Page 49 - A Course in Linear Algebra with Applications

2.1: Gaussian Elimination 33

xi + 3x 2 + 3x 3 + 2^4 = 1
^3 + §^4 = 1
6^3 + 2x4 = 6

Finally, operation (3) - 6(2) gives

Xi + 3x 2 + 3^3 + 2X4 = 1
1 x
£3 + k 3 4 = !
0 = 0

Here the third equation tells us nothing and can be ignored.
Now observe that we can assign arbitrary values c and d to
the unknowns X4 and x 2 respectively, and then use back sub-
stitution to find x 3 and x\. Hence the most general solution
of the linear system is

x± = — 2 — c — 3d, x 2 — d, X3 = 1 — , £4 = c.
-

Since c and d can be given arbitrary values, the linear system
has infinitely many solutions.

What has been learned from these three examples? In
the first place, the number of solutions of a linear system can
be 0, 1 or infinity. More importantly, we have seen that there
is a systematic method of eliminating some of the unknowns
from all equations of the system beyond a certain point, with
the result that a linear system is reached which is of such a
simple form that it is possible either to conclude that no solu-
tions exist or else to find all solutions by the process of back
substitution. This systematic procedure is called Gaussian
elimination; it is now time to give a general account of the
way in which it works.