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 $(-2, 1, 3)$.