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2.3 Consistency of Linear Systems 53
2.3 CONSISTENCY OF LINEAR SYSTEMS
A system of m linear equations in n unknowns is said to be a consistent sys-
tem if it possesses at least one solution. If there are no solutions, then the system
is called inconsistent. The purpose of this section is to determine conditions
under which a given system will be consistent.
Stating conditions for consistency of systems involving only two or three
unknowns is easy. A linear equation in two unknowns represents a line in 2-space,
and a linear equation in three unknowns is a plane in 3-space. Consequently, a
linear system of m equations in two unknowns is consistent if and only if the m
lines defined by the m equations have at least one common point of intersection.
Similarly, a system of m equations in three unknowns is consistent if and only
if the associated m planes have at least one common point of intersection.
However, when m is large, these geometric conditions may not be easy to verify
visually, and when n> 3, the generalizations of intersecting lines or planes are
impossible to visualize with the eye.
Rather than depending on geometry to establish consistency, we use Gaus-
sian elimination. If the associated augmented matrix [A|b] is reduced by row
operations to a matrix [E|c] that is in row echelon form, then consistency—or
lack of it—becomes evident. Suppose that somewhere in the process of reduc-
ing [A|b]to[E|c] a situation arises in which the only nonzero entry in a row
appears on the right-hand side, as illustrated below:
∗∗∗∗∗∗ ∗
000 ∗∗∗ ∗
0000 ∗∗ ∗
Row i −→ 000000 α ←− α
=0.
•••••• •
•••••• •
If this occurs in the i th row, then the i th equation of the associated system is
0x 1 +0x 2 + ··· +0x n = α.
For α
=0, this equation has no solution, and hence the original system must
also be inconsistent (because row operations don’t alter the solution set). The
converse also holds. That is, if a system is inconsistent, then somewhere in the
elimination process a row of the form
(0 0 ··· 0 | α ) , α
=0 (2.3.1)
must appear. Otherwise, the back substitution process can be completed and
a solution is produced. There is no inconsistency indicated when a row of the
form (0 0 ··· 0 | 0) is encountered. This simply says that 0 = 0, and although