Question

Solve the following system of equations by converting it to an augmented matrix and writing it in row echelon form. What is the value of z?

2x+y−2z=−1
2
𝑥
+
𝑦
−
2
𝑧
=
−
1

3x−3y−z=5
3
𝑥
−
3
𝑦
−
𝑧
=
5

x−2y+3z=6

Answer 1

We can start by converting the system of equations into an augmented matrix. The given equations are:

1. \( 2x + y - 2z = -1 \)
2. \( 3x - 3y - z = 5 \)
3. \( x - 2y + 3z = 6 \)

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:
\[
\boxed{1}
\]