College Algebra, Chapter 7, 7.1, Section 7.1, Problem 20

The system of linear equations

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

x + y + 6z =& 3
\\
x + y + 3z =& 3
\\
x + 2y + 4z =& 7

\end{aligned}
\end{equation}
\right.
$
has a unique solution.

Determine the solution using Gaussian Elimination or Gauss-Jordan Elimination.

We use Gauss-Jordan Elimination

Augmented Matrix

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

$R_2 - R_3 \to R_2$


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

$R_3 - R_1 \to R_3$

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

$R_3 + R_2 \to R_3$

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

$\displaystyle \frac{-1}{3} R_3$

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

$-R_2$

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

$R_1 - R_2 \to R_1$

$\left[ \begin{array}{cccc}
1 & 0 & 5 & -1 \\
0 & 1 & 1 & 4 \\
0 & 0 & 1 & 0
\end{array} \right]$

$R_1 - 5R_3 \to R_1$

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

$R_2 - R_3 \to R_2$

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

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
\\
y =& 4
\\
z =& 0

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


Hence we immediately arrive at the solution $(-1,4,0)$.