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Summary Solving systems of linear equations
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Mathematics - Solving systems of linear equations
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1. Systems of Linear Equations1.1 Solutions and Elementary Operations
Practical problems in many fields of study—such as biology, business, chemistry, computer science, eco-
nomics, electronics, engineering, physics and the social sciences—can often be reduced to solving a sys-
tem of linear equations. Linear algebra arose from attempts to find systematic methods for solving these
systems, so it is natural to begin this book by studying linear equations.
If a, b, and c are real numbers, the graph of an equation of the form
ax + by = c
is a straight line (if a and b are not both zero), so such an equation is called a linear equation in the
variables x and y. However, it is often convenient to write the variables as x1 , x2 , . . . , xn , particularly
when more than two variables are involved. An equation of the form
a1 x1 + a2 x2 + · · · + an xn = b
is called a linear equation in the n variables x1 , x2 , . . . , xn . Here a1 , a2 , . . . , an denote real numbers
(called the coefficients of x1 , x2 , . . . , xn , respectively) and b is also a number (called the constant term
of the equation). A finite collection of linear equations in the variables x1 , x2 , . . . , xn is called a system of
linear equations in these variables. Hence,
2x1 − 3x2 + 5x3 = 7
is a linear equation; the coefficients of x1 , x2 , and x3 are 2, −3, and 5, and the constant term is 7. Note that
each variable in a linear equation occurs to the first power only.
Given a linear equation a1 x1 + a2 x2 + · · · + an xn = b, a sequence s1 , s2 , . . . , sn of n numbers is called
a solution to the equation if
a1 s 1 + a2 s 2 + · · · + an s n = b
that is, if the equation is satisfied when the substitutions x1 = s1 , x2 = s2 , . . . , xn = sn are made. A
sequence of numbers is called a solution to a system of equations if it is a solution to every equation in
the system.
For example, x = −2, y = 5, z = 0 and x = 0, y = 4, z = −1 are both solutions to the system
x+y+ z=3
2x + y + 3z = 1
A system may have no solution at all, or it may have a unique solution, or it may have an infinite family of
solutions. For instance, the system x + y = 2, x + y = 3 has no solution because the sum of two numbers
cannot be 2 and 3 simultaneously. A system that has no solution is called inconsistent; a system with at
least one solution is called consistent. The system in the following example has infinitely many solutions.
1
,2 Systems of Linear Equations
Example 1.1.1
Show that, for arbitrary values of s and t,
x1 = t − s + 1
x2 = t + s + 2
x3 = s
x4 = t
is a solution to the system
x1 − 2x2 +3x3 +x4 = −3
2x1 − x2 +3x3 −x4 = 0
Solution. Simply substitute these values of x1 , x2 , x3 , and x4 in each equation.
x1 − 2x2 + 3x3 + x4 = (t − s + 1) − 2(t + s + 2) + 3s + t = −3
2x1 − x2 + 3x3 − x4 = 2(t − s + 1) − (t + s + 2) + 3s − t = 0
Because both equations are satisfied, it is a solution for all choices of s and t.
The quantities s and t in Example 1.1.1 are called parameters, and the set of solutions, described in
this way, is said to be given in parametric form and is called the general solution to the system. It turns
out that the solutions to every system of equations (if there are solutions) can be given in parametric form
(that is, the variables x1 , x2 , . . . are given in terms of new independent variables s, t, etc.). The following
example shows how this happens in the simplest systems where only one equation is present.
Example 1.1.2
Describe all solutions to 3x − y + 2z = 6 in parametric form.
Solution. Solving the equation for y in terms of x and z, we get y = 3x + 2z − 6. If s and t are
arbitrary then, setting x = s, z = t, we get solutions
x=s
y = 3s + 2t − 6 s and t arbitrary
z=t
Of course we could have solved for x: x = 13 (y − 2z + 6). Then, if we take y = p, z = q, the
solutions are represented as follows:
x = 13 (p − 2q + 6)
y = p p and q arbitrary
z = q
The same family of solutions can “look” quite different!
