Graphics Programs Reference
In-Depth Information
Solution
The augmented coefficient matrix is
⎡
⎣
⎤
⎦
6
−
4
1
−
14
22
−
4
6
−
4
36
−
18
−
1
4
6
6
7
The eliminationphase consists of the following two passes:
row2
←
row2
+
(2
/
3)
×
row1
row3
←
row3
−
(1
/
6)
×
row1
⎡
⎣
⎤
⎦
6
−
4
1
−
14
22
0
10
/
3
−
10
/
3
80
/
3
−
10
/
3
0
−
10
/
35
/
6
25
/
3 0
/
3
and
row3
←
row3
+
row2
⎡
⎣
⎤
⎦
6
−
4
1
−
14
22
0 10
/
3
−
10
/
3
80
/
3
−
10
/
3
00 5
/
2
35
0
In the solutionphase, we first compute
x
1
by back substitution:
35
5
X
31
=
2
=
14
/
80
/
3
+
(10
/
3)
X
31
80
/
3
+
(10
/
3)14
X
21
=
=
=
22
10
/
3
10
/
3
−
14
+
4
X
21
−
X
31
−
14
+
4(22)
−
14
X
11
=
=
=
10
6
6
Thus the first solutionvectoris
X
11
X
21
X
31
T
10 22 14
T
x
1
=
=
The second solutionvectoriscomputednext, also using back substitution:
X
32
=
0
−
10
/
3
+
(10
/
3)
X
32
−
10
/
3
+
0
X
22
=
=
=−
1
10
/
3
10
/
3
+
4
X
22
−
+
−
−
22
X
32
22
4(
1)
0
X
12
=
=
=
3
6
6
Therefore,
X
12
X
22
X
32
T
3
1 0
T
x
2
=
=
−
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