Page 67 - Matrix Analysis & Applied Linear Algebra

60 Chapter 2 Rectangular Systems and Echelon Forms

then we obtain the following reduced system:
x 1 +2x 2 + x 4 =0,
x 3 + x 4 =0.
Solving for the basic variables x 1 and x 3 in terms of x 2 and x 4 produces
exactly the same result as given in (2.4.3) and hence generates exactly the same
general solution as shown in (2.4.4).
Because it avoids the back substitution process, you may find it more con-
venient to use the Gauss–Jordan procedure to reduce A completely to E A
and then construct the general solution directly from the entries in E A . This
approach usually will be adopted in the examples and exercises.
As was previously observed, all homogeneous systems are consistent because
the trivial solution consisting of all zeros is always one solution. The natural
question is, “When is the trivial solution the only solution?” In other words,
we wish to know when a homogeneous system possesses a unique solution. The
form of the general solution (2.4.5) makes the answer transparent. As long as
there is at least one free variable, then it is clear from (2.4.5) that there will
be an infinite number of solutions. Consequently, the trivial solution is the only
solution if and only if there are no free variables. Because the number of free
variables is given by n − r, where r = rank (A), the previous statement can be
reformulated to say that a homogeneous system possesses a unique solution—the
trivial solution—if and only if rank (A)= n.
Example 2.4.1

The homogeneous system
x 1 +2x 2 +2x 3 =0,
2x 1 +5x 2 +7x 3 =0,
3x 1 +6x 2 +8x 3 =0,

has only the trivial solution because
   
122 122
A =  257  −→  013  = E
368 002
shows that rank (A)= n =3. Indeed, it is also obvious from E that applying
back substitution in the system [E|0] yields only the trivial solution.

Example 2.4.2
Problem: Explain why the following homogeneous system has infinitely many
solutions, and exhibit the general solution:
x 1 +2x 2 +2x 3 =0,
2x 1 +5x 2 +7x 3 =0,
3x 1 +6x 2 +6x 3 =0.