4 Consider the system Ax = b where A = b = [-2] A 16 -16 basis for the null space of A is given by {[]} {[1]} and x = [] is a particular solution. Use this information to find the row space solution x, to Ax = b. X₂ = -4 Ex: 1.2 and b
4 Consider the system Ax = b where A = b = [-2] A 16 -16 basis for the null space of A is given by {[]} {[1]} and x = [] is a particular solution. Use this information to find the row space solution x, to Ax = b. X₂ = -4 Ex: 1.2 and b
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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Hey there, I need some help with solving this problem. The steps are just not clicking for me, thanks.
![### Linear Algebra System Solution
Consider the linear system \( \mathbf{Ax} = \mathbf{b} \) where:
\[
\mathbf{A} = \begin{bmatrix} -4 & 4 \\ 16 & -16 \end{bmatrix} \quad \text{and} \quad \mathbf{b} = \begin{bmatrix} -4 \\ -2 \end{bmatrix}.
\]
A basis for the null space of \( \mathbf{A} \) is given by:
\[
\left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\}
\]
and
\[
\mathbf{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}
\]
is a particular solution.
Use this information to find the row space solution \( \mathbf{x_r} \) to \( \mathbf{Ax} = \mathbf{b} \).
\[
\mathbf{x_r} = \left[ \boxed{\text{Ex: 1.2}} \right]
\]
## Explanation
In the above problem, the matrix \( \mathbf{A} \) and vector \( \mathbf{b} \) are defined, and a particular solution \( \mathbf{x} \) as well as a basis for the null space of \( \mathbf{A} \) are given. You are required to use these pieces of information to find the row space solution, \( \mathbf{x_r} \).
### Steps to Solve:
1. **Understanding Null Space Basis**:
- The null space basis indicates solutions to the homogeneous equation \( \mathbf{Ax} = 0 \).
2. **Particular Solution**:
- The particular solution given is one solution to the non-homogeneous equation \( \mathbf{Ax} = \mathbf{b} \).
3. **Combining Solutions**:
- General solution to \( \mathbf{Ax} = \mathbf{b} \) can be written as the sum of the particular solution and any solution to the homogeneous equation \( \mathbf{Ax} = 0 \).
4. **Row Space Solution**:
- By substituting the particular solution and the null space basis into the formulation, you will find \( \mathbf{x_r} \).
**Note:** The exact value for \( \mathbf{x_r](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf8aa82a-31a1-4382-a30e-3ad6a3174f84%2Fee66d9d9-8359-43f5-8922-288a70f6a669%2F9g98y9p_processed.png&w=3840&q=75)
Transcribed Image Text:### Linear Algebra System Solution
Consider the linear system \( \mathbf{Ax} = \mathbf{b} \) where:
\[
\mathbf{A} = \begin{bmatrix} -4 & 4 \\ 16 & -16 \end{bmatrix} \quad \text{and} \quad \mathbf{b} = \begin{bmatrix} -4 \\ -2 \end{bmatrix}.
\]
A basis for the null space of \( \mathbf{A} \) is given by:
\[
\left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\}
\]
and
\[
\mathbf{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}
\]
is a particular solution.
Use this information to find the row space solution \( \mathbf{x_r} \) to \( \mathbf{Ax} = \mathbf{b} \).
\[
\mathbf{x_r} = \left[ \boxed{\text{Ex: 1.2}} \right]
\]
## Explanation
In the above problem, the matrix \( \mathbf{A} \) and vector \( \mathbf{b} \) are defined, and a particular solution \( \mathbf{x} \) as well as a basis for the null space of \( \mathbf{A} \) are given. You are required to use these pieces of information to find the row space solution, \( \mathbf{x_r} \).
### Steps to Solve:
1. **Understanding Null Space Basis**:
- The null space basis indicates solutions to the homogeneous equation \( \mathbf{Ax} = 0 \).
2. **Particular Solution**:
- The particular solution given is one solution to the non-homogeneous equation \( \mathbf{Ax} = \mathbf{b} \).
3. **Combining Solutions**:
- General solution to \( \mathbf{Ax} = \mathbf{b} \) can be written as the sum of the particular solution and any solution to the homogeneous equation \( \mathbf{Ax} = 0 \).
4. **Row Space Solution**:
- By substituting the particular solution and the null space basis into the formulation, you will find \( \mathbf{x_r} \).
**Note:** The exact value for \( \mathbf{x_r
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