Find explicitly the projective transformation of $ββ^2$ that takes the points $[1,0,0],[1,1,0],[0,0,1]$ and $[0,1,1]$ to $[1,2,3],[1,0,1],[0,1,0]$ and $[1,0,-1]$ respectively.
Solution.
First we scale the points to satisfy linear relations
\begin{gather*}[1,0,0]-[1,1,0]-[0,0,1]+[0,1,1]=0\cr[1,2,3]-2[1,0,1]-2[0,1,0]+[1,0,-1]=0\end{gather*}
To find a linear map $T\colonβ^2ββ^2$
\begin{align*}
T(1,0,0)&=(1,2,3)\\
T(-1,-1,0)&=(-2,0,-2)\\
T(0,0,-1)&=(0,-2,0)
\end{align*}
by row reduction
\begin{align*}
T(1,0,0)&=(1,2,3)\\
T(0,1,0)&=(1,-2,-1)\\
T(0,0,1)&=(0,2,0)
\end{align*}
so the projective transformation $\tilde{T}$ induced by the linear map
\[T(x,y,z)=(x+y,2x-2y+2z,3x-y)\]
satisfies required conditions.
Remark. For $T:V_1βV_2$,
dual map $T^*:V_2^*βV_1^*$