A = [3,2,1,x;1,-1,-1,y;1,1,1,z] find the rref

We are given the matrix \( A = \begin{bmatrix} 3 & 2 & 1 & x \\ 1 & -1 & -1 & y \\ 1 & 1 & 1 & z \end{bmatrix} \), and we need to find its Reduced Row Echelon Form (RREF). The process involves applying elementary row operations to transform the matrix into its RREF.

### Steps to calculate the RREF:
1. **Step 1:** Make the first pivot element (in the top-left corner) equal to 1 by dividing the first row by 3.
   
   \[
   \frac{1}{3}R_1 \rightarrow R_1
   \]
   \[
   \begin{bmatrix} 
   1 & \frac{2}{3} & \frac{1}{3} & \frac{x}{3} \\ 
   1 & -1 & -1 & y \\ 
   1 & 1 & 1 & z 
   \end{bmatrix}
   \]

2. **Step 2:** Eliminate the entries below the first pivot (1) by subtracting the first row from both the second and third rows:
   \[
   R_2 - R_1 \rightarrow R_2 \quad \text{and} \quad R_3 - R_1 \rightarrow R_3
   \]
   After performing these operations, the matrix becomes:
   \[
   \begin{bmatrix} 
   1 & \frac{2}{3} & \frac{1}{3} & \frac{x}{3} \\ 
   0 & -\frac{5}{3} & -\frac{4}{3} & y - \frac{x}{3} \\ 
   0 & \frac{1}{3} & \frac{2}{3} & z - \frac{x}{3} 
   \end{bmatrix}
   \]

3. **Step 3:** Make the second pivot (in the second row, second column) equal to 1 by multiplying the second row by \( -\frac{3}{5} \):
   \[
   -\frac{3}{5} R_2 \rightarrow R_2
   \]
   \[
   \begin{bmatrix} 
   1 & \frac{2}{3} & \frac{1}{3} & \frac{x}{3} \\ 
   0 & 1 & \frac{4}{5} & \frac{3x - 5y}{5} \\ 
   0 & \frac{1}{3} & \frac{2}{3} & z - \frac{x}{3} 
   \end{bmatrix}
   \]

4. **Step 4:** Eliminate the entry above and below the second pivot by subtracting appropriate multiples of the second row from the first and third rows:
   \[
   R_1 - \frac{2}{3}R_2 \rightarrow R_1 \quad \text{and} \quad R_3 - \frac{1}{3}R_2 \rightarrow R_3
   \]
   After performing these operations, the matrix becomes:
   \[
   \begin{bmatrix} 
   1 & 0 & -\frac{1}{5} & \frac{2x - 5y}{5} \\ 
   0 & 1 & \frac{4}{5} & \frac{3x - 5y}{5} \\ 
   0 & 0 & \frac{1}{5} & \frac{5z - x - 3y}{5} 
   \end{bmatrix}
   \]

5. **Step 5:** Make the third pivot (in the third row, third column) equal to 1 by multiplying the third row by 5:
   \[
   5R_3 \rightarrow R_3
   \]
   \[
   \begin{bmatrix} 
   1 & 0 & -\frac{1}{5} & \frac{2x - 5y}{5} \\ 
   0 & 1 & \frac{4}{5} & \frac{3x - 5y}{5} \\ 
   0 & 0 & 1 & 5z - x - 3y 
   \end{bmatrix}
   \]

6. **Step 6:** Eliminate the entries above the third pivot by adding appropriate multiples of the third row to the first and second rows:
   \[
   R_1 + \frac{1}{5} R_3 \rightarrow R_1 \quad \text{and} \quad R_2 - \frac{4}{5} R_3 \rightarrow R_2
   \]
   After performing these operations, the final RREF is:
   \[
   \begin{bmatrix} 
   1 & 0 & 0 & x - y + z \\ 
   0 & 1 & 0 & y - z \\ 
   0 & 0 & 1 & 5z - x - 3y 
   \end{bmatrix}
   \]

Thus, the matrix in RREF form is:

\[
\boxed{
\begin{bmatrix} 
1 & 0 & 0 & x - y + z \\ 
0 & 1 & 0 & y - z \\ 
0 & 0 & 1 & 5z - x - 3y 
\end{bmatrix}
}
\]

Would you like further details on any step? 

Here are some follow-up questions for deeper understanding:
1. What are elementary row operations?
2. How is RREF different from regular row echelon form?
3. How do you interpret solutions from RREF in the context of systems of equations?
4. What is the significance of the pivot columns in RREF?
5. Can a matrix have more than one RREF?

**Tip:** In any matrix, the number of non-zero rows in the RREF gives you the rank of the matrix.

Explore the step-by-step process of finding the Reduced Row Echelon Form (RREF) of the matrix [3, 2, 1, x; 1, -1, -1, y; 1, 1, 1, z]. Learn essential linear algebra concepts, including elementary row operations and matrix transformation. Suitable for undergraduate students studying linear algebra.

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