Graphics Programs Reference
In-Depth Information
Now the first pass of Gauss eliminationiscarried out (the arrowpoints to the pivot
row), yielding
⎡
⎣
⎤
⎦
⎡
⎣
⎤
⎦
−
2 4 3 0
0 29 16
0 65
4
6
8
A
b
=
s
=
/
2
−
1
The potential pivot elements for the nexteliminationpass are
A
22
and
A
32
.We
determine the “winner” from
⎡
⎣
⎤
⎦
=
⎡
⎣
⎤
⎦
=
⎡
⎣
⎤
⎦
r
22
r
32
∗
1
|
A
22
|
/
s
2
/
3
|
A
32
|
/
s
3
3
/
4
Note that
r
12
is irrelevant, since row1already actedas the pivot row. Therefore, it is
excluded from further consideration. As
r
32
islarger than
r
22
, the third rowis the better
pivot row. Afterinterchanging rows 2 and 3, wehave
⎡
⎣
⎤
⎦
←
⎡
⎣
⎤
⎦
−
2 4
3
0
4
8
6
A
b
=
/
−
s
=
0 65
1
0 29 16
2
The second eliminationpass nowyields
⎡
⎣
⎤
⎦
−
2 4
3
0
A
b
U
c
=
=
0 65
/
2
−
1
0049
/
6
49
/
3
Thiscompletes the eliminationphase. It shouldbe noted that
U
is the matrix that
wouldresult in the LU decomposition of the following row-wise permutation of
A
(the
ordering of rows is the same as achievedbypivoting):
⎡
⎣
⎤
⎦
−
2
4 3
−
184
2
−
26
Since the solution of
Ux
c
by back substitutionis not affected by pivoting, weskip
the detailed computation. The result is
x
T
=
1
12
.
=
−
Alternate Solution
It it not necessary to physically exchangeequations during piv-
oting. Wecouldaccomplish Gauss eliminationjust as well by keeping the equations
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