Game Development Reference
In-Depth Information
Example 3.13. Calculate the inverse of the 3
×
3
matrix M given by
238
60 3
13 2
M
=
.
(3.35)
Solution. Concatenating the identity matrix to M , we have
238100
60 3010
13 2 001
M
=
.
(3.36)
We now apply steps C through F of the algorithm for
=
1
.
j
10 0 0
238 100
13 2 001
10
1
1
2
6
Exchange rows 1 and 2
Multiply new row 1 by
⎯⎯⎯⎯⎯⎯⎯→
1
6
1
0
1
0
2
6
⎯⎯⎯⎯⎯⎯⎯→
Add
−×
2
row 1 to row 2
03 9 1
0
(3.37)
1
3
Add row 1 to row 3
03
3
0
1
1
2
6
=
2
Applying the same steps for
gives us the following.
j
10
1
0
1
0
2
6
1
3
Multiply row 2 by
⎯⎯⎯⎯⎯⎯→
01 3
0
1
1
3
9
03
3
0
1
1
2
6
10
0
0
1
1
2
6
Add
−×
3
row 2 to row 3
⎯⎯⎯⎯⎯⎯⎯→
01 3
1
1
0
(3.38)
3
9
00
−−
1
1
15
1
2
2
Finally, we apply the algorithm for
=
3
.
j
10
1
0
1
0
2
6
2
15
Multiply row 3 by
⎯⎯⎯⎯⎯⎯ →
01 3
1
1
0
3
9
00 1
2
−−
1
2
15
15
15
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