Page 68 - Matrix Analysis & Applied Linear Algebra

2.4 Homogeneous Systems 61

Solution:
   
122 122
A =  257  −→  013  = E
366 000
shows that rank (A)=2 <n =3. Since the basic columns lie in positions
one and two, x 1 and x 2 are the basic variables while x 3 is free. Using back
substitution on [E|0] to solve for the basic variables in terms of the free variable
produces x 2 = −3x 3 and x 1 = −2x 2 − 2x 3 =4x 3 , so the general solution is
   
4
x 1
−3 , where x 3 is free.
x 2 = x 3
   
1
x 3
 
4
That is, every solution is a multiple of the one particular solution h 1 =  −3  .
1
Summary
Let A m×n be the coefficient matrix for a homogeneous system of m
linear equations in n unknowns, and suppose rank (A)= r.
• The unknowns that correspond to the positions of the basic columns
(i.e., the pivotal positions) are called the basic variables, and the
unknowns corresponding to the positions of the nonbasic columns
are called the free variables.
• There are exactly r basic variables and n − r free variables.
• To describe all solutions, reduce A to a row echelon form using
Gaussian elimination, and then use back substitution to solve for
the basic variables in terms of the free variables. This produces the
general solution that has the form

h n−r ,
x = x f 1 h 1 + x f 2 h 2 + ··· + x f n−r
are the free variables and where
where the terms x f 1 ,x f 2 ,...,x f n−r
h 1 , h 2 ,..., h n−r are n × 1 columns that represent particular solu-
tions of the homogeneous system. The h i ’s are independent of which
row echelon form is used in the back substitution process. As the free
range over all possible values, the general solution gen-
variables x f i
erates all possible solutions.
• A homogeneous system possesses a unique solution (the trivial solu-
tion) if and only if rank (A)= n —i.e., if and only if there are no
free variables.