anchor.diff:  
   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}={\left( \begin{array}{} A & B \\ C & D \\ \end{array} \right)}^{-1}={\left( \begin{array}{} X & Y \\ Z & W \\ \end{array} \right)}].Let’s start with the top left corner.
  +Let stem:[M^{-1}={\left( \begin{matrix} A & B \\ C & D \\ \end{matrix} \right)}^{-1}={\left( \begin{matrix} X & Y \\ Z & W \\ \end{matrix} \right)}].Let’s start with the top left corner.
   
   [stem]
   ++++
  
positive.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 good news 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.
  
negative.code: 
  [stem]
  ++++
  M^{-1}=\left( \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} \right) = \left( \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} \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].
  