Consider the system of equations \[ \begin{cases} \begin{array}{ccccc} 3x & - & 2y & = & 3 \\ 6x & + & k\cdot y & = & 4 \\ \end{array} \end{cases} \] Find the value of \(k\) so that the system has no solutions.

**Solution:**

Lets try to do Gauss-Jordan elimination. First, do \(R_2 \mapsto R_2 - 2\cdot R_1\).
\[
\begin{cases}
\begin{array}{ccccc}
3x & - & 2y & = & 3 \\
& & (k + 4)y & = & -2 \\
\end{array}
\end{cases}
\]
What happens if \(k = -4\)? We would get the system
\[
\begin{cases}
\begin{array}{ccccc}
3x & - & 2y & = & 3 \\
& & 0 & = & -2 \\
\end{array}
\end{cases}
\]
The bottom row says \(0 = -2\), which is nonsense. This means the system has NO SOLUTION.

Now lets look at this problem graphically. Use the slider to change the value of k.

What happens when \(k = -4\)? The two lines become parallel.

Keep in mind that

- finding the solution to a system of linear equations
- finding the point of intersection of a family of lines

Consider the system of equations \[ \begin{cases} \begin{array}{ccccc} 2x & - & y & = & 2 \\ 5x & + & k\cdot y & = & 5 \\ \end{array} \end{cases} \] Find the value of \(k\) so that the system has infinitely many solutions.

**Solution:**

We treat the "infinitely many solutions" case in much the same way as the "no solutions" case.

This time, lets begin with the geometry. Move the slider around to try to find the value of \(k\) which makes the two lines parallel.

When \(k = -2.5\), the lines are sort of parallel, in the sense that the have the same slope.
But not only do they have the same slope, they are actually the same line, and so the two lines intersect in infinitely many points.

Lets try to do this with Gauss-Jordan elimination. First, do \(R_2 \mapsto R_2 - 2.5\cdot R_1\).
\[
\begin{cases}
\begin{array}{ccccc}
2x & - & y & = & 2 \\
& & (k + 2.5)\cdot y & = & 0 \\
\end{array}
\end{cases}
\]
What happens if \(k = -2.5\)? We get the system
\[
\begin{cases}
\begin{array}{ccccc}
2x & - & y & = & 2 \\
& & 0 & = & 0 \\
\end{array}
\end{cases}
\]
The bottom row says \(0 = 0\), a true, but not particularly enlightening statement.
Thus, the "system of equations" is really just a single equation,
\[
2x - y = 2
\]
which has infinitely many solutions.

We apply the same approach to both questions

- Find \(k\) so that the system has no solution
- Find \(k\) so that the system has infinitely many solutions

When does this result in infinitely many solutions? No solutions?

- No solution: After applying row operations to make the lines parallel, one of the resulting equations will state that zero is equal to a nonzero number. This is a contradiction, and so the system has no solutions.
- Infinitely many solutions: After applying row operations to make the lines parallel, one of the resulting equations will become \(0 = 0\).

Two lines are parallel if they have the same slope.
However, our lines are usually written in General Form rather than point-slope or slope-intercept form.
We could try to compute the slopes and compare, but its better if we check if they are parallel directly from the General Form.

The two lines are parallel if the ratio of the \(x\) coefficients is equal to the ratio of the \(y\) coefficients.

In the first example, we need
\[
\frac{3}{6} = \frac{-2}{k},
\]
which gives us the \(k = -4\) we found several times already.

In the second example, we need
\[
\frac{2}{5} = \frac{-1}{k},
\]
which gives us the \(k = -2.5\) we found several times already.

Finally, we can also determine if there are no solutions or infinitely many solutions by looking at these ratios.

In the "infinitely many solutions" case, the "\(x\) to \(x\)", "\(y\) to \(y\)", and "right side to right side" ratios are all equal when \(k = -2.5\).
\[
\frac{2}{5} = \frac{-1}{2.5} = \frac{2}{5}
\]
In the "no solutions" case, the "\(x\) to \(x\)" and "\(y\) to \(y\)" ratios are equal when \(k = -4\), but the "right side to right side" ratio is different.
\[
\frac{3}{6} = \frac{-2}{4} \neq \frac{3}{4}
\]

Bye!

Paul Koester, 1 September 2013, Created with GeoGebra