One of the major goals of Chapter 2 is to develop systematic techniques for solving systems of equations. The Gauss-Jordan method is based on two ideas. The first is that certain systems of equations are very easy to solve.

Consider the system of equations \[ \begin{cases} \begin{array}{ccccc} x & & & = & 3 \\ & & y & = & 4 \\ \end{array} \end{cases} \] Find the solution to this system of equations.

**Solution:**
We really don't have to do anything. The solution is \((x,y) = (3,4)\).

Geometrically, we are finding the intersection of a horizontal line with a vertical line.

That's the first big idea with Gauss-Jordan elimination:
it is easy to solve systems of linear equations when each of the lines is parallel to one of the coordinate axes.

What if our system is not as simple as the above example. Consider this system:
\[
\begin{cases}
\begin{array}{ccccc}
x & - & y & = & -3 \\
2x & + & 3y & = & -5 \\
\end{array}
\end{cases}
\]

We will try to solve this system of equations by replacing our system with a new system, in a manner so that

- the new system and the old system are
*equivalent,*in the sense that they have the same solution set - the new system is
*easier to solve*than the old system.

**Change the order of the equations.**\[ \begin{cases} \begin{array}{ccccc} x & - & y & = & -3 \\ 2x & + & 3y & = & -5 \\ \end{array} \end{cases} \qquad \mbox{ and } \qquad \begin{cases} \begin{array}{ccccc} 2x & + & 3y & = & -5 \\ x & - & y & = & -3 \\ \end{array} \end{cases} \] are equivalent systems. They have the same solution sets. They have the same intersection points. All we've done is changed which line is "first" and which is "second." But whether we think of the red line as "the first line" and the purple line as "the second line", or the purple line as "the first line" and the red line as "the second line" doesn't change the above picture, and therefore won't change the point of intersection.**Replace any equation by a nonzero constant multiple of itself.**\[ \begin{cases} \begin{array}{ccccc} x & - & y & = & -3 \\ 2x & + & 3y & = & -5 \\ \end{array} \end{cases} \qquad \mbox{ and } \qquad \begin{cases} \begin{array}{ccccc} 2x & - & 2y & = & -6 \\ 2x & + & 3y & = & -5 \\ \end{array} \end{cases} \] are equivalent systems. They have the same solution sets. They have the same intersection points. Basic algebra tells us that \(x - y = -3\) and \(2x - 2y = -6\) are the same equation, and they both represent the same line. All we are doing here is changing the way we describe one of the lines. We haven't changed either of the lines, so the interesection point won't change.**Replace any equation by the sum of that equation and a constant multiple of any other equation.**Its a bit harder to justify why this doesn't change the intersection points, as in this case we are actually changing the lines.

According to this rule, \[ \begin{cases} \begin{array}{ccccc} x & - & y & = & -3 \\ 2x & + & 3y & = & -5 \\ \end{array} \end{cases} \qquad \mbox{ and } \qquad \begin{cases} \begin{array}{ccccc} x & - & y & = & -3 \\ 3x & + & 2y & = & -8 \\ \end{array} \end{cases} \] are equivalent systems. (The original second equation was replaced by the sum of the second and first equation.)

Use this applet to see what's going on. For the moment, keep \(k_2 = 0\), and slide \(k_1\) around. This will replace the second equation by the sum of the second equation and \(k_1\) times the first equation. The purple line moves around as we vary \(k_1\), so the system of equations is changing. However, notice the point of intersection does not change. Effectively, this row operation allows us to rotate one of the lines about the point of intersection.

As stated above, the Gauss-Jordan Method attempts to solve systems of equations by replacing the given system with a new system, in a manner so that

- the new system and the old system are
*equivalent,*in the sense that they have the same solution set - the new system is
*easier to solve*than the old system.

Solve the system of equations: \[ \begin{cases} \begin{array}{ccccc} x & - & y & = & -3 \\ 2x & + & 3y & = & -5 \\ \end{array} \end{cases} \]

We want to replace this system with an equivalent, but simpler system. The simpler system will hopefully be of the form
\(x = \) some number, \(y = \) some number. Geometrically, this means we want our two lines to be parallel to the coordinte axes.

First, lets try to get rid of the \(x\) in the bottom equation. In other words, lets vary \(k_1\) until the purple line becomes horizontal.

(We are doing the row operation \(R_2 \mapsto R_2 + k_1 \cdot R_1\) and we are trying to find a "good choice" of \(k_1\).)

A little trial and error, and we find \(k_1 = -2\) turns the purple line into the horizontal line \(y = 0.2\).
(In class, we'll learn how to do this without the fancy applet.) Our system is now \[ \begin{cases} \begin{array}{ccccc} x & - & y & = & -3 \\ & & y & = & 0.2 \\ \end{array} \end{cases} \] Now lets try to get rid of the \(y\) in the top equation. In other words, lets vary \(k_2\) until the red line becomes vertical.

(We are now doing the row operation \(R_1 \mapsto R_1 + k_2 \cdot R_2\) and we are trying to find a "good choice" of \(k_2\).)

A bit of trial and error, and we find that \(k_2 = 1\) turns the red line into the vertical line \(x = -2.8\).

We have now reduced our original, complex system into the much simpler \[ \begin{cases} \begin{array}{ccccc} x & & & = & -2.8 \\ & & y & = & 0.2 \\ \end{array} \end{cases} \] The solution is therefore \((x,y) = (-2.8, 0.2)\)

Paul Koester, 1 September 2013, Created with GeoGebra