anchor.diff:  
   [stem]
   ++++
  -A^{\#}={\bigl(\begin{matrix} A_0 & A_1 \\ A_2 & A_3 \\ \end{matrix}\bigr)}^{\#}=\bigl(\begin{matrix} A_3 & -A_1 \\ -A_2 & A_0 \\ \end{matrix}\bigr)
  +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^{\#}].
  
positive.code: 
  [stem]
  ++++
  M^{-1}=\bigl(\begin{matrix} \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{matrix}\bigr)= \bigl(\begin{matrix} \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{matrix}\bigr)
  ++++
  
  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]
  ++++
  M=\left[ \begin{array}{} \vec{X} & 0 \\ \vec{Y} & 0 \\ \vec{Z} & 0 \\ \vec{T} & 1 \\ \end{array} \right], M^{-1}=\left[ \begin{array}{} \frac{1}{{\left|\vec{X}\right|}^{2}}\vec{X} & \frac{1}{{\left|\vec{Y}\right|}^{2}}\vec{Y} & \frac{1}{{\left|\vec{Z}\right|}^{2}}\vec{Z} & \vec{0} \\ -\vec{T}\cdot\frac{1}{{\left|\vec{X}\right|}^{2}}\vec{X} & -\vec{T}\cdot\frac{1}{{\left|\vec{Y}\right|}^{2}}\vec{Y} & -\vec{T}\cdot\frac{1}{{\left|\vec{Z}\right|}^{2}}\vec{Z} & 1 \\ \end{array} \right]
  ++++
  
  Notice that for rescaling, we divide by squared size of scaled axis, instead of size, which is a good new for implementation. And if our transform is of unit scale, which is also common in games, our target becomes even simpler.
  
  [stem]
  ++++
  M^{-1}=\left[ \begin{array}{} \vec{X} & \vec{Y} & \vec{Z} & \vec{0} \\ -\vec{T}\cdot\vec{X} & -\vec{T}\cdot\vec{Y} & -\vec{T}\cdot\vec{Z} & 1 \\ \end{array} \right]
  ++++
  
  Alright, enough theory, let’s see some code. This is our matrix definition.
  