We can start by converting the system of equations into an augmented matrix. The given equations are:
- \( 2x + y - 2z = -1 \)
- \( 3x - 3y - z = 5 \)
- \( x - 2y + 3z = 6 \)
We will represent this system in an augmented matrix form:
\[ \begin{bmatrix} 2 & 1 & -2 & | & -1 \ 3 & -3 & -1 & | & 5 \ 1 & -2 & 3 & | & 6 \end{bmatrix} \]
Next, we will perform row operations to convert this matrix into row echelon form.
Step 1:
We can make the first column leading coefficient in the first row. Let's interchange Row 1 and Row 3 to obtain a leading 1 in the first row.
\[ \begin{bmatrix} 1 & -2 & 3 & | & 6 \ 3 & -3 & -1 & | & 5 \ 2 & 1 & -2 & | & -1 \end{bmatrix} \]
Step 2:
Now, we will eliminate the \(x\) terms from Row 2 and Row 3 using Row 1.
For Row 2, we can do the following operation: \[ R_2 \leftarrow R_2 - 3R_1 \] Calculating: \[ 3 - 3(1) = 0, \quad -3 - 3(-2) = 3, \quad -1 - 3(3) = -10, \quad 5 - 3(6) = -13 \] So, Row 2 becomes: \[ \begin{bmatrix} 0 & 3 & -10 & | & -13 \end{bmatrix} \]
For Row 3, we will do: \[ R_3 \leftarrow R_3 - 2R_1 \] Calculating: \[ 2 - 2(1) = 0, \quad 1 - 2(-2) = 5, \quad -2 - 2(3) = -8, \quad -1 - 2(6) = -13 \] So, Row 3 becomes: \[ \begin{bmatrix} 0 & 5 & -8 & | & -13 \end{bmatrix} \]
Now, we have: \[ \begin{bmatrix} 1 & -2 & 3 & | & 6 \ 0 & 3 & -10 & | & -13 \ 0 & 5 & -8 & | & -13 \end{bmatrix} \]
Step 3:
Now we will simplify Row 2 by dividing it by 3: \[ R_2 \leftarrow \frac{1}{3} R_2 \] This gives: \[ \begin{bmatrix} 0 & 1 & -\frac{10}{3} & | & -\frac{13}{3} \end{bmatrix} \] Now our matrix looks like: \[ \begin{bmatrix} 1 & -2 & 3 & | & 6 \ 0 & 1 & -\frac{10}{3} & | & -\frac{13}{3} \ 0 & 5 & -8 & | & -13 \end{bmatrix} \]
Next, let's eliminate the \(y\) term from Row 3: \[ R_3 \leftarrow R_3 - 5R_2 \] Calculating: \[ 0 - 5(0) = 0, \quad 5 - 5(1) = 0, \quad -8 - 5\left(-\frac{10}{3}\right) = -8 + \frac{50}{3} = \frac{-24 + 50}{3} = \frac{26}{3}, \quad -13 - 5\left(-\frac{13}{3}\right) = -13 + \frac{65}{3} = \frac{-39 + 65}{3} = \frac{26}{3} \] So Row 3 becomes: \[ \begin{bmatrix} 0 & 0 & \frac{26}{3} & | & \frac{26}{3} \end{bmatrix} \] Our augmented matrix now is: \[ \begin{bmatrix} 1 & -2 & 3 & | & 6 \ 0 & 1 & -\frac{10}{3} & | & -\frac{13}{3} \ 0 & 0 & \frac{26}{3} & | & \frac{26}{3} \end{bmatrix} \]
Back substitution:
From the last row, we see: \[ \frac{26}{3} z = \frac{26}{3} \] Thus, we have: \[ z = 1 \]
Therefore, the value of \( z \) is: \[ \boxed{1} \]
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