Systems of Linear Equations
2.1 Systems of Linear Equations and Elimination
A linear equation in n variables x1, •••, x11 is an equation that can be written in the form
(2.1) The numbers a1, ... , a,, are called the coefficients of the equation, and b is usually referred to as "the right-hand side," or "the constant term." The xi are the unknowns or variables to be solved for.
EXAMPLE 1
Definition
Solution
EXAMPLE2
The equations
X1 + 2x2 = 4 x1 - 3x2 +
../3x3
= JTX4(2.2) (2.3) are both linear equations since they both can be written in the form of equation (2. 1 ).
The equation
x�
-x2 = 1 is not a linear equation.A vector in !Rn is called a solution of equation (2.1) if the equation is satisfied when we make the substitution Sn x1 = s1, x2 = s2, ... , Xn = Sn.
A few solutions of equation (2.2) are
[n [ O�s l
and[
_�]
since2 + 2(1) = 4 3 + 2(0.5) = 4 6 + 2(- 1 ) = 4 0
The vector 0 0 is clearly a solution of x1 - 3x2 +
.../3x3
= JTX4.0
A general system of m linear equations inn variables is written in the form
a11X1 + a12x2 + · · · + a1nXn = b1 a21X1 + a22X2 + · · · + a2nXn = b2
Note that for each coefficient, the first index indicates in which equation the coefficient appears. The second index indicates which variable the coefficient multiplies. That is, au is the coefficient of Xj in the i-th equation. The indices on the right-hand side indicate which equation the constant appears in.
We want to establish a standard procedure for determining all the solutions of such a system-if there are any solutions! It will be convenient to speak of the solution set of a system to mean the set of all solutions of the system.
The standard procedure for solving a system of linear equations is elimination. By multiplying and adding some of the original equations, we can eliminate some of the
EXAMPLE 3
Section 2.1 Systems of Linear Equations and Elimination 65
variables from some of the equations. The result will be a simpler system of equations that has the same solution set as the original system, but is easier to solve.
We say that two systems of equations are equivalent if they have the same solution set. In elimination, each elimination step must be reversible and must leave the solution set unchanged. Every system produced during an elimination procedure is equivalent to the original system. We begin with an example and explain the general rules as we proceed.
Find all solutions of the system of linear equations
X1 + X2 - 2X3 = 4 X1 + 3x2 - X3 = 7 2x1 + x2 - Sx3 = 7
Solution: To solve this system by elimination, we begin by eliminating x1 from all equations except the first one.
Add (-1) times the first equation to the second equation. The first and third equations are unchanged, so the system is now
X1 + X2 - 2x3 = 4 2x2 + X3 = 3 2x1 + x2 - Sx3 = 7
Note two important things about this step. First, if x1, x2, x3 satisfy the original system, then they certainly satisfy the revised system after the step. This follows from the rule of arithmetic that if P = Q and R = S, then P + R = Q + S. So, when we add two equations and both are satisfied, the resulting sum equation is satisfied. Thus, the revised system is equivalent to the original system.
Second, the step is reversible: to get back to the original system, we just add (1) times the first equation to the revised second equation.
Add (-2) times the first equation to the third equation.
X1 + X2 - 2X3 = 4 2x2 + X3 = 3 -X2 - X3 = -1
Again, note that this step is reversible and does not change the solution set. Also note that x1 has been eliminated from all equations except the first one, so now we leave the first equation and turn our attention to x2.
Although we will not modify or use the first equation in the next several steps, we keep writing the entire system after each step. This is important because it leads to a good general procedure for dealing with large systems.
It is convenient, but not necessary, to work with an equation in which x2 has the coefficient 1. We could multiply the second equation by 1 /2, but to avoid fractions, follow the steps on the next page.
EXAMPLE3
(continued)
Interchange the second and third equations. This is another reversible step that does not change the solution set:
X1 + X2 - 2X3 = 4 -X2 - X3 = -1
2X2 + X3 = 3
Multiply the second equation by (-1). This step is reversible and does not change the solution set:
X1 + X2 - 2X3 = 4 X2 + X3 = 1 2X2 + X3 = 3
Add (-2) times the second equation to the third equation.
X1 + X2 - 2x3 = 4 X2 + X3 = } -X3 = 1
In the third equation, all variables except x3 have been eliminated; by elimination, we have solved for X3. Using similar steps, we could continue and eliminate X3 from the second and first equations and x2 from the first equation. However, it is often a much simpler task to complete the solution process by back-substitution.
