Describe all solutions of Axequals0 in parametric vector​ form, where A is row equivalent to the given matrix.
Start 2 By 4 Table 1st Row 1st Column 1 2nd Column 4 3rd Column 0 4st Column negative 2 2nd Row 1st Column 2 2nd Column 8 3rd Column 0 4st Column negative 4 EndTable 
 
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Part 1
xequalsx Subscript 2
select: v@Sub{2}
enter your response hereplusx Subscript 3
  
enter your response hereplusx Subscript 4
  
enter your response here
​(Type an integer or fraction for each matrix​ element.)

We are given a matrix \( A \) that is row equivalent to the following:

\[
\begin{pmatrix}
1 & 4 & 0 & -2 \\
2 & 8 & 0 & -4 \\
\end{pmatrix}
\]

We need to solve the homogeneous system \( A\mathbf{x} = 0 \), where \( \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \), and express the solution in parametric vector form.

### Step 1: Row Reduction
First, we reduce the given matrix \( A \) to row echelon form. Start with the matrix:

\[
\begin{pmatrix}
1 & 4 & 0 & -2 \\
2 & 8 & 0 & -4 \\
\end{pmatrix}
\]

Perform the row operation \( R_2 \rightarrow R_2 - 2R_1 \):

\[
\begin{pmatrix}
1 & 4 & 0 & -2 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}
\]

The second row is now entirely zeros, meaning we have one non-trivial equation.

### Step 2: Solve the System
The system of equations derived from the first row is:

\[
x_1 + 4x_2 - 2x_4 = 0
\]

Solving for \( x_1 \) in terms of the free variables \( x_2, x_3, \) and \( x_4 \):

\[
x_1 = -4x_2 + 2x_4
\]

Since \( x_2, x_3, \) and \( x_4 \) are free variables, we can express the solution in parametric form as follows:

\[
\mathbf{x} = x_2 \begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} 2 \\ 0 \\ 0 \\ 1 \end{pmatrix}
\]

### Final Solution:
The general solution in parametric vector form is:

\[
\mathbf{x} = x_2 \begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} 2 \\ 0 \\ 0 \\ 1 \end{pmatrix}
\]

### Summary:
- \( x_1 = -4x_2 + 2x_4 \)
- \( x_2, x_3, \) and \( x_4 \) are free variables.

Let me know if you need further clarification, or if you'd like to discuss any part of the process.

---

Here are five questions to deepen your understanding:
1. What is the geometric interpretation of the parametric vector solution?
2. How would the solution change if the matrix \( A \) had three rows instead of two?
3. What does it mean for a variable to be "free" in the context of linear algebra?
4. Could the system \( A\mathbf{x} = 0 \) have more than one solution form if the matrix were different?
5. How does row reduction help in solving systems of linear equations?

**Tip:** When solving homogeneous systems, always check the rank of the matrix, as it helps in identifying the number of free variables.

Describe all solutions of Ax=0 in parametric vector form, where A is row equivalent to the given matrix. Start 2 by 4 table: 1st Row [1, 4, 0, -2], 2nd Row [2, 8, 0, -4].

Learn how to solve the homogeneous system Ax=0 in parametric vector form. This example involves a 2x4 matrix row-reduced to echelon form. Suitable for college-level linear algebra students, it explores key concepts like free variables, row reduction, and parametric vector solutions.

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