Although Matlab solves systems like the one above with ease our aim here is to
develop a deeper understanding of Gaussian Elimination and so we
proceed by hand. This aim is motivated by a number of important considerations.
First, not all linear systems have solutions and even those that do do not
necessarily possess unique solutions. A careful look at Gaussian Elimination
will provide the general framework for not only classifying those systems that
possess unique solutions but also for providing detailed diagnoses of those
defective systems that lack solutions or possess too many.

In Gaussian Elimination one first uses linear combinations of preceding
rows to eliminate nonzeros below the main diagonal and then solves the
resulting triangular system via back-substitution. To firm up our
understanding let us take up the case where each
k
j
=1
k
j
1
and so Equation 4 takes the form

(
2-10
-12-1
0-12
)
x
1
x
2
x
3
=
f
1
f
2
f
3
2-10
-12-1
0-12
x
1
x
2
x
3
f
1
f
2
f
3

(5)
We eliminate the (2,1) (row 2, column 1) element by implementing

new row 2=old row 2+12row 1
new row 2
old row 2
1
2
row 1
bringing

(
2-10
032-1
0-12
)
x
1
x
2
x
3
=
f
1
f
2
+
f
1
2
f
3
2-10
0
3
2
-1
0-12
x
1
x
2
x
3
f
1
f
2
f
1
2
f
3
We eliminate the current (3,2) element by implementing

new row 3=old row 3+23row 2
new row 3
old row 3
2
3
row 2
bringing the upper-triangular system

(
2-10
032-1
0043
)
x
1
x
2
x
3
=
f
1
f
2
+
f
1
2
f
3
+2
f
2
3+
f
1
3
2-10
0
3
2
-1
00
4
3
x
1
x
2
x
3
f
1
f
2
f
1
2
f
3
2
f
2
3
f
1
3
One now simply reads off

x
3
=
f
1
+2
f
2
+3
f
3
4
x
3
f
1
2
f
2
3
f
3
4
This in turn permits the solution of the second equation

x
2
=2(
x
3
+
f
2
+
f
1
2)3=
f
1
+2
f
2
+
f
3
2
x
2
2
x
3
f
2
f
1
2
3
f
1
2
f
2
f
3
2
and, in turn,

x
1
=
x
2
+
f
1
2=3
f
1
+2
f
2
+
f
3
4
x
1
x
2
f
1
2
3
f
1
2
f
2
f
3
4
One must say that Gaussian Elimination has succeeded here.
For, regardless of the actual elements of

ff,
we have produced an

xx
for which

ATKAx=f
A
K
A
x
f
.