Find a least-squares solution of Ax = b by (a) constructing the normal equations for x and (b) solving for x. A = 2 2-3 3 - 1 -1 2
Find a least-squares solution of Ax = b by (a) constructing the normal equations for x and (b) solving for x. A = 2 2-3 3 - 1 -1 2
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 a Least-Squares Solution for the Equation \(Ax = b\)
To solve the problem of finding a least-squares solution for the equation \(Ax = b\), follow these steps:
## Problem Statement
Given the matrix \(A\) and vector \(b\):
\[ A = \begin{bmatrix} -1 & 2 \\ 2 & -3 \\ -1 & 3 \\ \end{bmatrix}, \quad b = \begin{bmatrix} 4 \\ 1 \\ 2 \\ \end{bmatrix} \]
We are to:
1. Construct the normal equations for \(\hat{x}\).
2. Solve for \(\hat{x}\).
## Steps to Solve the Problem
### Step 1: Constructing the Normal Equations
The normal equations are given by:
\[ A^T A \hat{x} = A^T b \]
1. Compute \(A^T\) (the transpose of A):
\[ A^T = \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & 3 \\ \end{bmatrix} \]
2. Compute \(A^T A\):
\[ A^T A = \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & 3 \\ \end{bmatrix} \begin{bmatrix} -1 & 2 \\ 2 & -3 \\ -1 & 3 \\ \end{bmatrix} = \begin{bmatrix} 2 & -7 \\ -7 & 22 \\ \end{bmatrix} \]
3. Compute \(A^T b\):
\[ A^T b = \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & 3 \\ \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 2 \\ \end{bmatrix} = \begin{bmatrix} -4 \\ 16 \\ \end{bmatrix} \]
4. The normal equations are:
\[ \begin{bmatrix} 2 & -7 \\ -7 & 22 \\ \end{bmatrix} \hat{x} = \begin{bmatrix} -4 \\ 16 \\ \end{bmatrix} \]
### Step 2: Solving for \(\hat{x}\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5246d2a-5796-4784-a543-e596d3b5542c%2F69fa3b2e-6552-4b71-a6a4-465cde0c26fa%2Fj5rnnf_processed.png&w=3840&q=75)
Transcribed Image Text:# Finding a Least-Squares Solution for the Equation \(Ax = b\)
To solve the problem of finding a least-squares solution for the equation \(Ax = b\), follow these steps:
## Problem Statement
Given the matrix \(A\) and vector \(b\):
\[ A = \begin{bmatrix} -1 & 2 \\ 2 & -3 \\ -1 & 3 \\ \end{bmatrix}, \quad b = \begin{bmatrix} 4 \\ 1 \\ 2 \\ \end{bmatrix} \]
We are to:
1. Construct the normal equations for \(\hat{x}\).
2. Solve for \(\hat{x}\).
## Steps to Solve the Problem
### Step 1: Constructing the Normal Equations
The normal equations are given by:
\[ A^T A \hat{x} = A^T b \]
1. Compute \(A^T\) (the transpose of A):
\[ A^T = \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & 3 \\ \end{bmatrix} \]
2. Compute \(A^T A\):
\[ A^T A = \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & 3 \\ \end{bmatrix} \begin{bmatrix} -1 & 2 \\ 2 & -3 \\ -1 & 3 \\ \end{bmatrix} = \begin{bmatrix} 2 & -7 \\ -7 & 22 \\ \end{bmatrix} \]
3. Compute \(A^T b\):
\[ A^T b = \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & 3 \\ \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 2 \\ \end{bmatrix} = \begin{bmatrix} -4 \\ 16 \\ \end{bmatrix} \]
4. The normal equations are:
\[ \begin{bmatrix} 2 & -7 \\ -7 & 22 \\ \end{bmatrix} \hat{x} = \begin{bmatrix} -4 \\ 16 \\ \end{bmatrix} \]
### Step 2: Solving for \(\hat{x}\)
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