Graphics Programs Reference
In-Depth Information
At the conclusion of the eliminationphase the matrix has the form (do not confuse
d
,
e
and
f
with the originalcontents of
A
)
⎡
⎣
⎤
⎦
d
1
e
1
f
1
0
···
0
0
d
2
e
2
f
2
···
0
00
d
3
e
3
···
0
U
∗
=
.
.
.
.
.
.
.
.
00
···
0
d
n
−
1
e
n
−
1
00
···
00
d
n
Nextcomes the solutionphase. The equations
Ly
=
b
have the augmented coef-
ficient matrix
⎡
⎣
⎤
⎦
1
00 0
···
0
b
1
e
1
1
00
···
0
b
2
L
b
f
1
e
2
1
0
···
0
b
3
=
0
f
2
e
3
1
···
0
b
4
.
.
.
.
.
.
.
.
.
000
f
n
−
2
e
n
−
1
1
b
n
Solutionbyforward substitutionyields
y
1
=
b
1
y
2
=
b
2
−
e
1
y
1
(2.28)
.
y
k
=
b
k
−
f
k
−
2
y
k
−
2
−
e
k
−
1
y
k
−
1
,
k
=
3
,
4
,...,
n
The equationstobe solvedbyback substitution, namely
Ux
=
y
,have the augmented
coefficient matrix
⎡
⎣
⎤
⎦
d
1
d
1
e
1
d
1
f
1
0
···
0
y
1
0
d
2
d
2
e
2
d
2
f
2
···
0
y
2
U
y
00
d
3
d
3
e
3
···
0
y
3
=
.
.
.
.
.
.
.
.
.
00
···
0
d
n
−
1
d
n
−
1
e
n
−
1
y
n
−
1
00
···
0
0
d
n
y
n
the solution of which isobtainedbyback substitution:
x
n
=
y
n
/
d
n
x
n
−
1
=
y
n
−
1
/
d
n
−
1
−
e
n
−
1
x
n
x
k
=
y
k
/
d
k
−
e
k
x
k
+
1
−
f
k
x
k
+
2
,
k
=
n
−
2
,
n
−
3
,...,
1
(2.29)
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