Use the factorization A= QR to find the least-squares solution of Ax = b. 1 23 }}} A = | 2 -[] N|MN|M - 13 -/3 2/3 - 01 ... . x= (Simplify your answer.)
Use the factorization A= QR to find the least-squares solution of Ax = b. 1 23 }}} A = | 2 -[] N|MN|M - 13 -/3 2/3 - 01 ... . x= (Simplify your answer.)
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 the Least-Squares Solution Using QR Factorization**
To determine the least-squares solution of \( A\mathbf{x} = \mathbf{b} \) using the QR factorization, follow these steps:
1. **Given Matrices:**
- Matrix \( A \) and vector \( \mathbf{b} \) are given as:
\[
A = \begin{bmatrix}
2 & 3 \\
2 & 4 \\
1 & 1
\end{bmatrix}
=
\begin{bmatrix}
\frac{2}{3} & -\frac{1}{3} \\
\frac{2}{3} & \frac{2}{3} \\
\frac{1}{3} & -\frac{2}{3}
\end{bmatrix}
\begin{bmatrix}
3 & 5 \\
0 & 1
\end{bmatrix}
\]
\[
\mathbf{b} = \begin{bmatrix}
4 \\
2 \\
3
\end{bmatrix}
\]
Here, the matrix \( A \) is decomposed into an orthogonal matrix \( Q \) and an upper triangular matrix \( R \):
\[
A = QR
\]
2. **Objective:**
- The goal is to solve for the vector \( \mathbf{\hat{x}} \) that minimizes the difference between \( A\mathbf{x} \) and \( \mathbf{b} \).
3. **Solution Representation:**
- By using the QR factorization, we convert the problem \( A\mathbf{x} = \mathbf{b} \) into:
\[
QR\mathbf{x} = \mathbf{b}
\]
- Multiply both sides by the transpose of \( Q \) (\( Q^T \)) to exploit the orthogonality of \( Q \):
\[
Q^T Q R \mathbf{x} = Q^T \mathbf{b}
\]
- Since \( Q^T Q = I \) (identity matrix):
\[
R \mathbf{x} = Q^T \mathbf{b}
\]
4. **](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5246d2a-5796-4784-a543-e596d3b5542c%2F1049ca2c-867b-463b-bf89-2c97259f11c9%2Fwd3m5hs_processed.png&w=3840&q=75)
Transcribed Image Text:**Finding the Least-Squares Solution Using QR Factorization**
To determine the least-squares solution of \( A\mathbf{x} = \mathbf{b} \) using the QR factorization, follow these steps:
1. **Given Matrices:**
- Matrix \( A \) and vector \( \mathbf{b} \) are given as:
\[
A = \begin{bmatrix}
2 & 3 \\
2 & 4 \\
1 & 1
\end{bmatrix}
=
\begin{bmatrix}
\frac{2}{3} & -\frac{1}{3} \\
\frac{2}{3} & \frac{2}{3} \\
\frac{1}{3} & -\frac{2}{3}
\end{bmatrix}
\begin{bmatrix}
3 & 5 \\
0 & 1
\end{bmatrix}
\]
\[
\mathbf{b} = \begin{bmatrix}
4 \\
2 \\
3
\end{bmatrix}
\]
Here, the matrix \( A \) is decomposed into an orthogonal matrix \( Q \) and an upper triangular matrix \( R \):
\[
A = QR
\]
2. **Objective:**
- The goal is to solve for the vector \( \mathbf{\hat{x}} \) that minimizes the difference between \( A\mathbf{x} \) and \( \mathbf{b} \).
3. **Solution Representation:**
- By using the QR factorization, we convert the problem \( A\mathbf{x} = \mathbf{b} \) into:
\[
QR\mathbf{x} = \mathbf{b}
\]
- Multiply both sides by the transpose of \( Q \) (\( Q^T \)) to exploit the orthogonality of \( Q \):
\[
Q^T Q R \mathbf{x} = Q^T \mathbf{b}
\]
- Since \( Q^T Q = I \) (identity matrix):
\[
R \mathbf{x} = Q^T \mathbf{b}
\]
4. **
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