First, observe that x3 = -1. Substitute this value into the second equation and find that
X2 = 1 - X3 = 1 -( -1) = 2
Next, substitute these values back into the first equation to obtain X1 = 4 - X2 + 2x3 = 4 -2 + 2(-1) = 0
Thus, the on! y solution of this system is
[ �:]
=[
_H
Since the final system is equl valent to the original system, this solution is also the unique solution of the problem.
Observe that we can easily check that
[
_�]
satisfies the original system of equations:
0 + 2 - 2( -1) = 4 0 + 3(2) - (-1) = 7 2(0) + 2 - 5(-1) = 7
It is important to observe the form of the equations in our final system. The first variable with a non-zero coefficient in each equation, called a leading variable, does not appear in any equation below it. Also, the leading variable in the second equation
EXAMPLE4
Section 2.1 Systems of Linear Equations and Elimination 67
is to the right of the leading variable in the first, and the leading variable in the thfrd is to the right of the leading variable in the second.
The system solved in Example 3 is a particularly simple one. However, the solution procedure introduces all the steps that are needed in the process of elimination. They are worth reviewing.
Types of Steps in Elimination
(1) Multiply one equation by a non-zero constant.
(2) Interchange two equations.
(3) Add a multiple of one equation to another equation.
Warning! Do not combine steps of type ( 1) and type (3) into one step of the form
"Add a multiple of one equation to a multiple of another equation." Although such a combination would not lead to errors in this chapter, it would lead to errors when we apply these ideas in Cnapter 5.
Determine the solution set of the system of linear equations
Remark
X1 + 2X3 + X4 = 14 X1 + 3x3 + 3x4 = 19
Notice that neither equation contains x2. This may seem peculiar, but it happens in some applications that one of the variables of interest does not appear in the linear equations. If it truly is one of the variables of the problem, ignoring it is incorrect.
Rewrite the equations to make it explicit:
x1 + Ox2 + 2x3 + X4 = 14 Xi + Ox2 + 3x3 + 3x4 = 19
Solution: As in Example 3, we want our leading variable in the first equation to be to the left of the leading variable in the second equation, and we want the leading variable to be eliminated from the second equation. Thus, we use a type (3) step to eliminate x1 from the second equation.
Add ( -1) times the first equation to the second equation:
xi + Ox2 + 2x3 + X4 = 14 X3 + 2X4 = 5
Observe that X2 is not shown in the second equation because the leading variable must have a non-zero coefficient. Moreover, we have already finished our elimination pro
cedure as we have our desired form. The solution can now be completed by back
substitution.
Note that the equations do not completely determine both x3 and x4: one of them can be chosen arbitrarily, and the equations can still be satisfied. For consistency, we always choose the variables that do not appear as a leading variable in any equation to be the ones that will be chosen arbitrarily. We will call these free variables.
EXAMPLE4
(continued)
EXERCISE 1
Thus, in the revised system, we see that neither x2 nor x4 appears as a leading variable in any equation. Therefore, x2 and x4 are the free variables and may be chosen arbitrarily (for example, x4 =
t
E JR and x2 =s
E JR). Then the second equation can be solved for the leading variable X3:X3 = 5 - 2X4 = 5 - 2
t
Now, solve the first equation for its leading variable x1:
Xt = 14 -2X3 - X4 = 14 -2(5 -
2t) - t
= 4 +3t
Thus, the solution set of the system is
X1 4 +
3t s
5 - 2t '
s, t
E JRIn this case, there are infinitely many solutions because for each value of s and each value oft that we choose, we get a different solution. We say that this equation is the general solution of the system, and we call s and
t
the parameters of the general solution. For many purposes, it is useful to recognize that this solution can be split into a constant part, a part int,
and a part ins:
XJ 4 0
3
X2 0 1 0
X3 = 5 + S 0 +
t
-2X4 0 0
This will be the standard format for displaying general solutions.
It is acceptable to leave x2 in the place ofs
and x4 in the place oft,
but then youmust
say x2, X4 EJR. Observe that one immediate advantage of this form is that we can instantly see the geometric interpretation of the solution. The intersection of the two hyperplanes x1 + 2x3 + X4 = 14 and x1 + 3x3 + 3x4 = 19 in JR4 is a plane in JR4 that passes through P(4, 0, 5, 0).
The solution procedure we have introduced is known as Gaussian elimination with back-substitution. A slight variation of this procedure is introduced in the next section.
Find the general solution to the system of linear equations
2x1 + 4x2 + Ox3 = 12 X1 + 2X2 - X3 = 4
Use the general solution to find three different solutions of the system.
EXAMPLES
Section 2.1 Systems of Linear Equations and Elimination 69