Game Development Reference
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2
4
3
5
2
5
−3
(b)
5 −1
2
1
7
1
−2 −1
4
=
(5)(2)+(−1)(1)+(2)(−2)
(5)(5)+(−1)(7)+(2)(−1)
(5)(−3)+(−1)(1)+(2)(4)
=
10+(−1)+(−4)
25+(−7)+(−2) −15+(−1)+8
=
5
16 −8
2
4
3
5
2
4
3
5
2
4
3
5
2
4
3
5
2
4
−1
3
5
2
5
−3
−1
2
(2)(5)+(5)(−1)+(−3)(2)
(1)(5)+(7)(−1)+(1)(2)
(−2)(5)+(−1)(−1)+(4)(2)
10+(−5)+(−6)
5+(−7)+2
−10+1+8
1
7
1
−1
=
=
=
−2 −1
4
2
4
3
5
1 7 2
7 0 −3
2 −3 −1
(c)
5 −1
2
=
(5)(1)+(−1)(7)+(2)(2)
(5)(7)+(−1)(0)+(2)(−3)
(5)(2)+(−1)(−3)+(2)(−1)
2
=
5+(−7)+4
35+0+(−6)
10+3+(−2)
=
29
11
2
4
3
5
2
4
3
5
2
4
3
5
2
4
3
5
2
4
3
5
1 7 2
7 0 −3
2 −3 −1
−1
2
(1)(5)+(7)(−1)+(2)(2)
(7)(5)+(0)(−1)+(−3)(2)
(2)(5)+(−3)(−1)+(−1)(2)
5+(−7)+4
35+0+(−6)
10+3+(−2)
2
29
11
=
=
=
2
4
3
5
0
−4
3
(d)
5 −1
2
4
0
−1
−3
1
0
=
(5)(0)+(−1)(4)+(2)(−3)
(5)(−4)+(−1)(0)+(2)(1)
(5)(3)+(−1)(−1)+(2)(0)
=
0 + (−4) + (−6)
(−20) + 0 + 2
15 + 1 + 0
=
−10 −18
16
2
4
3
5
2
4
3
5
2
4
3
5
2
4
3
5
2
4
3
5
0
−4
3
−1
2
(0)(5)+(−4)(−1)+(3)(2)
(4)(5)+(0)(−1)+(−1)(2)
(−3)(5)+(1)(−1)+(0)(2)
0+4+6
20+0+(−2)
−15+(−1)+0
10
18
−16
4
0
−1
=
=
=
−3
1
0
T
T
A
T
= A
T
6.
(a)
T
T
(B)
T
BA
T
CD
T
A
T
CD
T
AB
T
CD
T
= AB
T
CD
T
(b)
=
=
T
T
T
T
T
D
T
C
T
(AB)
T
(AB)
T
D
T
C
T
C
T
D
T
(c)
=
= (AB)
= (AB) (CD) = ABCD
T
T
T
(AB)
T
(CDE)
T
(CDE)
T
(AB)
T
(d)
=
= (CDE) (AB)
= CDEAB
7. For each of the matrices M, interpret the rows of M as basis vectors after transformation.
(a) The basis vectors [ 1, 0 ] and [ 0, 1 ] are transformed to [ 0, −1 ] and [ 1, 0 ], respectively.
Thus, M performs a 90
◦
clockwise rotation.
(b) The basis vectors [ 1, 0 ] and [ 0, 1 ] are transformed to [
√
√
] and [−
√
√
2
2
,
2
2
2
2
,
2
2
],
respectively. Thus, M performs a 45
◦
counterclockwise rotation.
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