We opened up this chapter talking about systems of equations
and we spent a couple of sections on them and then we moved away from them and
haven’t really talked much about them since.
It’s time to come back to systems and see how some of the ideas we’ve
been talking about since then can be used to help us solve systems. We’ll also take a quick look at a couple of
other ideas about systems that we didn’t look at earlier.
First let’s recall that any system of n equations and m
unknowns,
can be written in matrix form as follows.
In the matrix form A
is called the coefficient matrix and
each row contains the coefficients of the corresponding equations, x is a column matrix that contains all
the unknowns from the system of equations and finally b is a column matrix containing the constants on the right of the
equal sign.
Now, let’s see how inverses can be used to solve
systems. First, we’ll need to assume
that the coefficient matrix is a square 
matrix.
In other words there are the same number of equations as unknowns in our
system. Let’s also assume that A is invertible. In this case we actually saw in the proof of Theorem 3 in the section
on finding inverses that the solution to is unique (i.e. only a single solution exists) and that it’s given by,
So, if we’ve got the inverse of the coefficient matrix in
hand (not always an easy thing to find of course…) we can get the solution
based on a quick matrix multiplication.
Let’s see an example of this.
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Example 1 Use
the inverse of the coefficient matrix to solve the following system.
Solution
Okay, let’s first write down the matrix form of this
system.
Now, we found the inverse of the coefficient matrix back
in Example 2 of the
Finding Inverses section so here is the coefficient matrix and its inverse.
The solution to the system in matrix form is then,
Now since each of the entries of x are one of the unknowns in the original system above the system
to the original system is then,
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So, provided we have a square coefficient matrix that is
invertible and we just happen to have our hands on the inverse of the coefficient
matrix we can find the solution to the system fairly easily.
Next, let’s look at how the topic of the previous section (LU-Decompositions) can be used to solve systems
of equations. First let’s recall how
LU-Decompositions work. If we have a
square matrix, A, (so we’ll again be
working the same number of equations as unknowns) then if we can reduce it to
row-echelon form without using any row interchanges then we can write it as where L
is a lower triangular matrix and U is
an upper triangular matrix.
So, let’s start with a system where the coefficient matrix, A, is an square and has an LU-Decomposition of . Now, substitute this into the system for A to get,
Next, let’s just take a look at Ux. This will be an column matrix and let’s call it y.
So, we’ve got .
So, just what does this do for us? Well let’s write the system in the following
manner.
As we’ll see it’s very easy to solve for y and once we know y it
will be very easy to solve for x
which will be the solution to the original system.
It’s probably easiest to see how this method works with an
example so let’s work one.
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Example 2 Use
the LU-Decomposition method to find the solution to the following system of
equations.
Solution
First let’s write down the matrix form of the system.
Now, we found an LU-Decomposition to this coefficient
matrix in Example 1 of the
previous section. From that example we
see that,
According to the method outlined above this means that we
actually need to solve the following two systems.
in order.
So, let’s get started on the first one. Notice that we don’t really need to do
anything other than write down the equations that are associated with this
system and solve using forward
substitution. The first equation
will give us for free and once we know that the second
equation will give us . Finally, with these two values in hand the
third equation will give us . Here is that work.
The second system that we need to solve is then,
Again, notice that to solve this all we need to do is
write down the equations and do back
substitution. The third equation
will give us for free and plugging this into the second
equation will give us ,
etc. Here’s the work for this.
The solution to the original system is then shown
above. Notice that while the final
answers where a little messy the work was nothing more than a little
arithmetic and wasn’t terribly difficult.
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Let’s work one more of these since there’s a little more
work involved in this than the inverse matrix method of solving a system.
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Example 3 Use
the LU-Decomposition method to find a solution to the following system of
equations.
Solution
Once again, let’s first get the matrix form of the system.
Now let’s get an LU-Decomposition for the coefficient
matrix. Here’s the work that will
reduce it to row-echelon form.
Remember that the result of this will be U.
