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To solve the given system of linear equations using the Gauss-Jordan method, we first write down the augmented matrix corresponding to the system of equations:

\[
\begin{aligned}
&\text{Equations:}\\
&3x + 2y + z = 1 \quad \text{(Equation 1)} \\
&5x + 3y + 4z = 2 \quad \text{(Equation 2)} \\
&x + y - z = 1 \quad \text{(Equation 3)}
\end{aligned}
\]

This system can be represented by the following augmented matrix:

\[
\begin{pmatrix}
3 & 2 & 1 & | & 1 \\
5 & 3 & 4 & | & 2 \\
1 & 1 & -1 & | & 1
\end{pmatrix}
\]

### Step 1: Make the first element of the first column (pivot) equal to 1.
To do this, we divide the third row by 1 (since the first element of the third row is already 1):

\[
\begin{pmatrix}
3 & 2 & 1 & | & 1 \\
5 & 3 & 4 & | & 2 \\
1 & 1 & -1 & | & 1
\end{pmatrix}
\quad \rightarrow \quad
\begin{pmatrix}
1 & 1 & -1 & | & 1 \\
3 & 2 & 1 & | & 1 \\
5 & 3 & 4 & | & 2
\end{pmatrix}
\]

### Step 2: Eliminate the elements below the pivot in the first column.
Subtract 3 times the first row from the second row and 5 times the first row from the third row:

\[
R2 = R2 - 3R1, \quad R3 = R3 - 5R1
\]

\[
\begin{pmatrix}
1 & 1 & -1 & | & 1 \\
0 & -1 & 4 & | & -2 \\
0 & -2 & 9 & | & -3
\end{pmatrix}
\]

### Step 3: Make the second pivot element equal to 1.
To do this, we divide the second row by -1:

\[
\begin{pmatrix}
1 & 1 & -1 & | & 1 \\
0 & 1 & -4 & | & 2 \\
0 & -2 & 9 & | & -3
\end{pmatrix}
\]

### Step 4: Eliminate the elements above and below the second pivot.
Subtract the first row from the second row, and add 2 times the second row to the third row:

\[
R1 = R1 - R2, \quad R3 = R3 + 2R2
\]

\[
\begin{pmatrix}
1 & 0 & 3 & | & -1 \\
0 & 1 & -4 & | & 2 \\
0 & 0 & 1 & | & 1
\end{pmatrix}
\]

### Step 5: Make the third pivot element equal to 1 (already done).
The third pivot is already 1, so now we just eliminate the elements above the third pivot:

Subtract 3 times the third row from the first row, and add 4 times the third row to the second row:

\[
R1 = R1 - 3R3, \quad R2 = R2 + 4R3
\]

\[
\begin{pmatrix}
1 & 0 & 0 & | & -4 \\
0 & 1 & 0 & | & 6 \\
0 & 0 & 1 & | & 1
\end{pmatrix}
\]

### Conclusion:
The solution to the system of equations is:

\[
x = -4, \quad y = 6, \quad z = 1
\]

Would you like any additional details or explanations? Here are five related questions to further explore:

1. How can the Gauss-Jordan method be applied to systems with more variables?
2. What are the differences between the Gauss elimination method and the Gauss-Jordan method?
3. How can matrix inversion be used to solve systems of linear equations?
4. What happens if a system of linear equations is inconsistent when using the Gauss-Jordan method?
5. Can the Gauss-Jordan method be used for non-square matrices (more equations than variables)?

**Tip:** When performing row operations, always double-check each step to avoid errors in calculation, as these can easily lead to incorrect results in larger systems.

Learn how to solve systems of linear equations using the Gauss-Jordan method with a detailed step-by-step guide. This problem involves applying linear algebra concepts to solve for variables in equations. Suitable for university-level students and professionals.

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