2 1 Let A = -4 -7 2 36 nd v: Compute Au and Av, and compare them with b. Is it 4 3 possible that at least one of u or v could be a least-squares solution of Ax=b? (Answer this without computing a least-squares solution.)
2 1 Let A = -4 -7 2 36 nd v: Compute Au and Av, and compare them with b. Is it 4 3 possible that at least one of u or v could be a least-squares solution of Ax=b? (Answer this without computing a least-squares solution.)
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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![### Linear Algebra Problem: Least-Squares Solution
---
**Problem Statement:**
Given the following matrices and vectors:
\[ A = \begin{bmatrix} 2 & 1 \\ -4 & -7 \\ 4 & 3 \end{bmatrix}, \quad b = \begin{bmatrix} 2 \\ 36 \\ -4 \end{bmatrix}, \quad u = \begin{bmatrix} 7 \\ -8 \end{bmatrix}, \quad v = \begin{bmatrix} 3 \\ -8 \end{bmatrix} \]
1. Compute \( Au \) and \( Av \), and compare them with \( b \).
2. Is it possible that at least one of \( u \) or \( v \) could be a least-squares solution of \( Ax = b \)?
(Answer this without computing a least-squares solution.)
---
**Step-by-Step Solution:**
1. **Matrix-Vector Multiplication:**
To find \( Au \) and \( Av \), perform the matrix-vector multiplications:
\[ Au = A \times u = \begin{bmatrix} 2 & 1 \\ -4 & -7 \\ 4 & 3 \end{bmatrix} \times \begin{bmatrix} 7 \\ -8 \end{bmatrix} \]
Compute each element of the resulting vector \( Au \):
\[ Au = \begin{bmatrix} (2 \times 7) + (1 \times -8) \\ (-4 \times 7) + (-7 \times -8) \\ (4 \times 7) + (3 \times -8) \end{bmatrix} = \begin{bmatrix} 14 - 8 \\ -28 + 56 \\ 28 - 24 \end{bmatrix} = \begin{bmatrix} 6 \\ 28 \\ 4 \end{bmatrix} \]
Similarly:
\[ Av = A \times v = \begin{bmatrix} 2 & 1 \\ -4 & -7 \\ 4 & 3 \end{bmatrix} \times \begin{bmatrix} 3 \\ -8 \end{bmatrix} \]
Compute each element of the resulting vector \( Av \):
\[ Av = \begin{b](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5246d2a-5796-4784-a543-e596d3b5542c%2F9dca52fd-10c2-4b52-ba06-6698c7eb2da8%2Fbcdayz6_processed.png&w=3840&q=75)
Transcribed Image Text:### Linear Algebra Problem: Least-Squares Solution
---
**Problem Statement:**
Given the following matrices and vectors:
\[ A = \begin{bmatrix} 2 & 1 \\ -4 & -7 \\ 4 & 3 \end{bmatrix}, \quad b = \begin{bmatrix} 2 \\ 36 \\ -4 \end{bmatrix}, \quad u = \begin{bmatrix} 7 \\ -8 \end{bmatrix}, \quad v = \begin{bmatrix} 3 \\ -8 \end{bmatrix} \]
1. Compute \( Au \) and \( Av \), and compare them with \( b \).
2. Is it possible that at least one of \( u \) or \( v \) could be a least-squares solution of \( Ax = b \)?
(Answer this without computing a least-squares solution.)
---
**Step-by-Step Solution:**
1. **Matrix-Vector Multiplication:**
To find \( Au \) and \( Av \), perform the matrix-vector multiplications:
\[ Au = A \times u = \begin{bmatrix} 2 & 1 \\ -4 & -7 \\ 4 & 3 \end{bmatrix} \times \begin{bmatrix} 7 \\ -8 \end{bmatrix} \]
Compute each element of the resulting vector \( Au \):
\[ Au = \begin{bmatrix} (2 \times 7) + (1 \times -8) \\ (-4 \times 7) + (-7 \times -8) \\ (4 \times 7) + (3 \times -8) \end{bmatrix} = \begin{bmatrix} 14 - 8 \\ -28 + 56 \\ 28 - 24 \end{bmatrix} = \begin{bmatrix} 6 \\ 28 \\ 4 \end{bmatrix} \]
Similarly:
\[ Av = A \times v = \begin{bmatrix} 2 & 1 \\ -4 & -7 \\ 4 & 3 \end{bmatrix} \times \begin{bmatrix} 3 \\ -8 \end{bmatrix} \]
Compute each element of the resulting vector \( Av \):
\[ Av = \begin{b
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