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 .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
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
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.
Why is there a non-zero provision in Part item:constantmult of Definition def:elemrowops? Why is there not a
non-zero provision in Part item:addrow?
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?
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)?