Another method for the solution of non-homogeneous, linear equations involves the introduction of homogenous variables. In (26.1), set
x1 = t1/t1n+1, x2 = t2/t1n+1, ··· , x1 = tn/t1n+1
and multiply by t1n+1. Then (26.1) becomes the homogeneous system
a11t1
+ ··· + a1ntn
- c1tn+1 =
0,
····························································(29.1)
ap1t1 + ··· +
apntn -
cptn+1 =
0.
Every solution of
(29.1) in which tn+1
0
yields a solution of (26.1). However, you must find out whether
(29.1) has solutions with tn+1
0.
According to Theorem 14, the number of linearly independent solutions of (29.1) is n + 1 - r1. According to the same Theorem,
a11t1
+ ··· + a1ntn
= 0,
····························································(29.2)
ap1t1 + ··· +
apntn =
0
has exactly n -
r linearly independent solutions. Every solution of
(29.1) with tn+1=0 yields a solution
of (29.2) and every solution of (29.2) a solution of (29.1) with tn+1
= 0, whence (29.1) has exactly then solutions with tn+1
0, if
n + 1 - r1 > n - r
or r1 < r + 1. On the other hand, certainly r1 = r or r1 = r + 1, so that is r1<r+ 1 is equivalent to r1 = r. Hence you have arrived again at the statement of Theorem 15.