Question

Solve the following system of equations by using row operations to convert an augmented matrix to reduced row echelon form. What is the value of z?

⎧⎩⎨⎪⎪2x+y−z=−54x−2y+z=−3x−y+5z=−13

Answer 1

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 \(z\):

1. \(z = -\frac{25}{24}\)

Therefore, the value of \(z\) is:

\[
\boxed{-\frac{25}{24}}
\]