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 $(-2,3,3)$.