Find a least-squares solution of Ax=b by (a) constructing the normal equations for x and (b) solving for X. 1 -4 -1 4 2 9 A = 0 4 b= 50 1 N 50
Find a least-squares solution of Ax=b by (a) constructing the normal equations for x and (b) solving for X. 1 -4 -1 4 2 9 A = 0 4 b= 50 1 N 50
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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Question
![To find a least-squares solution of \(Ax = b\), follow these steps:
**Given:**
\[ A = \begin{bmatrix}
1 & -4 \\
-1 & 4 \\
0 & 2 \\
4 & 9
\end{bmatrix}, \quad b = \begin{bmatrix}
5 \\
1 \\
-2 \\
5
\end{bmatrix} \]
**Steps:**
**(a) Construct the normal equations for \(\hat{x}\):**
1. Compute \(A^T\), the transpose of matrix \(A\):
\[ A^T = \begin{bmatrix}
1 & -1 & 0 & 4 \\
-4 & 4 & 2 & 9
\end{bmatrix} \]
2. Compute \(A^T A\):
\[ A^T A = \begin{bmatrix}
1 & -1 & 0 & 4 \\
-4 & 4 & 2 & 9
\end{bmatrix} \begin{bmatrix}
1 & -4 \\
-1 & 4 \\
0 & 2 \\
4 & 9
\end{bmatrix} = \begin{bmatrix}
18 & 29 \\
29 & 101
\end{bmatrix} \]
3. Compute \(A^T b\):
\[ A^T b = \begin{bmatrix}
1 & -1 & 0 & 4 \\
-4 & 4 & 2 & 9
\end{bmatrix} \begin{bmatrix}
5 \\
1 \\
-2 \\
5
\end{bmatrix} = \begin{bmatrix}
18 \\
90
\end{bmatrix} \]
4. The normal equations are:
\[ (A^T A) \hat{x} = A^T b \]
Substitute \(A^T A\) and \(A^T b\):
\[ \begin{bmatrix}
18 & 29 \\
29 & 101
\end{bmatrix} \hat{x} = \begin{bmatrix}
18 \\
90
\end{bmatrix} \]
**(b) Solve for \(\hat{x}\):**
To find \(\hat{x}\), solve the system of linear equations \( (](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5246d2a-5796-4784-a543-e596d3b5542c%2Ff0924f38-ad9f-43ba-af54-228d53d23584%2Fqn790s_processed.png&w=3840&q=75)
Transcribed Image Text:To find a least-squares solution of \(Ax = b\), follow these steps:
**Given:**
\[ A = \begin{bmatrix}
1 & -4 \\
-1 & 4 \\
0 & 2 \\
4 & 9
\end{bmatrix}, \quad b = \begin{bmatrix}
5 \\
1 \\
-2 \\
5
\end{bmatrix} \]
**Steps:**
**(a) Construct the normal equations for \(\hat{x}\):**
1. Compute \(A^T\), the transpose of matrix \(A\):
\[ A^T = \begin{bmatrix}
1 & -1 & 0 & 4 \\
-4 & 4 & 2 & 9
\end{bmatrix} \]
2. Compute \(A^T A\):
\[ A^T A = \begin{bmatrix}
1 & -1 & 0 & 4 \\
-4 & 4 & 2 & 9
\end{bmatrix} \begin{bmatrix}
1 & -4 \\
-1 & 4 \\
0 & 2 \\
4 & 9
\end{bmatrix} = \begin{bmatrix}
18 & 29 \\
29 & 101
\end{bmatrix} \]
3. Compute \(A^T b\):
\[ A^T b = \begin{bmatrix}
1 & -1 & 0 & 4 \\
-4 & 4 & 2 & 9
\end{bmatrix} \begin{bmatrix}
5 \\
1 \\
-2 \\
5
\end{bmatrix} = \begin{bmatrix}
18 \\
90
\end{bmatrix} \]
4. The normal equations are:
\[ (A^T A) \hat{x} = A^T b \]
Substitute \(A^T A\) and \(A^T b\):
\[ \begin{bmatrix}
18 & 29 \\
29 & 101
\end{bmatrix} \hat{x} = \begin{bmatrix}
18 \\
90
\end{bmatrix} \]
**(b) Solve for \(\hat{x}\):**
To find \(\hat{x}\), solve the system of linear equations \( (
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