Describe all solutions of Ax = 0 in parametric vector form, where A is row equivalent to the given matrix.
{{1,4,0,-3},{4,16,0,-12}}

To describe all solutions of the equation \( A\mathbf{x} = 0 \) in parametric vector form, where \( A \) is a matrix row equivalent to the given matrix, we proceed with the following steps:

### Given Matrix
The matrix \( A \) is row equivalent to the given matrix:
\[
A = \begin{pmatrix}
1 & 4 & 0 & -3 \\
4 & 16 & 0 & -12
\end{pmatrix}
\]

### Step 1: Row Reduction
First, we row reduce the matrix to its Row Echelon Form (REF).

1. The first row is already in the desired form. We keep it as it is:
   \[
   \begin{pmatrix}
   1 & 4 & 0 & -3 \\
   4 & 16 & 0 & -12
   \end{pmatrix}
   \]

2. Subtract 4 times the first row from the second row to eliminate the entry below the leading 1 in the first column:
   \[
   \text{Row 2} \rightarrow \text{Row 2} - 4 \times \text{Row 1}
   \]
   \[
   \begin{pmatrix}
   1 & 4 & 0 & -3 \\
   0 & 0 & 0 & 0
   \end{pmatrix}
   \]

Now, the matrix is in Row Echelon Form (REF).

### Step 2: Set Up the System of Equations
The system of linear equations corresponding to the row-reduced matrix is:
\[
1x_1 + 4x_2 + 0x_3 - 3x_4 = 0
\]
\[
0 = 0
\]

This simplifies to:
\[
x_1 + 4x_2 - 3x_4 = 0
\]

### Step 3: Express Leading Variables in Terms of Free Variables
Here, \( x_1 \) is a leading variable, and \( x_2 \), \( x_3 \), and \( x_4 \) are free variables. We can express \( x_1 \) in terms of the free variables \( x_2 \) and \( x_4 \):

\[
x_1 = -4x_2 + 3x_4
\]

### Step 4: Parametric Vector Form
We express the solution in parametric vector form. Let \( x_2 = t \), \( x_3 = s \), and \( x_4 = u \), where \( t \), \( s \), and \( u \) are parameters.

The solution can be written as:
\[
\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} -4t + 3u \\ t \\ s \\ u \end{pmatrix}
\]

This can be decomposed into:
\[
\mathbf{x} = t\begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \end{pmatrix} + s\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + u\begin{pmatrix} 3 \\ 0 \\ 0 \\ 1 \end{pmatrix}
\]

### Final Answer
The parametric vector form of the solution is:
\[
\mathbf{x} = t\begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \end{pmatrix} + s\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + u\begin{pmatrix} 3 \\ 0 \\ 0 \\ 1 \end{pmatrix}
\]
where \( t \), \( s \), and \( u \) are parameters that can take any real values.

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Would you like more details or have any questions?

Here are 8 follow-up questions to explore this further:

1. What is the geometric interpretation of the solution set in this case?
2. How would the solution change if the original matrix had a different row rank?
3. What are leading and free variables, and how are they identified?
4. How does the rank of the matrix relate to the number of free variables?
5. What if \( A \) had more rows than columns, how would that affect the solution?
6. How can the parametric form be used to describe all possible solutions?
7. How would you verify that the parametric vector form is correct?
8. What would the solution look like if \( A \) had more than one row of zeros?

**Tip:** When solving systems of equations, identifying free and leading variables is crucial for writing the solution in parametric form.

Explore the parametric vector form for solutions to Ax = 0, where A is a matrix row equivalent to the given matrix. Understand how to express solutions using free variables and leading variables. Suitable for advanced high school students and beyond.

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