[求助]怎么对这个线性变化编程进行参数估计
 

两个方程：
X（c1x+c2y+c3)=a1x+a2y+a3
Y (c1x+c2y+c3)=b1x+b2y+b3
要估计a1~c3九个参数
现在已知 x,y 和 X，Y一共有34对数据点 我用nlinfit报错 说：

Warning: Divide by zero.
> In inlineeval at 13
In inline.subsref at 25
In nlinfit at 111
In asdf at 72
Warning: Divide by zero.
> In inlineeval at 13
In inline.subsref at 25
In nlinfit>LMfit at 271
In nlinfit at 168
In asdf at 72
Warning: Divide by zero.
> In inlineeval at 13
In inline.subsref at 25
In nlinfit>getjacobian at 474
In nlinfit>LMfit at 289
In nlinfit at 168
In asdf at 72
Warning: Divide by zero.
> In inlineeval at 13
In inline.subsref at 25
In nlinfit>getjacobian at 474
In nlinfit>LMfit at 289
In nlinfit at 168
In asdf at 72
Warning: Divide by zero.
> In inlineeval at 13
In inline.subsref at 25
In nlinfit>getjacobian at 474
In nlinfit>LMfit at 289
In nlinfit at 168
In asdf at 72
Warning: Divide by zero.
> In inlineeval at 13
In inline.subsref at 25
In nlinfit>getjacobian at 474
In nlinfit>LMfit at 289
In nlinfit at 168
In asdf at 72
Warning: Divide by zero.
> In inlineeval at 13
In inline.subsref at 25
In nlinfit>getjacobian at 474
In nlinfit>LMfit at 289
In nlinfit at 168
In asdf at 72
Warning: Divide by zero.
> In inlineeval at 13
In inline.subsref at 25
In nlinfit>getjacobian at 474
In nlinfit>LMfit at 289
In nlinfit at 168
In asdf at 72
Warning: Divide by zero.
> In inlineeval at 13
In inline.subsref at 25
In nlinfit>getjacobian at 474
In nlinfit>LMfit at 289
In nlinfit at 168
In asdf at 72
Warning: Divide by zero.
> In inlineeval at 13
In inline.subsref at 25
In nlinfit>getjacobian at 474
In nlinfit>LMfit at 289
In nlinfit at 168
In asdf at 72
Warning: Matrix is singular to working precision.
> In nlinfit>LMfit at 296
In nlinfit at 168
In asdf at 72
Warning: The model is overparameterized, and model parameters are not
identifiable. You will not be able to compute confidence or prediction
intervals, and you should use caution in making predictions.
> In nlinfit at 220
In asdf at 72

兄弟们帮帮我啊