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问题描述

此处为了简化问题，在2维空间中处理，按以下方式调用，AcGeMatrix2d::alignCoordSys是如何求出一个矩阵的呢，这里提供一个实现思路（但效率不保证好）

AcGeMatrix2d matTrans = AcGeMatrix2d::alignCoordSys(AcGePoint2d(150, 125),-AcGeVector2d::kYAxis,AcGeVector2d::kXAxis,AcGePoint2d(1000, 1000),AcGeVector2d::kXAxis,AcGeVector2d::kYAxis);

求解过程

AcGeMatrix2d::alignCoordSys函数的功能，即如何求出矩阵：

a11 a12 a13
a21 a22 a23
0   0    1

已知变量如下：

fromOriginX,fromOriginY =(150,125)
fromAxisXx, fromAxisXy = (0,-1)
fromAxisYx, fromAxisYy = (1,0)
toOriginX,toOriginY =(1000,1000)
toAxisXx, toAxisXy =(1,0)
toAxisYx, toAxisYy =(0,1)

根据对齐映射关系，得到如下等式

a11*fromOriginX + a12*fromOriginY + a13 = toOriginX
a21*fromOriginX + a22*fromOriginY + a23 = toOriginY
a11*fromAxisXx + a12*fromAxisXy + a13 = toAxisXx
a21*fromAxisXx + a22*fromAxisXy + a23 = toAxisXy
a11*fromAxisYx + a12*fromAxisYy + a13 = toAxisYx
a21*fromAxisYx + a22*fromAxisYy + a23 = toAxisYy

所以得到如下线性方程组

a11*150 + a12*125 + a13 = 1000
a21*150 + a22*125 + a23 = 1000
a11*0 + a12*-1 + a13 = 1
a21*0 + a22*-1 + a23 = 0
a11*1 + a12*0 + a13 = 0
a21*1 + a22*0 + a23 = 1

使用消元法求解即可（即用矩阵的初等变换将系数矩阵化为阶梯型矩阵）