, 1.1. Solutions and Elementary Operations 3
y When only two variables are involved, the solutions to systems of lin-
ear equations can be described geometrically because the graph of a lin-
x−y = 1 ear equation ax + by = c is a straight line if a and b are not both zero.
Moreover, a point P(s, t) with coordinates s and t lies on the line if and
x+y = 3 only if as + bt = c—that is when x = s, y = t is a solution to the equa-
tion. Hence the solutions to a system of linear equations correspond to the
P(2, 1)
points P(s, t) that lie on all the lines in question.
x
In particular, if the system consists of just one equation, there must
be infinitely many solutions because there are infinitely many points on a
(a) Unique Solution line. If the system has two equations, there are three possibilities for the
(x = 2, y = 1) corresponding straight lines:
y
1. The lines intersect at a single point. Then the system has a unique
solution corresponding to that point.
x+y = 4
2. The lines are parallel (and distinct) and so do not intersect. Then
the system has no solution.
x+y = 2
x 3. The lines are identical. Then the system has infinitely many
solutions—one for each point on the (common) line.
(b) No Solution
y These three situations are illustrated in Figure 1.1.1. In each case the
graphs of two specific lines are plotted and the corresponding equations are
indicated. In the last case, the equations are 3x−y = 4 and −6x+2y = −8,
−6x + 2y = −8 which have identical graphs.
With three variables, the graph of an equation ax + by + cz = d can be
3x − y = 4
shown to be a plane (see Section 4.2) and so again provides a “picture”
of the set of solutions. However, this graphical method has its limitations:
x
When more than three variables are involved, no physical image of the
graphs (called hyperplanes) is possible. It is necessary to turn to a more
(c) Infinitely many solutions “algebraic” method of solution.
(x = t, y = 3t − 4)
Before describing the method, we introduce a concept that simplifies
Figure 1.1.1 the computations involved. Consider the following system
3x1 + 2x2 − x3 + x4 = −1
2x1 − x3 + 2x4 = 0
3x1 + x2 + 2x3 + 5x4 = 2
of three equations in four variables. The array of numbers1
3 2 −1 1 −1
2 0 −1 2 0
3 1 2 5 2
occurring in the system is called the augmented matrix of the system. Each row of the matrix consists
of the coefficients of the variables (in order) from the corresponding equation, together with the constant
1A rectangular array of numbers is called a matrix. Matrices will be discussed in more detail in Chapter 2.
, 4 Systems of Linear Equations
term. For clarity, the constants are separated by a vertical line. The augmented matrix is just a different
way of describing the system of equations. The array of coefficients of the variables
3 2 −1 1
2 0 −1 2
3 1 2 5
−1
is called the coefficient matrix of the system and 0 is called the constant matrix of the system.
2
Elementary Operations
The algebraic method for solving systems of linear equations is described as follows. Two such systems
are said to be equivalent if they have the same set of solutions. A system is solved by writing a series of
systems, one after the other, each equivalent to the previous system. Each of these systems has the same
set of solutions as the original one; the aim is to end up with a system that is easy to solve. Each system
in the series is obtained from the preceding system by a simple manipulation chosen so that it does not
change the set of solutions.
As an illustration, we solve the system x + 2y = −2, 2x + y = 7 in this manner. At each stage, the
corresponding augmented matrix is displayed. The original system is
x + 2y = −2 1 2 −2
2x + y = 7 2 1 7
First, subtract twice the first equation from the second. The resulting system is
x + 2y = −2 1 2 −2
− 3y = 11 0 −3 11
which is equivalent to the original (see Theorem 1.1.1). At this stage we obtain y = − 11
3 by multiplying
1
the second equation by − 3 . The result is the equivalent system
x + 2y = −2 1 2 −2
y = − 11
3 0 1 − 113
Finally, we subtract twice the second equation from the first to get another equivalent system.
16 16
x= 3 1 0 3
11 11
y=− 3 0 1 −3
Now this system is easy to solve! And because it is equivalent to the original system, it provides the
solution to that system.
Observe that, at each stage, a certain operation is performed on the system (and thus on the augmented
matrix) to produce an equivalent system.
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