anchor.diff:  \vec{P'}=P_0a\vec{X}+P_1b\vec{Y}+P_2c\vec{Z}+\vec{T}
   ++++
   
  -This is the same as extend stem:[\vec{P}] to a 4 component vector stem:[\vec{P}(P_0,P_1,P_2,1)] and multiply by matrix stem:[M]. Now what does inverse matrix stem:[M^{-1}] mean? In this case it represents a world to local transform, so if we multiply stem:[\vec{P'}] by stem:[M^{-1}], we should get stem:[\vec{P}] back. How do we transform the point stem:[\vec{P'}] in world space back in local space? We subtract the local space origin (aka the translation stem:[\vec{T}]), then dot each axes to get its local space coordinate and rescale it:
  +This is the same as extend stem:[\vec{P}] to a 4 component vector stem:[\vec{P}(P_0,P_1,P_2,1)] and multiply by matrix stem:[M]. Now what does inverse matrix stem:[M^{-1}] mean? In this case it represents a world to local transform, so if we multiply stem:[\vec{P'}] by stem:[M^{-1}], we should get stem:[\vec{P}] back. How do we transform the point stem:[\vec{P'}] from world space back in local space? We subtract the local space origin (aka the translation stem:[\vec{T}]), then dot each axes to get its local space coordinate and rescale it:
   
   [stem]
   ++++
  
positive.code: ++++
  
  Notice that we put parentheses around stem:[D^{\#}C] and stem:[A^{\#}B], and you will see the reason soon.
  Now let’s do the top right corner, and make use of the result of top left corner stem:[X]:
  
  [stem]
  
negative.code: &=-\frac{1}{\left|M\right|\left|D\right|}(\left|D\right|A-B(D^{\#}C))^{\#}(BD^{\#})\\
  &=-\frac{1}{\left|M\right|\left|D\right|}(\left|D\right|A-B(D^{\#}C))^{\#}(DB^{\#})^{\#}\\
  &=-\frac{1}{\left|M\right|\left|D\right|}(\left|D\right|DB^{\#}A-DB^{\#}B(D^{\#}C))^{\#}\\
  &=-\frac{1}{\left|M\right|\left|D\right|}(\left|D\right|D(A^{\#}B)^{\#}-\left|D\right|\left|B\right|C))^{\#}\\
  &=\frac{1}{\left|M\right|}(\left|B\right|C-D(A^{\#}B)^{\#})^{\#}
  \end{align*}
  ++++
  