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Mathematical Expression Editor

We solve systems of equations in two and three variables and interpret the results
geometrically.

SYS-0010: Introduction to Systems of Linear Equations

You have certainly studied linear equations for many years now. Perhaps the easiest
way to characterize linear equations is that they are polynomial equations where each
term is either a constant or has degree 1.

A linear equation in variables is an equation that can be written in the
form
where and are constants.

An -tuple is a solution to the equation provided that it turns the equation into a
true statement. The set of all -tuples that are solutions to a given equation is called
the graph of the equation. The graph of a linear equation in two variables is a line in .
The graph of a linear equation in three variables is a plane in . In for , we say that
the graph of a linear equation is a hyperplane. A hyperplane cannot be visualized,
but we can still talk about intersections of hyperplanes and their other attributes in
algebraic terms.

In linear algebra, we often look for solutions to systems of linear equations or linear
systems. A linear system of equations and unknowns is typically written as
follows

A solution to a system of linear equations in variables is an -tuple that satisfies every
equation in the system. All solutions to a system of equations, taken together, form a
solution set. We will focus on algebraic methods for finding solution sets, but
we will also consider the geometric aspect of systems to gain additional
insights.

Algebra of Linear Systems

You are probably familiar with two algebraic methods for solving systems of linear
equations. One method requires us to solve for one variable in terms of the other(s),
then substitute. The second method involves adding multiples of one equation to
another equation in order to eliminate one of the variables. The second method will
form the foundation for an algorithm we will develop for solving linear systems and
performing other computations related to systems. Exploration Problem init:systwoeqs1 illustrates
how the second method works.

The purpose of this problem is to formalize what you already know (perhaps under a
different name) about elementary row operations as means of solving systems of
linear equations. Consider the system
We will begin by adding twice the first row to the second row and replacing the
second row with the sum. This gives us
Note that this step eliminates from the second equation. Next we divide both sides
of the second equation by . This gives us
Our next goal is to eliminate from the first equation. To this end, we subtract twice
the second row from the first row and replace the first row with the difference. This
gives us
Next we multiply both sides of the first equation by . This gives us

Finally, we can switch the order of equations in order to display in the top row. This
gives us

This solution can be written as an ordered pair .

To obtain the solution to Exploration Problem init:systwoeqs1 we utilized three elementary row
operations. These operations are:

Switching the order of two equations

Multiplying both sides of an equation by the same non-zero constant

Adding a multiple of one equation to another

At each stage of the process, the system of equations looked different from the
original system, but a quick check will convince you that all six systems have the
same solution: . Systems (eq:step1)-(eq:step6) are said to be equivalent.

It turns out that if a system of equations is transformed into another system
through a sequence of elementary row operations, the new system will be
equivalent to the original system, in other words, both systems will have the
same solution set. We will formalize this statement in the last section of this
module.

As we perform elementary row operations, we describe our steps using the following
notation.

Switching row and row :

Multiplying both sides of equation by the same non-zero constant , and
replacing equation with the result:

Adding times row to row , and replacing row with the result:

Solve the system of equations using elementary row operations.

It may be daunting to think about how to begin. But keep in mind the desired
end-result. What we want is to use elementary row operations to transform the given
system into something like this

We will accomplish this by using a convenient variable in one row to “wipe out” this
variable from the other two rows. For example, we can use in the third equation to
wipe out in the first equation and in the second equation. To do this, multiply the
third row by and add it to the top row, then multiply the third row by and add it
to the second row. We now have:
In the previous step was a convenient variable to use because the coefficient in front
of was 1. We no longer have a variable with coefficient 1. We could create a
coefficient of 1 using division, but that would lead to fractions, making computations
cumbersome. Instead, we will subtract twice the second row from the first row. This
gives us:

Next we add seven times the first row to the second row, and subtract four times the
first row from the third row.

Now we divide both sides of the second row by .

Adding times the second row to the first row and subtracting times the second row
from the third row gives us

Finally, rearranging the rows gives us

Thus the system has a unique solution .

At this point you may be wondering whether it will always be possible to take a
system of three equations and three unknowns and use elementary row operations to
transform it to a system of the form
The short answer to this question is no. The existence of an equivalent system of this
form implies that the original system has a unique solution. However, it is possible
for a system to have no solutions or to have infinitely many solutions. We will study
these different possibilities from an algebraic perspective in subsequent modules. For
now, we will attempt to gain insight into existence and uniqueness of solutions
through geometry.

Geometry of Linear Systems in Two Variables

Exploration Problem init:systwoeqs1 offers an example of a linear system of two equations and two
unknowns with a unique solution.

Geometrically, the graph of each equation is a line in . The point is a solution to
both equations, so it must lie on both lines. The graph below shows the two lines
intersecting at .

Given a system of two equations with two unknowns, there are three possible
geometric outcomes. First, the graphs of the two equations intersect at a point. If this
is the case, the system has exactly one solution. We say that the system is consistent
and has a unique solution.

