College Algebra, Chapter 7, 7.1, Section 7.1, Problem 46

Find the complete solution of the system $
\left\{
\begin{equation}
\begin{aligned}

3x-y + 2z =& -1
\\
4x - 2y + z =& -7
\\
-x + 3y - 2z =& -1

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


We transform the system into reduced row-echelon form

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

$\displaystyle \frac{1}{3} R_1$

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

$\displaystyle R_2 - 4R_1 \to R_2$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{-1}{3} & \displaystyle \frac{2}{3} & \displaystyle \frac{-1}{3} \\
0 & \displaystyle \frac{-2}{3} & \displaystyle \frac{-5}{3} & \displaystyle \frac{-17}{3} \\
-1 & 3 & -2 & -1
\end{array} \right]$

$\displaystyle R_3 + R_1 \to R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{-1}{3} & \displaystyle \frac{2}{3} & \displaystyle \frac{-1}{3} \\
0 & \displaystyle \frac{-2}{3} & \displaystyle \frac{-5}{3} & \displaystyle \frac{-17}{3} \\
0 & \displaystyle \frac{8}{3} & \displaystyle \frac{-4}{3} & \displaystyle \frac{-4}{3}
\end{array} \right]$

$\displaystyle \frac{-3}{2} R_2$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{-1}{3} & \displaystyle \frac{2}{3} & \displaystyle \frac{-1}{3} \\
0 & 1 & \displaystyle \frac{5}{2} & \displaystyle \frac{17}{2} \\
0 & \displaystyle \frac{8}{3} & \displaystyle \frac{-4}{3} & \displaystyle \frac{-4}{3}
\end{array} \right]$

$\displaystyle R_3 - \frac{8}{3} R_2 \to R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{-1}{3} & \displaystyle \frac{2}{3} & \displaystyle \frac{-1}{3} \\
0 & 1 & \displaystyle \frac{5}{2} & \displaystyle \frac{17}{2} \\
0 & 0 & -8 & -24
\end{array} \right]$

$\displaystyle \frac{-1}{8} R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{-1}{3} & \displaystyle \frac{2}{3} & \displaystyle \frac{-1}{3} \\
0 & 1 & \displaystyle \frac{5}{2} & \displaystyle \frac{17}{2} \\
0 & 0 & 1 & 3
\end{array} \right]$

$\displaystyle R_2 - \frac{5}{2} R_3 \to R_2$

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

$\displaystyle R_1 - \frac{2}{3} R_3 \to R_1$

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

$\displaystyle R_1 + \frac{1}{3} R_2 \to R_1$

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

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


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

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

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


We can write the solution as the ordered triple $(-2, 1, 3)$.