My question is what matrix operations are used to get to from eq.(1) to eq.(3)?
I'm pretty clear with eq.(1), solving for the unknown image pts [u,v].
Now I want to solve for an unknown XYZ world point with known image [u,v]. This will require moving matrices from the right to the left side, eq.(3). How do I move matrices K, [R|T] to the left to isolate $W_{xyz}$ like eq.(3)?
eq.(1) $s\cdot\underbrace{\begin{bmatrix}
u \\
v \\
1 \\
\end{bmatrix}
}_\text{Cam pts}
= \underbrace{\begin{bmatrix}fx & 0 & cx\\0 & fy & cy\\0 & 0 & 1\end{bmatrix}}_\text{K}
\underbrace{\begin{bmatrix}r_{11} & r_{12} & r_{13} & t_x\\r_{21} & r_{22} & r_{23} & t_y\\r_{31} & r_{32} & r_{33} & t_x\end{bmatrix}}_\text{RT(rotation/translation)}
\underbrace{\begin{bmatrix}
X \\
Y \\
Z \\
1 \\
\end{bmatrix}
}_\text{World Coordinates}
$
This simplifies to:
eq.(2) $s\cdot C_{uv}$ = $K\cdot[R|t]\cdot W_{xyz}$
eq.(3)$\underbrace{ \Biggl(
s
\begin{bmatrix}
u \\
v \\
1 \\
\end{bmatrix} K^{-1}-t\Biggr) R^{-1} = \begin{bmatrix}
X \\
Y \\
Z \\
1 \\
\end{bmatrix} }_\text{Matrices moved leftwise to solve for World Points X,Y,Z} $
Matrices do not commute, you need to multiply them on the left hand side: $sR^{-1}\left ( K^{-1}C-t\right )=W_{xyz}$