To solve the system of equations using row operations, we first write the augmented matrix corresponding to the system:
\[ \begin{bmatrix} 2 & 1 & -1 & | & -5 \ 4 & -2 & 1 & | & -3 \ -3 & -1 & 5 & | & -13 \end{bmatrix} \]
Now we will perform row operations to convert this matrix into reduced row echelon form.
Step 1: Making the leading coefficient of the first row 1.
First, we can divide the first row by 2 to make the leading coefficient of the first row equal to 1:
\[ R_1 \leftarrow \frac{1}{2} R_1 \]
This gives:
\[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & -\frac{5}{2} \ 4 & -2 & 1 & | & -3 \ -3 & -1 & 5 & | & -13 \end{bmatrix} \]
Step 2: Eliminate the first column entries in Rows 2 and 3.
Next, we will eliminate the first column entries in \(R_2\) and \(R_3\).
For Row 2:
\[ R_2 \leftarrow R_2 - 4R_1 \]
Calculating \(R_2\):
\[ \begin{bmatrix} 4 & -2 & 1 & | & -3 \end{bmatrix} - 4 \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & -\frac{5}{2} \end{bmatrix} = \begin{bmatrix} 0 & -4 & 3 & | & 7 \end{bmatrix} \]
For Row 3:
\[ R_3 \leftarrow R_3 + 3R_1 \]
Calculating \(R_3\):
\[ \begin{bmatrix} -3 & -1 & 5 & | & -13 \end{bmatrix} + 3 \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & -\frac{5}{2} \end{bmatrix} = \begin{bmatrix} 0 & \frac{1}{2} & \frac{7}{2} & | & -\frac{29}{2} \end{bmatrix} \]
So now our matrix looks like:
\[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & -\frac{5}{2} \ 0 & -4 & 3 & | & 7 \ 0 & \frac{1}{2} & \frac{7}{2} & | & -\frac{29}{2} \end{bmatrix} \]
Step 3: Make the leading coefficient of Row 2 equal to 1.
Now, divide \(R_2\) by -4:
\[ R_2 \leftarrow -\frac{1}{4} R_2 \]
This gives:
\[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & -\frac{5}{2} \ 0 & 1 & -\frac{3}{4} & | & -\frac{7}{4} \ 0 & \frac{1}{2} & \frac{7}{2} & | & -\frac{29}{2} \end{bmatrix} \]
Step 4: Eliminate the second column entry in Row 3.
We will now eliminate the second column entry in \(R_3\):
\[ R_3 \leftarrow R_3 - \frac{1}{2} R_2 \]
Calculating \(R_3\):
\[ R_3 - \frac{1}{2} \begin{bmatrix} 0 & 1 & -\frac{3}{4} & | & -\frac{7}{4} \end{bmatrix} = \begin{bmatrix} 0 & 0 & 6 & | & -\frac{25}{4} \end{bmatrix} \]
So now the matrix becomes:
\[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & -\frac{5}{2} \ 0 & 1 & -\frac{3}{4} & | & -\frac{7}{4} \ 0 & 0 & 6 & | & -\frac{25}{4} \end{bmatrix} \]
Step 5: Make the leading coefficient in Row 3 equal to 1.
Divide \(R_3\) by 6:
\[ R_3 \leftarrow \frac{1}{6} R_3 \]
This gives:
\[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & -\frac{5}{2} \ 0 & 1 & -\frac{3}{4} & | & -\frac{7}{4} \ 0 & 0 & 1 & | & -\frac{25}{24} \end{bmatrix} \]
Step 6: Back substitution to get rows in reduced echelon form.
We will perform back substitution to eliminate above rows:
For \(R_1\): \[ R_1 \leftarrow R_1 + \frac{1}{2} R_3 \]
Calculating \(R_1\):
\[ R_1 + \frac{1}{2} \begin{bmatrix} 0 & 0 & 1 & | & -\frac{25}{24} \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{2} & 0 & | & -\frac{55}{48} \end{bmatrix} \]
For \(R_2\): \[ R_2 \leftarrow R_2 + \frac{3}{4} R_3 \]
Calculating \(R_2\):
\[ R_2 + \frac{3}{4} \begin{bmatrix} 0 & 0 & 1 & | & -\frac{25}{24} \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & | & -\frac{47}{32} \end{bmatrix} \]
The final matrix in reduced row echelon form is:
\[ \begin{bmatrix} 1 & 0 & 0 & | & -\frac{55}{48} \ 0 & 1 & 0 & | & -\frac{47}{32} \ 0 & 0 & 1 & | & -\frac{25}{24} \end{bmatrix} \]
Step 7: Identify the values of the variables.
From the final matrix, we can read off the values of \(x\), \(y\), and \(z\):
- \(z = -\frac{25}{24}\)
Therefore, the value of \(z\) is:
\[ \boxed{-\frac{25}{24}} \]