given a system of linear equations we use the algorithm to solve the system.
Example:
$$ \begin{array}{rrrrrrrrr} 7x_1 & + & 14x_2 & + & 70x_3 & - & 6x_4 & = & -11 \\ x_1 & + & 2x_2 & + & 10x_3 & - & x_4 & = & -2 \\ 3x_1 & + & 7x_2 & + & 33x_3 & - & 3x_4 & = & -8 \end{array} $$
Step 1:
$$
\left[ \begin{array}{rrrrrrrrr}
7 & 14 & 70 & -6 & | & -11 \\
1 & 2 & 10 & -1 & | & -2 \\
3 & 7 & 33 & -3 & | & -8
\end{array} \right]
$$
Step 2: we skip solving for it for time
$$
\left[ \begin{array}{rrrrrrrrr}
1 & 0 & 4 & 0 & | & -11 \\
0 & 1 & 3 & 0 & | & -2 \\
0 & 0 & 0 & 1 & | & -8
\end{array} \right]
$$
Step 3:
$$
\begin{array}{rrrrrrrrr}
x_1 & + & 4x_3 & = & 5 \\
x_2 & + & 3x_3 & = & -2 \\
x_4 & = & 3
\end{array}
$$
$$ \begin{array}{rrrrrrrrr} x_1 = & 5 - 4x_3 \\ x_2 = & -3x_3 - 2 \\ x_3 = & \text{ True? } \\ x_4 = & 3 \end{array} $$
$$ \left(\begin{array}{r} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right) = \left(\begin{array}{r} 5 - 4x_3 \\ -2 - 3x_3 \\ x_3 \\ 3 \end{array}\right) = \left(\begin{array}{r} 5 \\ -2 \\ 0 \\ 3 \end{array}\right) + x_3 \left(\begin{array}{r} -4 \\ -3 \\ 1 \\ 0 \end{array}\right) $$
$ \therefore x_3 \in \mathbb{R} \Rightarrow$ there are infinite solutions
Example: Solve
$$ \left[ \begin{array}{rrrrrrrrr} 2 & -5 & 2 & | & 9 \\ 3 & -8 & 2 & | & -13 \\ 0 & 1 & 2 & | & 0 \end{array} \right] $$
Step 2:
$ R_2 \cdot \frac{1}{2}, R_2 \cdot \frac{1}{3}, R_2 - R_1, R_2 - R_3, R_2 \leftrightarrow r_3, R_1 + R_2 =$
$$ \left[ \begin{array}{rrrrrrrrr} 1 & 0 & 6 & | & \frac{9}{2} \\ 0 & 1 & 2 & | & 0 \\ 0 & 0 & 0 & | & 1 \end{array} \right] $$
Step 3:
$ x_1 + 6x_3 = \frac{9}{2}$
$x_2 + 2x_3 = 0$
$1 = 0$
$\therefore$ NO SOLUTION!
Example: Solve
$$ \left[ \begin{array}{rrrrrrrrr} 2 & -5 & 13 & | & 1000 \\ 3 & -9 & 3 & | & -0 \\ -3 & 6 & 8 & | & -600 \end{array} \right] $$
$ R_2 \cdot \frac{1}{3}, R_1 \leftrightarrow R_2, R_2 - 2R_1, R_3 + 5R_1, R_2 \cdot \frac{1}{11}, R_3 \cdot \frac{-1}{9}, R_3 - R_2, R_3 \cdot \frac{-11}{8}, R_1 - R_3, R_2 + \frac{15}{11}R_3, R_1 + 3R_2 =$
$$ \left[ \begin{array}{rrrrrrrrr} 1 & 0 & 0 & | & 1200 \\ 0 & 1 & 0 & | & 500 \\ 0 & 0 & 1 & | & 300 \end{array} \right] $$
$ x_1 = 1200$
$ x_2 = 500$
$ x_3 = 300$
$\therefore$ There is a unique solution.
Theorem: every linear system can have infinite solutions, no solution, a unique solution or some combination.