Previously, we solved the following using the method of substitution:
\begin{eqnarray}
2x_1 - x_2 & = 0 &~~~~~~\text{(1)}\\
-x_1 + x_2 - 2x_3 & = 4 &~~~~~~\text{(2)} \\
3x_1 - 2x_2 + x_3 & = -2 &~~~~~~\text{(3)}
\end{eqnarray}
Recall that we first used \((1)\) to obtain that \(x_1 = \frac{1}{2}x_2\).
Then we replaced every occurrence of \(x_1\) in \((2)\) and \((3)\) with
\(\frac{1}{2}x_2\). As a result, we obtained the equivalent system:
\begin{eqnarray}
x_1 & = & \frac{1}{2}x_2 \\
\frac{1}{2} x_2 - 2x_3 & = & 4 \\
- \frac{1}{2}x_2 + x_3 & = & -2 \\
\end{eqnarray}
Rewriting gives:
\begin{eqnarray}
x_1 - \frac{1}{2}x_2 & = 0 ~~~~~~\text{(4)}\\
\frac{1}{2} x_2 - 2x_3 & = 4~~~~~~\text{(5)} \\
- \frac{1}{2}x_2 + x_3 & = -2~~~~~~\text{(6)} \\
\end{eqnarray}
We now see how this could be obtained from a different point of view.
Note that one can obtain \((4)\) by multipying both sides of \((1)\) by
\(\frac{1}{2}\). We use the notation \((4) = \frac{1}{2} \times (1)\)
to denote this operation.

One can obtain \((5)\) by adding \(\frac{1}{2}\) of the left-hand side of
\((1)\) to the left-hand side of \((2)\) and \(\frac{1}{2}\) of
the right-hand side of \((1)\) to the right-hand side of \((2)\).
We use the notation \((5) = (2) + \frac{1}{2} \times (1)\) to denote
this operation.

Similarly, we can say \((6) = (3) - \frac{3}{2} \times (1)\).

The system consisting of \((4), (5),\) and \((6)\) is equivalent to the
original system in the sense that any solution to it is a solution to the
original system and vice versa. That is, the operations we have performed
do not result in a gain or loss of solutions. In fact, the operations are
reversible.

Elementary operations

The types of operations we used in the example above turn out to be sufficient
to solve any system of linear equations defined over a field.
More precisely, the elementary operations are:

Multiplying both sides of an equation by the same nonzero constant.

Adding a constant multiple of an equation to another equation.

Swapping the positions of a pair of equations.

The third operation is not really necessary and is not employed in the example
above. But it is rather convenient to have around when we solve
equations by transforming matrices as we will soon see.

Use elementary operations to transform
\(\begin{array}{r}
x - y = 1 \\
2x + y = 5
\end{array}\)
to
\(\begin{array}{r}
3x = 6 \\
x - y = 1
\end{array}\).

First, add the first equation to the second equation. Then
swap the positions of the two equations.

Use elementary operations to transform
\(\begin{array}{r}
x - y + z = 1 \\
2x + y - z = 2 \\
-x + 2 y - 2z = -1
\end{array}\)
to
\(\begin{array}{r}
x - y + z = 1 \\
x + 2 y - 2z = 1 \\
y - z = 0
\end{array}\).

First, add \(-1\) times the first equation to the second equation. Then
add the first equation to the third equation.