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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
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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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 \( (
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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