So, U is then,
Now, to get L
remember that we start off with a general lower triangular matrix and on the
main diagonals we put the reciprocal of the scalar used in the work above to
get a one in that spot. Then, in the
entries below the main diagonal we put the negative of the multiple used to
get a zero in that spot above. L is then,
We’ll leave it to you to verify that . Now let’s solve the system. This will mean we need to solve the
following two systems.
Here’s the work for the first system.
Now let’s get the actual solution by solving the second
system.
Here is the substitution work for this system.
So there’s the solution to this system.
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Before moving onto the next topic of this section we should
probably address why we even bothered with this method. It seems like a lot of work to solve a system
of equations and when solving systems by hand it can be a lot of work. However, because the method for finding L and U is a fairly straightforward process and once those are found the
method for solving the system is also very straightforward this is a perfect
method for use in computer systems when programming the solution to
systems. So, while it seems like a lot
of work, it is a method that is very easy to program and so is a very useful
method.
The remaining topics in this section don’t really rely on
previous sections as the first part of this section has. Instead we just need to look at a couple of
ideas about solving systems that we didn’t have room to put into the section on
solving systems of equations.
First we want to take a look at the following scenario. Suppose that we need so solve a system of
equations only there are two or more sets of the ’s that we need to look at. For instance suppose we wanted to solve the
following systems of equations.
Again, the coefficient matrix is the same for all these
systems and the only thing that is different is the ’s.
We could use any of the methods looked at so far to solve these
systems. However, each of the methods
we’ve looked at so far would require us to do each system individually and that
could potentially lead to a lot of work.
There is one method however that can be easily extended to
solve multiple systems simultaneously provided they all have the same
coefficient matrix. In fact the method
is the very first one we looked at. In
that method we solved systems by adding the column matrix b, onto the coefficient matrix and then reducing it to row-echelon
or reduced row-echelon form. For the
systems above this would require working with the following augmented matrices.
However, if you think about it almost the whole reduction
process revolves around the columns in the augmented matrix that are associated
with A and not the b column. So, instead of doing these individually let’s
add all of them onto the coefficient matrix as follows.
All we need to do this is reduce this to reduced row-echelon
form and we’ll have the answer to each of the systems. Let’s take a look at an example of this.
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Example 4 Find
the solution to each of the following systems.
Solution
So, we’ve got two systems with the same coefficient matrix
so let’s form the following matrix.
Note that we’ll leave the vertical bars in to make sure we remember
the last two columns are really b’s
for the systems we’re solving.
Now, we just need to reduce this to reduced row-echelon
form. Here is the work for that.
Okay the solution to the first system is in the
fourth column since that is the b
for the first system and likewise the solution to the second system is in the
fifth column. Therefore, the solution
to the first system is,
and the solution to the second system is,
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The remaining topic to discuss in this section gives us a
method for answering the following question.
Given an matrix A
determine all the matrices, b,
for which is consistent, that is has at least one solution. This is a question that can arise fairly
often and so we should take a look at how to answer it.
Of course if A is
invertible (and hence square) this answer is that is consistent for all b as we saw in an earlier section.
However, what if A isn’t
square or isn’t invertible? The method
we’re going to look at doesn’t really care about whether or not A is invertible but it really should be
pointed out that we do know the answer for invertible matrices.
It’s easiest to see how these work with an example so let’s
jump into one.
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Example 5 Determine
the conditions (if any) on ,
,
and in order for the following system to be
consistent.
Solution
Okay, we’re going to use the augmented matrix method we
first looked at here and reduce the matrix down to reduced row-echelon
form. The final form will be a little
messy because of the presence of the ’s but other than that the work is
identical to what we’ve been doing to this point.
Here is the work.
Okay, just what does this all mean? Well go back to equations and let’s see
what we’ve got.
So, what this says is that no matter what our choice
of ,
,
and we can find a solution using the general
solution above and in fact there will always be exactly one solution to the
system for a given choice of b.
Therefore, there are no conditions on ,
,
and in order for the system to be consistent.
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Note that the result of the previous example shouldn’t be
too surprising given that the coefficient matrix is invertible.
Now, we need to see what happens if the coefficient matrix
is singular (i.e.not invertible).