Intermediate Algebra, Chapter 4, Review Exercises, Section Review Exercises, Problem 28

Use row operations to solve the system $
\begin{equation}
\begin{aligned}

x + 2y - z =& 1 \\
3x + 4y + 2z =& -2 \\
-2x - y + z =& -1

\end{aligned}
\end{equation}
$
.

Augmented Matrix

$\displaystyle \left[
\begin{array}{ccc|c}
1 & 2 & -1 & 1 \\
3 & 4 & 2 & -2 \\
-2 & -1 & 1 & -1
\end{array}
\right]$

$\displaystyle R_2 - 3R_1 \to R_2$

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & -1 & 1 \\
0 & -2 & 5 & -5 \\
-2 & -1 & 1 & -1
\end{array}
\right]$

$\displaystyle R_3 + 2R_1 \to R_3$

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & -1 & 1 \\
0 & -2 & 5 & -5 \\
0 & 3 & -1 & 1
\end{array}
\right]$

$\displaystyle - \frac{1}{2} R_2$

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & -1 & 1 \\
0 & 1 & \displaystyle - \frac{5}{2} & \displaystyle \frac{5}{2} \\
0 & 3 & -1 & 1
\end{array}
\right]$

$R_3 - 3R_2 \to R_3$

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & -1 & 1 \\
0 & 1 & \displaystyle - \frac{5}{2} & \displaystyle \frac{5}{2} \\
0 & 0 & \displaystyle \frac{13}{2} & \displaystyle - \frac{13}{2}
\end{array}
\right]$

$\displaystyle \frac{2}{13} R_3$

$\displaystyle \left[
\begin{array}{cccc}
1 & 2 & -1 & 1 \\
0 & 1 & \displaystyle - \frac{5}{2} & \displaystyle \frac{5}{2} \\
0 & 0 & 1 & -1
\end{array}
\right]$

This augmented matrix leads to the system of equations.


$
\begin{equation}
\begin{aligned}

x + 2y - z =& 1
\\
\\
y - \frac{5}{2}z =& \frac{5}{2}
\\
\\
z =& -1

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

y - \frac{5}{2} (-1) =& \frac{5}{2}
&& \text{Substitute $z = -1$ in equation 2}
\\
\\
y + \frac{5}{2} =& \frac{5}{2}
&& \text{Multiply}
\\
\\
y =& 0
&& \text{Subtract each side by } \frac{5}{2}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

x + 2(0) - (-1) =& 1
&& \text{Substitute $y = 0$ and $z = -1$ in equation 1}
\\
x + 0 + 1 =& 1
&& \text{Multiply}
\\
x =& 0
&& \text{Subtract each side by $1$}

\end{aligned}
\end{equation}
$


The solution set of the system $\{ (0,0,-1) \}$.