College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 8

Find the complete solution of the system
$\left\{\begin{equation} \begin{aligned} x - y + z =& 2 \\ x + y + 3z =& 6 \\ 2y + 3z =& 5 \end{aligned} \end{equation} \right.$
using Gaussian Elimination.

For this system we have

$\displaystyle \left[ \begin{array}{cccc} 1 & -1 & 1 & 2 \\ 1 & 1 & 3 & 6 \\ 0 & 2 & 3 & 5 \end{array} \right]$

$R_2 - R_1 \to R_2$

$\displaystyle \left[ \begin{array}{cccc} 1 & -1 & 1 & 2 \\ 0 & 2 & 2 & 4 \\ 0 & 2 & 3 & 5 \end{array} \right]$

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

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

$R_3 - 2R_2 \to R_3$

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

Now we have equivalent matrix in row-echelon form and the corresponding system is


$\left\{ \begin{equation} \begin{aligned} x - y + z =& 2 \\ y + z =& 2 \\ z =& 1 \end{aligned} \end{equation} \right.$


Then we back-substitute $z = 1$ into the second equation and solve for $y$


$\begin{equation} \begin{aligned} y + 1 =& 2 && \text{Back-substitute } z= 1 \\ y =& 1 && \text{Subtract } 1 \end{aligned} \end{equation}$


Now we back-substitute $y=1$ and $z=1$ into the first equation and solve for $x$


$\begin{equation} \begin{aligned} x -1 + 1 =& 2 && \text{Back-substitute } y = 1 \text{ and } z = 1 \\ x =& 2 && \text{Simplify} \end{aligned} \end{equation}$