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The augmented matrix provided is:

\[
\begin{bmatrix}
3 & -4 & 2 & 0 \\
-9 & 12 & -6 & 0 \\
-6 & 8 & -4 & 0
\end{bmatrix}
\]

### Step 1: Row Reduction to Row Echelon Form (REF)

To find the pivot positions and columns, we will perform Gaussian elimination to reduce the matrix to row echelon form (REF).

**Matrix:**
\[
\begin{bmatrix}
3 & -4 & 2 & 0 \\
-9 & 12 & -6 & 0 \\
-6 & 8 & -4 & 0
\end{bmatrix}
\]

#### Row Operations:

- **Row 2:** \( R_2 \rightarrow R_2 + 3R_1 \)
  \[
  \begin{bmatrix}
  3 & -4 & 2 & 0 \\
  0 & 0 & 0 & 0 \\
  -6 & 8 & -4 & 0
  \end{bmatrix}
  \]

- **Row 3:** \( R_3 \rightarrow R_3 + 2R_1 \)
  \[
  \begin{bmatrix}
  3 & -4 & 2 & 0 \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0
  \end{bmatrix}
  \]

### Step 2: Identify Pivot Positions and Columns

After row reduction, we observe that:

- The first pivot position is in the first column, first row.
- There are no nonzero entries in other columns, meaning only the first column contains a pivot.

### Step 3: General Solution

We write the system of equations corresponding to the original matrix:

\[
\begin{aligned}
3x_1 - 4x_2 + 2x_3 &= 0 \\
0x_1 + 0x_2 + 0x_3 &= 0 \\
0x_1 + 0x_2 + 0x_3 &= 0
\end{aligned}
\]

The solution can be expressed in terms of free variables \(x_2\) and \(x_3\) (since only \(x_1\) is dependent on them):

From the first equation:
\[
x_1 = \frac{4}{3}x_2 - \frac{2}{3}x_3
\]

Thus, the general solution is:
\[
\begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}
=
\begin{pmatrix}
\frac{4}{3}x_2 - \frac{2}{3}x_3 \\
x_2 \\
x_3
\end{pmatrix}
\]

### Step 4: Pivot Columns

The pivot columns are: **Column 1**.

---

Do you have any questions or would you like further details on a specific part?

Here are some related questions:
1. How is row echelon form (REF) different from reduced row echelon form (RREF)?
2. What conditions must be met for a system to have a unique solution?
3. Can a system of linear equations have more free variables than equations?
4. How does Gaussian elimination relate to matrix rank?
5. What is the significance of a pivot column in the context of linear dependence?

**Tip:** When row-reducing a matrix, always aim for a leading 1 in pivot positions for easier calculations.

Find the general solutions of the systems whose augmented matrix is given. Circle the pivot positions in the final matrix and the original matrix, and list the pivot columns.

This problem involves finding the general solutions for a system of equations using Gaussian elimination to reduce the augmented matrix to row echelon form. It highlights how to identify pivot columns and positions through row reduction and provides insights into solving linear systems of equations.

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