College Algebra, Chapter 7, 7.1, Section 7.1, Problem 40

Find the complete solution of the system

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

2x - 3y + 5z =& 14
\\
4x - y - 2z =& -17
\\
-x - y + z =& 3

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


We first write the augmented matrix of the system and using Gauss-Jordan Elimination.

$\left[ \begin{array}{cccc}
2 & -3 & 5 & 14 \\
4 & -1 & -2 & -17 \\
-1 & -1 & 1 & 3
\end{array} \right]$

$\displaystyle \frac{1}{2} R_1$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{-3}{2} & \displaystyle \frac{5}{2} & 7 \\
4 & -1 & -2 & -17 \\
-1 & -1 & 1 & 3
\end{array} \right]$

$\displaystyle R_3 + R_1 \to R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{-3}{2} & \displaystyle \frac{5}{2} & 7 \\
4 & -1 & -2 & -17 \\
0 & \displaystyle \frac{-5}{2} & \displaystyle \frac{7}{2} & 10
\end{array} \right]$

$\displaystyle R_2 - 4 R_1 \to R_2$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{-3}{2} & \displaystyle \frac{5}{2} & 7 \\
0 & 5 & -12 & -45 \\
0 & \displaystyle \frac{-5}{2} & \displaystyle \frac{7}{2} & 10
\end{array} \right]$

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

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{-3}{2} & \displaystyle \frac{5}{2} & 7 \\
0 & 5 & -12 & -45 \\
0 & 1 & \displaystyle \frac{-7}{5} & -4
\end{array} \right]$

$\displaystyle \frac{1}{5}
R_2$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{-3}{2} & \displaystyle \frac{5}{2} & 7 \\
0 & 1 & \displaystyle \frac{-12}{5} & -9 \\
0 & 1 & \displaystyle \frac{-7}{5} & -4
\end{array} \right]$

$\displaystyle R_3 - R_2 \to R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{-3}{2} & \displaystyle \frac{5}{2} & 7 \\
0 & 1 & \displaystyle \frac{-12}{5} & -9 \\
0 & 0 & 1 & 5
\end{array} \right]$

$\displaystyle R_1 + \frac{3}{2} R_2 \to R_1$

$\left[ \begin{array}{cccc}
1 & 0 & \displaystyle \frac{-11}{10} & \displaystyle \frac{-13}{2} \\
0 & 1 & \displaystyle \frac{-12}{5} & -9 \\
0 & 0 & 1 & 5
\end{array} \right]$

$\displaystyle R_2 + \frac{12}{5} R_3 \to R_2$

$\left[ \begin{array}{cccc}
1 & 0 & \displaystyle \frac{-11}{10} & \displaystyle \frac{-13}{2} \\
0 & 1 & 0 & 3 \\
0 & 0 & 1 & 5
\end{array} \right]$

$\displaystyle R_1 + \frac{11}{10} R_3 \to R_1$

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

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


$
\begin{equation}
\begin{aligned}

x =& -1
\\
\\
y =& 3
\\
\\
z =& 5

\end{aligned}
\end{equation}
$


We can write the solution as the ordered triple $(-1, 3, 5)$.