anchor.diff:  [stem]
   ++++
   \begin{align*}
  -{\bigl(\begin{matrix} A & B \\ C & D \\ \end{matrix}\bigr)}^{-1}&=\bigl(\begin{matrix} A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1} & -A^{-1}B(D-CA^{-1}B)^{-1} \\ -(D-CA^{-1}B)^{-1}CA^{-1} & (D-CA^{-1}B)^{-1} \\ \end{matrix}\bigr)\\
  -&=\bigl(\begin{matrix} (A-BD^{-1}C)^{-1} & -(A-BD^{-1}C)^{-1}BD^{-1} \\ -D^{-1}C(A-BD^{-1}C)^{-1} & D^{-1}+D^{-1}C(A-BD^{-1}C)^{-1}BD^{-1} \\ \end{matrix}\bigr)
  +{\left[ \begin{array}{} A & B \\ C & D \\ \end{array} \right]}^{-1}&=\left[ \begin{array}{} A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1} & -A^{-1}B(D-CA^{-1}B)^{-1} \\ -(D-CA^{-1}B)^{-1}CA^{-1} & (D-CA^{-1}B)^{-1} \\ \end{array} \right]\\
  +&=\left[ \begin{array}{} (A-BD^{-1}C)^{-1} & -(A-BD^{-1}C)^{-1}BD^{-1} \\ -D^{-1}C(A-BD^{-1}C)^{-1} & D^{-1}+D^{-1}C(A-BD^{-1}C)^{-1}BD^{-1} \\ \end{array} \right]
   \end{align*}
   ++++
   
  
positive.code: 
  [stem]
  ++++
  M^{-1}={\bigl(\begin{matrix} A & B \\ C & D \\ \end{matrix}\bigr)}^{-1}=\frac{1}{\left|M\right|}{\begin{matrix} \begin{array}{} (\left|D\right|A-B(D^{\#}C))^{\#} & (\left|B\right|C-D(A^{\#}B)^{\#})^{\#} \\ (\left|C\right|B-A(D^{\#}C)^{\#})^{\#} & (\left|A\right|D-C(A^{\#}B))^{\#} \\ \end{array} \right]}
  ++++
  
  Now it is clear what kind of calculation we need. We need 2x2 matrix multiply and multiply by adjugate: stem:[AB], stem:[A^{\#}B] and stem:[AB^{\#}]. We already know how to do adjugate, but in this case, adjugate can be combined with multiplication so we don’t waste instructions. Just expand the result and rearrange the order, for example:
  
negative.code: 
  [stem]
  ++++
  M=\left[ \begin{array}{} a\vec{X} & 0 \\ b\vec{Y} & 0 \\ c\vec{Z} & 0 \\ \vec{T} & 1 \\ \end{array} \right] = \left[ \begin{array}{} aX_0 & aX_1 & aX_2 & 0 \\ bY_0 & bY_1 & bY_2 & 0 \\ cZ_0 & cZ_1 & cZ_2 & 0 \\ T_0 & T_1 & T_2 & 1 \\ \end{array} \right]
  ++++
  
  The first 3 component of the last row is the translation stem:[\vec{T}]. The top left 3x3 sub-matrix is the scaled rotation matrix, with each row as a scaled axis. We have stem:[\vec{X}\cdot\vec{Y}=\vec{X}\cdot\vec{Z}=\vec{Y}\cdot\vec{Z}=0], and stem:[\left|\vec{X}\right|=\left|\vec{Y}\right|=\left|\vec{Z}\right|=1]. And the scale is stem:[(a,b,c)]. 
  