In the first section in this chapter we saw that there were
some special cases in the solution to systems of two equations. We saw that there didn’t have to be a
solution at all and that we could in fact have infinitely many solutions. In this section we are going to generalize
this out to general systems of equations and we’re going to look at how to deal
with these cases when using augmented matrices to solve a system.
Let’s first give the following fact.
Fact
|
Given any system of equations there are exactly three
possibilities for the solution.
- There
will not be a solution.
- There
will be exactly one solution.
- There
will be infinitely many solutions.
|
This is exactly what we found the possibilities to be when
we were looking at two equations. It
just turns out that it doesn’t matter how many equations we’ve got. There are still only these three
possibilities.
Now, let’s see how we can identify the first and last
possibility when we are using the augmented matrix method for solving. In the previous section we stated that we
wanted to use the row operations to convert the augmented matrix into the
following form,
depending upon the number of equations present in the
system. It turns out that we should have
added the qualifier, “if possible” to this instruction, because it isn’t always
possible to do this. In fact, if it
isn’t possible to put it into one of these forms then we will know that we are
in either the first or last possibility for the solution to the system.
Before getting into some examples let’s first address how we
knew what the solution was based on these forms of the augmented matrix. Let’s work with the two equation case.
Since,
is an augmented matrix we can always convert back to
equations. Each row represents an equation and the first column is the coefficient of x in the equation while the second column is the coefficient of the
y in the equation. The final column is the constant that will be
on the right side of the equation.
So, if we do that for this case we get,
and this is exactly what we said the solution was in the
previous section.
This idea of turning an augmented matrix back into equations
will be important in the following examples.
Speaking of which, let’s go ahead and work a couple of
examples. We will start out with the two
systems of equations that we looked at in the first section that gave the
special cases of the solutions.
|
Example 1 Use
augmented matrices to solve each of the following systems.
(a)  [Solution]
(b) [Solution]
Solution
(a)
Now, we’ve already
worked this one out so we know that there is no solution to this system. Knowing that let’s see what the augmented
matrix method gives us when we try to use it.
We’ll start with the augmented matrix.
Notice that we’ve already got a 1 in the upper left corner
so we don’t need to do anything with that.
So, we next need to make the -2 into a 0.
Now, the next step should be to get a 1 in the lower right
corner, but there is no way to do that without changing the zero in the lower
left corner. That’s a problem, because
we must have a zero in that spot as well as a one in the lower right corner. What this tells us is that it isn’t
possible to put this augmented matrix form.
Now, go back to equations and see what we’ve got in this
case.
The first row just converts back into the first
equation. The second row however
converts back to nonsense. We know
this isn’t true so that means that there is no solution. Remember, if we reach a point where we have
an equation that just doesn’t make sense we have no solution.
Note that if we’d gotten
we would have been okay since the last row would return
the equation so don’t get confused between this case and
what we actually got for this system.
[Return to Problems]
(b)
In this case we know
from the first section that there are infinitely many solutions to this
system. Let’s see what we get when we
use the augmented matrix method for the solution.
Here is the augmented matrix for this system.
In this case we’ll need to first get a 1 in the upper left
corner and there isn’t going to be any easy way to do this that will avoid
fractions so we’ll just divide the first row by 2.
Now, we can get a zero in the lower left corner.
Now, as with the first part we are never going to be able
to get a 1 in place of the red zero without changing the first zero in that
row. However, this isn’t the nonsense
that the first part got. Let’s convert
back to equations.
That last equation is a true equation and so there isn’t
anything wrong with this. In this case
we have infinitely many solutions.
Recall that we still need to do a little work to get the
solution. We solve one of the equations
for one of the variables. Note
however, that if we use the equation from the augmented matrix this is very
easy to do.
We then write the solution as,
We get solutions by picking t and plugging this into the equation for x. Note that this is NOT
the same set of equations we got in the first section. That is okay. When there are infinitely many solutions
there are more than one way to write the equations that will describe all the
solutions.
[Return to Problems]
|
Let’s summarize what we learned in the previous set of
examples. First, if we have a row in
which all the entries except for the very last one are zeroes and the last
entry is NOT zero then we can stop and the system will have no solution.
Next, if we get a row of all zeroes then we will have
infinitely many solutions. We will then
need to do a little more work to get the solution and the number of equations
will determine how much work we need to do.
Now, let’s see how some systems with three equations
work. The no solution case will be
identical, but the infinite solution case will have a little work to do.
|
Example 2 Solve
the following system of equations using augmented matrices.
Solution
Here’s the augmented matrix for this system.
We can get a 1 in the upper left corner by dividing by the
first row by a 3.
Next we’ll get the two numbers under this one to be
zeroes.
And we can stop.
The middle row is all zeroes except for the final entry which isn’t
zero. Note that it doesn’t matter what
the number is as long as it isn’t zero.
Once we reach this type of row we know that the system
won’t have any solutions and so there isn’t any reason to go any farther.
|
Okay, let’s see how we solve a system of three equations
with an infinity number of solutions with the augmented matrix method. This example will also illustrate an
interesting idea about systems.
|
Example 3 Solve
the following system of equations using augmented matrices.
Solution
Notice that this system is almost identical to the system
in the previous example. The only
difference is the number to the right of the equal sign in the second
equation. In this system it is -2 and
in the previous example it was 10.
Changing that one number completely changes the type of solution that
we’re going to get. Often this kind of
simple change won’t affect the type of solution that we get, but in some rare
cases it can.
Since the first two steps of the process are identical to
the previous part we won’t discuss them.
Here they are.
We’ve got a row of all zeroes so we instantly know that
we’ve got infinitely many solutions.
Unlike the two equation case we aren’t going to stop however. It looks like with a couple of row
operations we can make the second column look like it is supposed to in the
final form so let’s do that.
In this case we were able to make the second column look
like it’s supposed to and the third column will never look correct. However, it is possible that the situation
could be reversed and it would be the third column that we can make look
correct and the second wouldn’t look correct.
Every system is different.
Once we reach this point we go back to equations.
Now, both of these equations contain a z and so we’ll move that to the other
side in each equation.
This means that we get to pick the value of z for free and we’ll write the
solution as,
Since there are an infinite number of ways to choose t there are an infinite number of
solutions to this system.
|