College Algebra, Chapter 7, 7.1, Section 7.1, Problem 24

The system of linear equations

$
\left\{
\begin{array}{ccccc}
& 2y & + z & = & 4 \\
x & + y & & = & 4 \\
3x & + 3y & - z & = & 10
\end{array}
\right.
$
has a unique solution.

We use Gauss-Jordan Elimination

Augmented Matrix

$\left[ \begin{array}{cccc}
0 & 2 & 1 & 4 \\
1 & 1 & 0 & 4 \\
3 & 3 & -1 & 10
\end{array} \right]$

$R_3 \leftrightarrow R_1$


$\left[ \begin{array}{cccc}
3 & 3 & -1 & 10 \\
1 & 1 & 0 & 4 \\
0 & 2 & 1 & 4
\end{array} \right]$

$3R_2 - R_1 \to R_2$

$\left[ \begin{array}{cccc}
3 & 3 & -1 & 10 \\
0 & 0 & 1 & 2 \\
0 & 2 & 1 & 4
\end{array} \right]$

$R_3 \leftrightarrow R_2$

$\left[ \begin{array}{cccc}
3 & 3 & -1 & 10 \\
0 & 2 & 1 & 4 \\
0 & 0 & 1 & 2
\end{array} \right]$

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

$\left[ \begin{array}{cccc}
3 & 3 & -1 & 10 \\
0 & 1 & \displaystyle \frac{1}{2} & 2 \\
0 & 0 & 1 & 2
\end{array} \right]$

$\displaystyle \frac{1}{3}R_1$

$\left[ \begin{array}{cccc}
1 & 1 & \displaystyle \frac{-1}{3} & \displaystyle \frac{10}{3} \\
0 & 1 & \displaystyle \frac{1}{2} & 2 \\
0 & 0 & 1 & 2
\end{array} \right]$

$R_1 - R_2 \to R_1$

$\left[ \begin{array}{cccc}
1 & 0 & \displaystyle \frac{-5}{6} & \displaystyle \frac{4}{3} \\
0 & 1 & \displaystyle \frac{1}{2} & 2 \\
0 & 0 & 1 & 2
\end{array} \right]$

$\displaystyle R_1 + \frac{5}{6} R_3 \to R_1$

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

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

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

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


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

x =& 3
\\
y =& 1
\\
z =& 2

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


Hence we immediately arrive at the solution $(3,1,2)$.