmatrix} \) and \( \mathbf{b} = \begin{bmatrix} 6 \\ 0 \\ 6 \end{bmatrix} \). Given that \( A \) has an \( LU \) factorization 

   \[
   \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 3 & -1 & 1 \end{bmatrix}
   \begin{bmatrix} 2 & -4 & 2 \\ 0 & -3 & 6 \\ 0 & 0 & 1 \end{bmatrix},
   \]

   use this \( LU \) factorization to solve \( A\mathbf{x} = \mathbf{b} \).

2. Find the \( LU \) factorizations of 

   \[
   \begin{bmatrix} 2 & 0 & 5 & 2 \\ -6 & 3 & -13 & -3 \\ 4 & 9 & 16 & 17 \end{bmatrix}
   \]

   and

   \[
   \begin{bmatrix} 2 & 3 & 2 \\ 4 & 13 & 9 \\ -6 & 5 & 4 \end{bmatrix}
   \]
Transcribed Image Text:**Section 2.5** 1. Let \( A = \begin{bmatrix} 2 & -4 & 2 \\ -4 & 5 & 2 \\ 6 & -9 & 1 \end{bmatrix} \) and \( \mathbf{b} = \begin{bmatrix} 6 \\ 0 \\ 6 \end{bmatrix} \). Given that \( A \) has an \( LU \) factorization \[ \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 3 & -1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -4 & 2 \\ 0 & -3 & 6 \\ 0 & 0 & 1 \end{bmatrix}, \] use this \( LU \) factorization to solve \( A\mathbf{x} = \mathbf{b} \). 2. Find the \( LU \) factorizations of \[ \begin{bmatrix} 2 & 0 & 5 & 2 \\ -6 & 3 & -13 & -3 \\ 4 & 9 & 16 & 17 \end{bmatrix} \] and \[ \begin{bmatrix} 2 & 3 & 2 \\ 4 & 13 & 9 \\ -6 & 5 & 4 \end{bmatrix} \]
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