The Fundamental Theorem of Linear Algebra
express the general solution of Ax = b where A = [-3 -6; -9 -18]
and b = [-6; -18] as x = x_r + cx_n where x_r is the row space solution and x_n is a basis for null(A)

To express the general solution of the system \( Ax = b \) in the form \( x = x_r + cx_n \), where \( x_r \) is the row space solution and \( x_n \) is a basis for the null space of \( A \), we proceed with the following steps.

### 1. Matrix and Vector Definition

Given:

\[
A = \begin{pmatrix} -3 & -6 \\ -9 & -18 \end{pmatrix}, \quad b = \begin{pmatrix} -6 \\ -18 \end{pmatrix}
\]

### 2. Check Consistency of the System \( Ax = b \)

First, let's verify if the system \( Ax = b \) is consistent. This involves checking if the vector \( b \) lies in the column space of \( A \).

\[
\text{Rank}(A) = 1 \quad \text{and} \quad \text{Rank}([A|b]) = 1
\]

Since the ranks are equal, the system is consistent, and \( b \) lies in the column space of \( A \).

### 3. Row Reduction to Row Echelon Form (REF)

Let's reduce the augmented matrix \( [A|b] \) to its row echelon form (REF):

\[
\begin{pmatrix} -3 & -6 & -6 \\ -9 & -18 & -18 \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 2 \\ 0 & 0 & 0 \end{pmatrix}
\]

### 4. Particular Solution \( x_r \) (Row Space Solution)

From the row-reduced form, we obtain the following equation:

\[
x_1 + 2x_2 = 2
\]

We can express the particular solution \( x_r \) as:

\[
x_r = \begin{pmatrix} 2 \\ 0 \end{pmatrix} \quad \text{(set } x_2 = 0 \text{)}
\]

### 5. Null Space Solution \( x_n \)

To find the null space \( \text{null}(A) \), solve \( Ax = 0 \):

\[
\begin{pmatrix} -3 & -6 \\ -9 & -18 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

This leads to the equation:

\[
x_1 + 2x_2 = 0 \quad \text{or} \quad x_1 = -2x_2
\]

Thus, the null space solution \( x_n \) can be written as:

\[
x_n = \begin{pmatrix} -2 \\ 1 \end{pmatrix}
\]

### 6. General Solution

The general solution to \( Ax = b \) is given by:

\[
x = x_r + c x_n
\]

Substituting the values of \( x_r \) and \( x_n \):

\[
x = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + c \begin{pmatrix} -2 \\ 1 \end{pmatrix}
\]

Where \( c \) is any scalar.

### Final Answer

The general solution is:

\[
x = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + c \begin{pmatrix} -2 \\ 1 \end{pmatrix}
\]

Would you like any more details or have any questions? 

Here are 5 related questions to consider:

1. How do you find the null space of a matrix?
2. What is the significance of the row space and null space in linear algebra?
3. How can you verify if a solution is indeed a solution to a given system?
4. What happens when \( b \) does not lie in the column space of \( A \)?
5. Can a matrix have a null space with more than one dimension?

**Tip:** Always check the consistency of a system \( Ax = b \) by comparing the ranks of \( A \) and \( [A|b] \) before proceeding with solutions.

Explore the general solution of Ax = b using matrix operations and the Fundamental Theorem of Linear Algebra. Learn how to find the row space solution and the basis for the null space, with detailed step-by-step explanations. Suitable for undergraduate students studying linear algebra.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation