College Algebra, Chapter 7, 7.1, Section 7.1, Problem 26

The system of linear equations

$\left\{ \begin{array}{ccccc} 2x_1 & + x_2 & & = & 7 \\ 2x_1 & - x_2 & + x_3 & = & 6 \\ 3x_1 & - 2x_2 & + 4x_3 & = & 11 \end{array} \right.$
has a unique solution.

We use Gauss-Jordan Elimination

Augmented Matrix

$\left[ \begin{array}{cccc} 2 & 1 & 0 & 7 \\ 2 & -1 & 1 & 6 \\ 3 & -2 & 4 & 11 \end{array} \right]$

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


$\left[ \begin{array}{cccc} 1 & \displaystyle \frac{1}{2} & 0 & \displaystyle \frac{7}{2} \\ 2 & -1 & 1 & 6 \\ 3 & -2 & 4 & 11 \end{array} \right]$

$R_3 - 3R_1 \to R_3$

$\left[ \begin{array}{cccc} 1 & \displaystyle \frac{1}{2} & 0 & \displaystyle \frac{7}{2} \\ 2 & -1 & 1 & 6 \\ 0 & \displaystyle \frac{-7}{2} & 4 & \displaystyle \frac{1}{2} \end{array} \right]$

$R_2 - 2R_1 \to R_2$

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

$\displaystyle R_3 - \frac{7}{4} R_2 \to R_3$

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

$\displaystyle \frac{4}{9} R_3$

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

$\displaystyle \frac{-1}{2} R_2$

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

$\displaystyle R_1 - \frac{1}{2} R_2 \to R_1$

$\left[ \begin{array}{cccc} 1 & 0 & \displaystyle \frac{1}{4} & \displaystyle \frac{13}{4} \\ 0 & 1 & \displaystyle \frac{-1}{2} & \displaystyle \frac{1}{2} \\ 0 & 0 & 1 & 1 \end{array} \right]$

$\displaystyle R_1 - \frac{1}{4} R_3 \to R_1$

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

$\displaystyle R_2 + \frac{1}{2} R_3 \to R_2$

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

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


$\left\{ \begin{equation} \begin{aligned} x_1 =& 3 \\ x_2 =& 1 \\ x_3 =& 1 \end{aligned} \end{equation} \right.$


Hence we immediately arrive at the solution $(3,1,1)$.