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Mathematical Expression Editor
Introduction to Systems of Linear Equations
You are probably familiar with the concept of a system of linear equations and with
some methods for solving such systems. In this section, we will look at the algebra
and geometry of finding and interpreting solutions of systems of linear equations. We
will start with two-variable and three-variable systems, then move on to systems
involving more variables.
Algebra of Linear Systems
When you were first introduced to systems of equations, you learned to solve for one
variable in terms of the other(s), then substitute. Here, we will introduce another
method. This alternative method involves adding multiples of one equation to
another equation in order to eliminate one of the variables. This method will form
the foundation for an algorithm we will develop for solving linear systems and
performing other computations related to systems. Exploration init:systwoeqs1 illustrates how the
second method works.
The purpose of this problem is to formalize what you may 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.
Note that this step eliminates from the second equation.
Next we divide both sides of the second equation by .
We now know what is. 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.
Next we multiply both sides of the first equation by .
Finally, we can switch the order of equations in order to display in the top
row.
This solution can be written as an ordered pair .
In Exploration init:systwoeqs1 we introduced elementary row operations and the notation associated
with them. 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 equations (rows) and :
(b)
Multiplying both sides of equation (row) by the same non-zero constant,
, and replacing equation with the result:
(c)
Adding times equation (row) to equation (row) , and replacing equation
with the result:
As we applied elementary row operations to the system in Exploration init:systwoeqs1, the system
changed, but a quick check will convince you that all six systems have the same
solution: . The six systems 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(s). We will formalize this statement as Theorem th:elemRowOpsEquivSys at the end of this
section.
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
sections. 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 (variables) 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. Adding twice the first equation to the second
equation gives us
This is where we run into a problem: there are no values of and that 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, any combination of and satisfies the second
equation. So, 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. In terms of the number of solutions, there are three
possibilities.
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 common to all of them. If
this is the case, the system is inconsistent.
Finally, it is possible for all of the lines to coincide, giving us infinitely many
solutions.
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 in 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.
General Systems of Linear Equations
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.
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.
Two systems of linear equations are said to be equivalent if they have the same
solution set.
Recall that to solve systems of equations in this section, we utilized three elementary
row operations. These operations are:
(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
Given a system of linear equations, any of the three elementary row operations
performed on the system produces an equivalent system.
Proof
Clearly, the order of equations 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. To see that item:addrow produces
an equivalent system, note that 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. (If you can figure out how
to carry the second system to this one, you should be able to reverse the
process.)
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)?
