1.缩放矩阵
几何图示
对应公式
x ′ = − x y ′ = y \begin{aligned} &x^{\prime}=-x \\ &y^{\prime}=y \end{aligned} x′=−xy′=y
[ x ′ y ′ ] = [ s x 0 0 s y ] [ x y ] \left[\begin{array}{l} x^{\prime} \\ y^{\prime} \end{array}\right]=\left[\begin{array}{cc} s_{x} & 0 \\ 0 & s_{y} \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] [x′y′]=[sx00sy][xy]
带入数据
[ x ′ y ′ ] = [ 0.5 0 0 1 ] [ x y ] \left[\begin{array}{l} x^{\prime} \\ y^{\prime} \end{array}\right]=\left[\begin{array}{cc} 0.5 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] [x′y′]=[0.5001][xy]
2.反转矩阵
几何图示
对应公式
x ′ = − x y ′ = y \begin{aligned} &x^{\prime}=-x \\ &y^{\prime}=y \end{aligned} x′=−xy′=y
[ x ′ y ′ ] = [ − 1 0 0 1 ] [ x y ] \left[\begin{array}{l} x^{\prime} \\ y^{\prime} \end{array}\right]=\left[\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] [x′y′]=[−1001][xy]
3.切片（剪切）矩阵
几何图示
对应公式
[ x ′ y ′ ] = [ 1 a 0 1 ] [ x y ] \left[\begin{array}{l} x^{\prime} \\ y^{\prime} \end{array}\right]=\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] [x′y′]=[10a1][xy]
4.旋转矩阵
几何图示
对应公式
R θ = [ cos θ − sin θ sin θ cos θ ] \mathbf{R}_{\theta}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] Rθ=[cosθsinθ−sinθcosθ]
推导：
- 取点 ( 1 , 0 ) → ( c o s θ , s i n θ ) (1,0)→(cosθ,sinθ) (1,0)→(cosθ,sinθ)
( x ′ y ′ ) = ( A B C D ) ( x y ) ( cos θ sin θ ) = ( A B C D ) ( 1 0 ) \begin{aligned} &\left(\begin{array}{l} x^{\prime} \\ y^{\prime} \end{array}\right)=\left(\begin{array}{l} A & B \\ C & D \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)\\ &\left(\begin{array}{c} \cos \theta \\ \sin \theta \end{array}\right)=\left(\begin{array}{ll} A & B \\ C & D \end{array}\right)\left(\begin{array}{l} 1 \\ 0 \end{array}\right) \end{aligned} (x′y′)=(ACBD)(xy)(cosθsinθ)=(ACBD)(10)
得出 A = c o s θ , C = s i n θ A=cosθ,C=sinθ A=cosθ,C=sinθ
- 取点
(
0
,
1
)
→
(
−
s
i
n
θ
,
c
o
s
θ
)
(0,1)→(-sinθ,cosθ)
(0,1)→(−sinθ,cosθ)
( − sin θ cos θ ) = ( A B C D ) ( 0 1 ) \begin{aligned} &\left(\begin{array}{c} -\sin \theta \\ \cos \theta \end{array}\right)=\left(\begin{array}{ll} A & B \\ C & D \end{array}\right)\left(\begin{array}{l} 0 \\ 1 \end{array}\right) \end{aligned} (−sinθcosθ)=(ACBD)(01)
得出 B = − s i n θ , D = c o s θ B=-sinθ,D=cosθ B=−sinθ,D=cosθ
