Construct a 2 by 3 system Ax = b with particular solution xp = (2,4,0) and homogeneous solution xn = any multiple of (1, 1, 1).
Construct a 2 by 3 system Ax = b with particular solution xp = (2,4,0) and homogeneous solution xn = any multiple of (1, 1, 1).
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 16EQ
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![### Problem 10
**Task**:
Construct a \(2 \times 3\) system \( \mathbf{A} \mathbf{x} = \mathbf{b} \) with a particular solution \( \mathbf{x}_p = (2, 4, 0) \) and a homogeneous solution \( \mathbf{x}_n \) that is any multiple of \( (1, 1, 1) \).
**Solution**:
To construct a \(2 \times 3\) system \( \mathbf{A} \mathbf{x} = \mathbf{b} \) where the particular solution is \( \mathbf{x}_p = (2, 4, 0) \) and the homogeneous solution is \( \mathbf{x}_n = k(1, 1, 1) \) for any scalar \( k \), follow these steps:
1. **Write out the particular solution**:
\[
\mathbf{x}_p = \begin{pmatrix} 2 \\ 4 \\ 0 \end{pmatrix}
\]
2. **Homogeneous solution**:
\[
\mathbf{x}_n = k \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} k \\ k \\ k \end{pmatrix}
\]
This implies that if \(\mathbf{A}\mathbf{x} = \mathbf{0}\):
\[
\mathbf{A}
\begin{pmatrix}
1 \\ 1 \\ 1
\end{pmatrix} = \mathbf{0}
\]
3. **Form the system**:
Given the above, \(\mathbf{A} \mathbf{x}_p = \mathbf{b}\) becomes:
\[
\mathbf{A} \begin{pmatrix} 2 \\ 4 \\ 0 \end{pmatrix} = \mathbf{b}
\]
For the homogeneous solution to hold, the rows of \(\mathbf{A}\) must be orthogonal to \(\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).
Let’s pick \(\mathbf{A}\) as follows:
\[
\mathbf{A} =
\begin{pmatrix}
1 & -1 & 0 \\
1 &](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe243852a-5427-4fc7-bd56-ad30d22cf89a%2Fa2cf543d-a7fc-454d-831d-800dcb48dc0f%2Fo3wtati_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 10
**Task**:
Construct a \(2 \times 3\) system \( \mathbf{A} \mathbf{x} = \mathbf{b} \) with a particular solution \( \mathbf{x}_p = (2, 4, 0) \) and a homogeneous solution \( \mathbf{x}_n \) that is any multiple of \( (1, 1, 1) \).
**Solution**:
To construct a \(2 \times 3\) system \( \mathbf{A} \mathbf{x} = \mathbf{b} \) where the particular solution is \( \mathbf{x}_p = (2, 4, 0) \) and the homogeneous solution is \( \mathbf{x}_n = k(1, 1, 1) \) for any scalar \( k \), follow these steps:
1. **Write out the particular solution**:
\[
\mathbf{x}_p = \begin{pmatrix} 2 \\ 4 \\ 0 \end{pmatrix}
\]
2. **Homogeneous solution**:
\[
\mathbf{x}_n = k \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} k \\ k \\ k \end{pmatrix}
\]
This implies that if \(\mathbf{A}\mathbf{x} = \mathbf{0}\):
\[
\mathbf{A}
\begin{pmatrix}
1 \\ 1 \\ 1
\end{pmatrix} = \mathbf{0}
\]
3. **Form the system**:
Given the above, \(\mathbf{A} \mathbf{x}_p = \mathbf{b}\) becomes:
\[
\mathbf{A} \begin{pmatrix} 2 \\ 4 \\ 0 \end{pmatrix} = \mathbf{b}
\]
For the homogeneous solution to hold, the rows of \(\mathbf{A}\) must be orthogonal to \(\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).
Let’s pick \(\mathbf{A}\) as follows:
\[
\mathbf{A} =
\begin{pmatrix}
1 & -1 & 0 \\
1 &
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