anchor.diff:  
   [stem]
   ++++
  -{\left[ \begin{array}{} A & B \\ C & D \\ \end{array} \right]}^{-1}=\left[ \begin{array}{} (A-BD^{-1}C)^{-1} & -(A-BD^{-1}C)^{-1}BD^{-1} \\ -(D-CA^{-1}B)^{-1}CA^{-1} & (D-CA^{-1}B)^{-1} \\ \end{array} \right]
  +{\begin{matrix} A & B \\ C & D \\ \end{matrix}}^{-1}=\begin{matrix} (A-BD^{-1}C)^{-1} & -(A-BD^{-1}C)^{-1}BD^{-1} \\ -(D-CA^{-1}B)^{-1}CA^{-1} & (D-CA^{-1}B)^{-1} \\ \end{matrix}
   ++++
   
   This choice might not seem obvious. Take the first form for example, it seems we only need to calculate two 2x2 matrix inverse: stem:[A^{-1}] and stem:[(D-CA^{-1} B)^{-1}], however it can be further simplified by proper derivation. Since each corresponding sub-matrices equals to each other, it doesn’t matter which form you choose to work your math on. We just select the easier row from both forms.
  
positive.code: 
  [stem]
  ++++
  A^{\#}={\left[ \begin{array}{} A_0 & A_1 \\ A_2 & A_3 \\ \end{array} \right]}^{\#}=\left[ \begin{array}{} A_3 & -A_1 \\ -A_2 & A_0 \\ \end{array} \right]
  ++++
  
  Adjugate of 2x2 matrix has the following property: stem:[(AB)^{\#}=B^{\#}], stem:[A^{\#}],stem:[(A^{\#})^{\#}=A], stem:[(cA)^{\#}=cA^{\#}].
  
negative.code: 
  I only listed properties needed for derivation. If you are not familiar with these concepts, or want to know more about them, take a look at the wiki pages above.
  
  Let stem:[M^{-1}={\begin{matrix} A & B \\ C & D \\ \end{matrix}}^{-1}={\begin{matrix} X & Y \\ Z & W \\ \end{matrix}}].Let’s start with the top left corner.
  
  [stem]
  ++++
  