anchor.diff:  
   [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]
  +A^{\#}={\left( \begin{matrix} A_0 & A_1 \\ A_2 & A_3 \\ \end{matrix} \right)}^{\#}=\left( \begin{matrix} A_3 & -A_1 \\ -A_2 & A_0 \\ \end{matrix} \right)
   ++++
   
   Adjugate of 2x2 matrix has the following property: stem:[(AB)^{\#}=B^{\#}A^{\#}],stem:[(A^{\#})^{\#}=A], stem:[(cA)^{\#}=cA^{\#}].
  
positive.code: 
  [stem]
  ++++
  M=\left[ \begin{array}{} A & B \\ C & D \\ \end{array} \right]=\left[ \begin{array}{} A_0 & A_1 & B_0 & B_1 \\ A_2 & A_3 & B_2 & B_3 \\ C_0 & C_1 & D_0 & D_1 \\ C_2 & C_3 & D_2 & D_3 \\ \end{array} \right]
  ++++
  
  Matrix block-wise inverse is given by the following:
  
negative.code: 
  [stem]
  ++++
  M^{-1}={\left( \begin{matrix} A & B \\ C & D \\ \end{matrix} \right)}^{-1}=\frac{1}{\left|M\right|}{\left( \begin{matrix} (\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{matrix} \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:
  