Systems of Differential Equations (Part 2. Review : Systems of Equations)
Posted by Suhaivi Hamdan on 4:10 AM with No comments
Because we are going to be working almost exclusively with
systems of equations in which the number of unknowns equals the number of
equations we will restrict our review to these kinds of systems.
All of what we will be doing here can be easily extended to
systems with more unknowns than equations or more equations than unknowns if
need be.
Let’s start with the following system of n equations with the n unknowns, x_{1}, x_{2},…, x_{n}.

Note that in the subscripts on the coefficients in this
system, a_{ij}, the i corresponds to the equation that the
coefficient is in and the j
corresponds to the unknown that is multiplied by the coefficient.
To use linear algebra to solve this system we will first
write down the augmented matrix for
this system. An augmented matrix is
really just all the coefficients of the system and the numbers for the
right side of the system written in matrix form. Here is the augmented matrix for this system.


To solve this system we will use elementary row operations
(which we’ll define these in a bit) to rewrite the augmented matrix in
triangular form. The matrix will be in
triangular form if all the entries below the main diagonal (the diagonal
containing a_{11}, a_{22},
…,a_{nn}) are zeroes.
Once this is done we can recall that each row in the
augmented matrix corresponds to an equation.
We will then convert our new augmented matrix back to equations and at
this point solving the system will become very easy.
Before working an example let’s first define the elementary
row operations. There are three of them.
 Interchange
two rows. This is exactly what it
says. We will interchange row i with row j. The notation that
we’ll use to denote this operation is :
 Multiply
row i by a constant, c.
This means that every entry in row i will get multiplied by the constant c. The notation for
this operation is :
 Add a
multiply of row i to row j.
In our heads we will multiply row i by an appropriate constant and then add the results to row j and put the new row back into row
j leaving row i in the matrix unchanged. The notation for this operation is :
It’s always a little easier to understand these operations
if we see them in action. So, let’s
solve a couple of systems.
Example 1 Solve
the following system of equations.
Solution
The first step is to write down the augmented matrix for
this system. Don’t forget that
coefficients of terms that aren’t present are zero.
Now, we want the entries below the main diagonal to be
zero. The main diagonal has been
colored red so we can keep track of it during this first example. For reasons that will be apparent
eventually we would prefer to get the main diagonal entries to all be ones as
well.
We can get a one in the upper most spot by noticing that
if we interchange the first and second row we will get a one in the uppermost
spot for free. So let’s do that.
Now we need to get the last two entries (the 2 and 3) in
the first column to be zero. We can do
this using the third row operation.
Note that if we take 2 times the first row and add it to the second
row we will get a zero in the second entry in the first column and if we take
3 times the first row to the third row we will get the 3 to be a zero. We can do both of these operations at the
same time so let’s do that.
Before proceeding with the next step, let’s make sure that
you followed what we just did. Let’s
take a look at the first operation that we performed. This operation says to multiply an entry in
row 1 by 2 and add this to the corresponding entry in row 2 then replace the
old entry in row 2 with this new entry.
The following are the four individual operations that we performed to
do this.
Okay, the next step optional, but again is convenient to
do. Technically, the 5 in the second
column is okay to leave. However, it
will make our life easier down the road if it is a 1. We can use the second row operation to take
care of this. We can divide the whole
row by 5. Doing this gives,
The next step is to then use the third row operation to
make the 6 in the second column into a zero.
Now, officially we are done, but again it’s somewhat
convenient to get all ones on the main diagonal so we’ll do one last step.
We can now convert back to equations.
At this point the solving is quite easy. We get x_{3}
for free and once we get that we can plug this into the second equation and
get x_{2}. We can then use the first equation to get x_{1}. Note as well that having 1’s along the main
diagonal helped somewhat with this process.
The solution to this system of equation is

The process used in this example is called Gaussian Elimination. Let’s take a look at another example.
Example 2 Solve
the following system of equations.
Solution
First write down the augmented matrix.
We won’t put down as many words in working this
example. Here’s the work for this
augmented matrix.
We won’t go any farther in this example. Let’s go back to equations to see why.
The last equation should cause some concern. There’s one of three options here. First, we’ve somehow managed to prove that
0 equals 8 and we know that’s not possible.
Second, we’ve made a mistake, but after going back over our work it
doesn’t appear that we have made a mistake.
This leaves the third option. When we get something like the third
equation that simply doesn’t make sense we immediately know that there is no
solution. In other words, there is no
set of three numbers that will make all three of the equations true at the
same time.

Let’s work another example.
We are going to get the system for this new example by making a very
small change to the system from the previous example.
Example 3 Solve
the following system of equations.
Solution
So, the only difference between this system and the system
from the second example is we changed the 1 on the right side of the equal
sign in the third equation to a 7.
Now write down the augmented matrix for this system.
The steps for this problem are identical to the steps for
the second problem so we won’t write them all down. Upon performing the same steps we arrive at
the following matrix.
This time the last equation reduces to
and unlike the second example this is not a problem. Zero does in fact equal zero!
We could stop here and go back to equations to get a
solution and there is a solution in this case. However, if we go one more step and get a
zero above the one in the second column as well as below it our life will be
a little simpler. Doing this gives,
If we now go back to equation we get the following two
equations.
We have two equations and three unknowns. This means that we can solve for two of the
variables in terms of the remaining variable.
Since x_{3} is in
both equations we will solve in terms of that.
What this solution means is that we can pick the value of x_{3} to be anything that we’d
like and then find values of x_{1}
and x_{2}. In these cases we typically write the
solution as follows,
In this way we get an infinite number of solutions, one
for each and every value of t.

These three examples lead us to a nice fact about systems of
equations.
Fact
Given a system of equations, (1),
we will have one of the three possibilities for the number of solutions.
1. No
solution.
2. Exactly
one solution.
3. Infinitely
many solutions.

Before moving on to the next section we need to take a look
at one more situation. The system of
equations in (1)
is called a nonhomogeneous system if at least one of the b_{i}’s is not
zero. If however all of the b_{i}’s are zero we call the system homogeneous and the system will
be,


Now, notice that in the homogeneous case we are guaranteed
to have the following solution.


This solution is often
called the trivial solution.
For homogeneous systems the fact above can be modified to
the following.
Fact
Given a homogeneous system of equations, (2),
we will have one of the two possibilities for the number of solutions.

In the second possibility we can say nonzero solution
because if there are going to be infinitely many solutions and we know that one
of them is the trivial solution then all the rest must have at least one of the
x_{i}’s be nonzero and hence we get a nonzero solution.
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