anchor.diff:  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^{\#}].
  -
  -For determinant of 2x2 matrix, we will use the following properties: stem:[\left|A\right|={A_0}{A_3}-{A_1}{A_2}], stem:[\left|-A\right|=\left|A\right|], stem:[\left|AB\right|=\left|A\right|\left|B\right|], stem:[\left|A+B\right|=\left|A\right|+\left|B\right|+\operatorname{tr}(A^{\#}{B})].
  +Adjugate of 2x2 matrix has the following property: stem:[(AB)^{\#}=B^{\#}A^{\#}],stem:[(A^{\#})^{\#}=A], stem:[(cA)^{\#}=cA^{\#}].
  +
  +For determinant of 2x2 matrix, we will use the following properties: stem:[\left|A\right|={A_0}{A_3}-{A_1}{A_2}], stem:[\left|-A\right|=\left|A\right|], stem:[\left|AB\right|=\left|A\right|\left|B\right|], stem:[\left|A+B\right|=\left|A\right| + \left|B\right| + \operatorname{tr}(A^{\#}{B})].
   
   For trace of matrix we have stem:[\operatorname{tr}(AB)=\operatorname{tr}(BA)], stem:[\operatorname{tr}(-A)=-\operatorname{tr}(A)].
   
  
positive.code: 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]
  ++++
  
negative.code: &=-\frac{1}{\left|M\right|\left|D\right|}(\left|D\right|A-B(D^{\#}C))^{\#}(BD^{\#})\\
  &=-\frac{1}{\left|M\right|\left|D\right|}(\left|D\right|A-B(D^{\#}C))^{\#}(DB^{\#})^{\#}\\
  &=-\frac{1}{\left|M\right|\left|D\right|}(\left|D\right|DB^{\#}A-DB^{\#}B(D^{\#}C))^{\#}\\
  &=-\frac{1}{\left|M\right|\left|D\right|}(\left|D\right|D(A^{\#}B)^{\#}-\left|D\right|\left|B\right|C))^{\#}\\
  &=\frac{1}{\left|M\right|}(\left|B\right|C-D(A^{\#}B)^{\#})^{\#}
  \end{align*}
  ++++
  