anchor.diff:  
   [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]
  +M^{-1}=\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}= \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}
   ++++
   
   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].
  
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]
  ++++
  \begin{align*}
  {\begin{matrix} A & B \\ C & D \\ \end{matrix}}^{-1}&=\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}\\
  &=\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}
  \end{align*}
  ++++
  
  