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To solve this system of equations using the Gauss-Jordan method, we will transform the augmented matrix into reduced row echelon form. The system given is:

\[
\begin{cases}
x - 2y + z = 5 \\
x + y - 5z = -1 \\
3x - 5y + 2z = 16 
\end{cases}
\]

### Step 1: Write the Augmented Matrix
\[
\begin{bmatrix}
1 & -2 & 1 & | & 5 \\
1 & 1 & -5 & | & -1 \\
3 & -5 & 2 & | & 16
\end{bmatrix}
\]

### Step 2: Make the First Element in the First Column a Leading 1
The first row already has a 1 in the first column, so we’ll use it as our pivot.

### Step 3: Eliminate the First Element in the Other Rows
- Subtract Row 1 from Row 2 to make the first element of Row 2 zero:
\[
R_2 = R_2 - R_1 \Rightarrow [0, 3, -6, | -6]
\]
- Subtract 3 times Row 1 from Row 3:
\[
R_3 = R_3 - 3 \cdot R_1 \Rightarrow [0, 1, -1, | 1]
\]

The matrix is now:
\[
\begin{bmatrix}
1 & -2 & 1 & | & 5 \\
0 & 3 & -6 & | & -6 \\
0 & 1 & -1 & | & 1
\end{bmatrix}
\]

### Step 4: Make the Second Element in the Second Row a Leading 1
Divide Row 2 by 3:
\[
R_2 = \frac{1}{3} R_2 \Rightarrow [0, 1, -2, | -2]
\]

The matrix is now:
\[
\begin{bmatrix}
1 & -2 & 1 & | & 5 \\
0 & 1 & -2 & | -2 \\
0 & 1 & -1 & | 1
\end{bmatrix}
\]

### Step 5: Eliminate the Second Element in the Other Rows
- Add 2 times Row 2 to Row 1 to make the second element in Row 1 zero:
\[
R_1 = R_1 + 2 \cdot R_2 \Rightarrow [1, 0, -3, | 1]
\]
- Subtract Row 2 from Row 3:
\[
R_3 = R_3 - R_2 \Rightarrow [0, 0, 1, | 3]
\]

The matrix is now:
\[
\begin{bmatrix}
1 & 0 & -3 & | & 1 \\
0 & 1 & -2 & | & -2 \\
0 & 0 & 1 & | & 3
\end{bmatrix}
\]

### Step 6: Make the Third Element in the Third Row a Leading 1
Row 3 already has a leading 1 in the third column.

### Step 7: Eliminate the Third Element in the Other Rows
- Add 3 times Row 3 to Row 1:
\[
R_1 = R_1 + 3 \cdot R_3 \Rightarrow [1, 0, 0, | 10]
\]
- Add 2 times Row 3 to Row 2:
\[
R_2 = R_2 + 2 \cdot R_3 \Rightarrow [0, 1, 0, | 4]
\]

The matrix is now in reduced row echelon form:
\[
\begin{bmatrix}
1 & 0 & 0 & | & 10 \\
0 & 1 & 0 & | & 4 \\
0 & 0 & 1 & | & 3
\end{bmatrix}
\]

### Solution
The solution to the system is:
\[
x = 10, \quad y = 4, \quad z = 3
\]

Would you like a more detailed explanation of any of these steps, or do you have any questions?

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### Related Questions
1. How would you solve a system of equations using the Gaussian elimination method instead of Gauss-Jordan?
2. What are the differences between Gaussian elimination and Gauss-Jordan elimination?
3. How would you check if the solution obtained is correct for the given system of equations?
4. How can you interpret the results of a system of equations if you find infinitely many solutions?
5. How would the method change if one of the rows was entirely zeros?

### Tip
When solving a system with the Gauss-Jordan method, always aim to achieve a diagonal of 1s with zeros above and below each leading 1. This helps you reach the solution more systematically.

Use the Gauss-Jordan method to solve the system: 
\(x - 2y + z = 5\)
\(x + y - 5z = -1\)
\(3x - 5y + 2z = 16\)

Learn how to solve a system of linear equations using the Gauss-Jordan method. This problem demonstrates step-by-step row operations to achieve reduced row echelon form and find solutions for x, y, and z. Suitable for high school students in Grades 10-12.

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