Second, the two lines may have no points in common. If this is the case, the system
has no solutions. We say that the system is inconsistent.

Finally, the two lines may coincide. In this case, there are infinitely many points that
satisfy both equations simultaneously. We say that the system is consistent and has
infinitely many solutions.

Solve the system of equations and interpret your results geometrically.

We will use elementary row operations. Our goal is to eliminate from the first
equation and eliminate from the second. Observe that adding twice the first
equation to the second equation accomplishes both tasks.
It is clear, however, that there are no values of and that would satisfy the second
equation. We conclude that the system is inconsistent. Plotting the two lines in the
same coordinate plane shows that the two lines are parallel.

Solve the system of equations and interpret your results geometrically.

To eliminate from the second equation, we subtract one quarter of the first equation
from the second. This gives us

Unlike the situation in Example ex:systwoeqs2, there are values of and that satisfy the second
equation. In fact, any ordered pair that satisfies the first equation will satisfy the
second equation. Thus, the solution set for this system is the same as the set of all
solutions of .

When we plot the two equations of the original system, we find that the two lines
coincide.

Given a linear system in two variables and more than two equations, we have a
variety of geometric possibilities. Three of them are depicted below. First, it is
possible for the graphs of all equations in the system to intersect at a single point,
giving us a unique solution.

Second, it is possible for the graphs to have no points in common.

If this is the case, the system is inconsistent.

Geometry of Linear Systems in Three Variables

In Example ex:threeeqthreevars1 we solved the following linear system of three equations and three
unknowns
We found that the system has a unique solution . The graph of each equation is a
plane. The three planes intersect at a single point, as shown in the figure.

Given a linear system of three equations and three variables, there are three ways in
which the system can be consistent. First, the three planes could intersect at a single
point, giving us a unique solution.

Second, the three planes can intersect in a line, forming a paddle-wheel shape. In this
case, every point along the line of intersection is a solution to the system, giving us
infinitely many solutions.

Finally, the three planes can coincide. If this is the case, there are infinitely many
solutions.

There are four ways for a system to be inconsistent. They are depicted below.

Equivalent Systems and Elementary Row Operations

In Exploration Problem init:systwoeqs1 we introduced elementary row operations and equivalent
systems. We now make these definitions formal.

Elementary Row Operations The following three operations performed on a linear
system are called elementary row operations

(a)

Switching the order of two equations

(b)

Multiplying both sides of an equation by the same non-zero constant

(c)

Adding a multiple of one equation to another

Two systems of linear equations are said to be equivalent if they have the same
solution set.

It is not difficult to see that performing a sequence of elementary row operations on a
system of equations produces an equivalent system. We can justify this by
considering the row operations one at a time. Clearly, the order of equations down
does not affect the solution set, so item:rowswap produces an equivalent system. Next, you learned
years ago that multiplying both sides of an equation by a non-zero constant does not
change its solution set, which establishes that item:constantmult produces an equivalent system. It is
also true that item:addrow produces an equivalent system. To see this, note that a multiple
of an equation is still an equation, so if we add a multiple of an equation
to another equation in the system, we are adding the same thing to both
sides, which does not change the solution set of that equation, nor of the
system.

Practice Problems

Give a graphical illustration of each of the following scenarios for a system of three
equations and two unknowns:

(a)

The system of three equations is inconsistent, but a combination of any
two of the three equations forms a consistent system.

(b)

The system is consistent and has a unique solution.

(c)

The system is consistent and has infinitely many solutions.

(d)

The system is inconsistent and no two equations form a consistent system.

Solve each system of linear equations or demonstrate that a solution does not exist,
and interpret your results geometrically.

Solution:

Solution:

Consider the following system of equations.

Find all possible values of k such that this system has no solution.

Solution:

Find all possible values of k such that this system has infinitely many
solutions.

Suppose the following system was obtained from system by adding twice the second
row of to the first row.
Find system .

Answer:

The following figures show a geometric depiction of two equivalent systems. (The
systems are equivalent because they have the same solution set.) Can the first system
be transformed into the second system by elementary row operations? If so,
how?

Begin by carrying the first system to
Then carry this system to the second system.

Consider the system of equations

Show that if is a solution to this system, and if we apply elementary row operation item:addrow
to the system, then will be a solution to the new system of equations.

Demonstrate that elementary row operations are reversible by answering the
following questions. Be specific about the elementary row operation that you would
use.

(a)

Suppose we obtained system (B) from system (A) by swapping two
equations. How would we obtain system (A) from system (B)?

(b)

Suppose we obtained system (B) from system (A) by multiplying one of
the equations of (A) by a non-zero constant . How would we obtain system
(A) from system (B)?

(c)

Suppose we obtained system (B) from system (A) by adding a multiple of
one of the equations of (A) to another. How would we obtain system (A)
from system (B)?