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The task is to use row operations to transform the given matrix into its reduced form. The matrix provided is:

\[
\begin{bmatrix}
1 & 4 & -4 & -1 \\
0 & 3 & -6 & 1 \\
0 & -1 & 2 & -\frac{1}{3}
\end{bmatrix}
\]

To perform row operations and reduce the matrix, we aim to transform it into row echelon form (REF) and then reduced row echelon form (RREF). Here's a step-by-step approach:

### Step 1: Row Operations to Reach REF
1. **First Row Operation:** Normalize the second row to make the leading coefficient 1.
   - Divide Row 2 by 3:
   \[
   \text{R2} \rightarrow \frac{1}{3}\text{R2} \Rightarrow \begin{bmatrix} 0 & 1 & -2 & \frac{1}{3} \end{bmatrix}
   \]
   
   Matrix becomes:
   \[
   \begin{bmatrix}
   1 & 4 & -4 & -1 \\
   0 & 1 & -2 & \frac{1}{3} \\
   0 & -1 & 2 & -\frac{1}{3}
   \end{bmatrix}
   \]

2. **Second Row Operation:** Eliminate the 4 in Row 1, Column 2.
   - Row 1 - 4 × Row 2:
   \[
   \text{R1} \rightarrow \text{R1} - 4\text{R2} \Rightarrow \begin{bmatrix} 1 & 0 & 4 & - \frac{5}{3} \end{bmatrix}
   \]
   
   Matrix becomes:
   \[
   \begin{bmatrix}
   1 & 0 & 4 & -\frac{5}{3} \\
   0 & 1 & -2 & \frac{1}{3} \\
   0 & -1 & 2 & -\frac{1}{3}
   \end{bmatrix}
   \]

3. **Third Row Operation:** Eliminate the -1 in Row 3, Column 2.
   - Row 3 + Row 2:
   \[
   \text{R3} \rightarrow \text{R3} + \text{R2} \Rightarrow \begin{bmatrix} 0 & 0 & 0 & 0 \end{bmatrix}
   \]
   
   Matrix becomes:
   \[
   \begin{bmatrix}
   1 & 0 & 4 & -\frac{5}{3} \\
   0 & 1 & -2 & \frac{1}{3} \\
   0 & 0 & 0 & 0
   \end{bmatrix}
   \]

### Step 2: Reduced Row Echelon Form (RREF)
1. **Fourth Row Operation:** Eliminate the 4 in Row 1, Column 3.
   - Row 1 - 4 × Row 3:
   \[
   \text{R1} \rightarrow \text{R1} - 4\text{R3} \Rightarrow \begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}
   \]
   
   Matrix becomes:
   \[
   \begin{bmatrix}
   1 & 0 & 0 & 0 \\
   0 & 1 & -2 & \frac{1}{3} \\
   0 & 0 & 0 & 0
   \end{bmatrix}
   \]

This is the reduced row echelon form (RREF).

### Final Reduced Form
\[
\begin{bmatrix}
1 & 0 & 4 & -\frac{5}{3} \\
0 & 1 & -2 & \frac{1}{3} \\
0 & 0 & 0 & 0
\end{bmatrix}
\]

If you need more details or have any questions, feel free to ask!

### Related Questions
1. How do you determine the rank of a matrix after finding its RREF?
2. Can all matrices be reduced to RREF?
3. What are the conditions for a system of linear equations to have no solution, one solution, or infinitely many solutions?
4. How can you use RREF to find the inverse of a matrix?
5. What are the differences between row echelon form and reduced row echelon form?

**Tip:** Always ensure that each leading entry in the RREF is 1 and is the only non-zero entry in its column.

Use row operations to change the matrix below to reduced form.

This guide provides a step-by-step solution to reduce a matrix to its reduced row echelon form (RREF) using row operations. The problem involves applying elementary row operations to transform the matrix and achieve RREF, important for solving linear systems in Linear Algebra. Suitable for college-level students.

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