College Algebra, Chapter 7, 7.1, Section 7.1, Problem 22

The system of linear equations

$
\left\{
\begin{array}{ccccc}
x & + y & + z & = & 4 \\
-x & + 2y & + 3z & = & 17 \\
2x & - y & & = & -7
\end{array}
\right.
$
has a unique solution.

We use Gauss-Jordan Elimination

Augmented Matrix

$\left[ \begin{array}{cccc}
1 & 1 & 1 & 4 \\
-1 & 2 & 3 & 17 \\
2 & -1 & 0 & -7
\end{array} \right]$

$R_2 + R_1 \to R_2$


$\left[ \begin{array}{cccc}
1 & 1 & 1 & 4 \\
0 & 3 & 4 & 21 \\
2 & -1 & 0 & -7
\end{array} \right]$

$R_3 - 2R_1 \to R_3$

$\left[ \begin{array}{cccc}
1 & 1 & 1 & 4 \\
0 & 3 & 4 & 21 \\
0 & -3 & -2 & -15
\end{array} \right]$

$R_3 + R_2 \to R_3$

$\left[ \begin{array}{cccc}
1 & 1 & 1 & 4 \\
0 & 3 & 4 & 21 \\
0 & 0 & 2 & 6
\end{array} \right]$

$\displaystyle \frac{1}{2} R_3$

$\left[ \begin{array}{cccc}
1 & 1 & 1 & 4 \\
0 & 3 & 4 & 21 \\
0 & 0 & 1 & 3
\end{array} \right]$

$\displaystyle \frac{1}{3}R_2$

$\left[ \begin{array}{cccc}
1 & 1 & 1 & 4 \\
0 & 1 & \displaystyle \frac{4}{3} & 7 \\
0 & 0 & 1 & 3
\end{array} \right]$

$R_1 - R_2 \to R_1$

$\left[ \begin{array}{cccc}
1 & 0 & \displaystyle \frac{-1}{3} & -3 \\
0 & 1 & \displaystyle \frac{4}{3} & 7 \\
0 & 0 & 1 & 3
\end{array} \right]$

$\displaystyle R_1 + \frac{1}{3} R_3 \to R_1$

$\left[ \begin{array}{cccc}
1 & 0 & 0 & -2 \\
0 & 1 & \displaystyle \frac{4}{3} & 7 \\
0 & 0 & 1 & 3
\end{array} \right]$

$\displaystyle R_2 - \frac{4}{3} R_3 \to R_2$

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

We now have an equivalent matrix in reduced row-echelon form and the corresponding system of equations is


$
\left\{
\begin{equation}
\begin{aligned}

x =& -2
\\
y =& 3
\\
z =& 3

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


Hence we immediately arrive at the solution $(-2,3,3)$.