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augmented coefficient matrix becomes
⎡
⎣
⎤
⎦
0
−
110
A
b
=
(b)
−
12
−
1
0
2
−
1
0
1
Sincewe didnot change the equations(only their orderwas altered), the solutionisstill
x
1
=
1. However,Gauss elimination fails immediatelydue to the presence
of the zero pivot element (the element
A
11
).
The aboveexample demonstrates that it issometimes essentialtoreorder the
equations during the eliminationphase. The reordering,or
row pivoting
, is also re-
quiredif the pivot element is not zero, but very small in comparison to other elements
in the pivot row, as demonstratedbythe following set of equations:
x
2
=
x
3
=
⎡
⎣
⎤
⎦
ε
−
110
A
b
=
−
12
−
1
0
(c)
2
−
1
0
1
These equations are the same as Eqs. (b), exceptthat the small number
ε
replaces the
zero element
A
11
in Eq. (b). Therefore, if we let
0, the solutionsofEqs. (b) and (c)
shouldbecome identical.After the first phase of Gauss elimination, the augmented
coefficient matrix becomes
ε
→
⎡
⎣
⎤
⎦
ε
−
1
1
0
A
b
=
0
2
−
1
/ε
−
1
+
1
/ε
0
(d)
0
−
1
+
2
/ε
−
2
/ε
1
Because the computerworks with afixedword length, all numbers are rounded off
to a finite number of significant figures. If
ε
is very small, then1
/ε
ishuge, and an
elementsuch as 2
−
1
/ε
is rounded to
−
1
/ε
. Therefore, for sufficiently small
ε
, the
Eqs. (d) are actually storedas
⎡
⎣
⎤
⎦
ε
−
1
1
0
A
b
=
0
−
1
/ε
1
/ε
0
0
2
/ε
−
2
/ε
1
Because the second and third equationsobviously contradict each other, the solution
process fails again. This problemwouldnot arise if the first and second,or the first
and the third,equations were interchangedinEqs. (c) before the elimination.
The last example illustrates the extremecase where
wassosmall that roundoff
errors resultedintotalfailure of the solution. If we weretomake
ε
somewhat bigger
so that the solutionwouldnot “bomb” any more, the roundoff errors might still be
largeenoughtorender the solution unreliable. Again,this difficulty couldbe avoided
by pivoting.
ε
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