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 $(-1, 3, 5)$.