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^{-1}=\left[ \begin{array}{} \frac{1}{a}\vec{X} & \frac{1}{b}\vec{Y} & \frac{1}{c}\vec{Z} & \vec{0} \\ -\vec{T}\cdot\frac{1}{a}\vec{X} & -\vec{T}\cdot\frac{1}{b}\vec{Y} & -\vec{T}\cdot\frac{1}{c}\vec{Z} & 1 \\ \end{array} \right] = \left[ \begin{array}{} \frac{1}{a}X_0 & \frac{1}{b}Y_0 & \frac{1}{c}Z_0 & 0 \\ \frac{1}{a}X_1 & \frac{1}{b}Y_1 & \frac{1}{c}Z_1 & 0 \\ \frac{1}{a}X_2 & \frac{1}{b}Y_2 & \frac{1}{c}Z_2 & 0 \\ -\vec{T}\cdot\frac{1}{a}\vec{X} & -\vec{T}\cdot\frac{1}{b}\vec{Y} & -\vec{T}\cdot\frac{1}{c}\vec{Z} & 1 \\ \end{array} \right]
  ++++
  
  The nice thing about transform matrix stem:[M] is its first 3 rows are perpendicular to each other. Its inverse form is basically transpose the 3x3 rotation matrix, and rescale it, and change translation part by doing dot product with 3 rescaled axes. It should be easy to confirm stem:[MM^{-1}=I].
  
negative.code: [stem]
  ++++
  \begin{align*}
  A^{\#}B&={\left( \begin{matrix} A_3 & -A_1 \\ -A_2 & A_0 \\ \end{matrix} \right)}{\left( \begin{array}{} B_0 & B_1 \\ B_2 & B_3 \\ \end{array} \right)}\\
  &={\left( \begin{array}{} {A_3}{B_0}-{A_1}{B_2} &{A_3}{B_1}-{A_1}{B_3} \\ {A_0}{B_2}-{A_2}{B_0} & {A_0}{B_3}-{A_2}{B_1} \\ \end{array} \right)}
  \end{align*}
  ++++
  
  