Find the affine transformation of space that maps: 1^°…

Find the affine transformation of space that maps: $1^\circ$ $M_1(0,0,0)$, $M_2(1,0,0)$, $M_3(0,1,0)$, $M_4(0,0,1)$ to $M_1'(3,5,-2)$, $M_2'(2,-1,-1)$, $M_3'(4,0,1)$, $M_4'(1,0,2)$; $2^\circ$ $M_1(3,5,-2)$, $M_2(-3,2,1)$, $M_3(0,3,0)$, $M_4(4,-2,0)$ to $M_1'(1,1,1)$, $M_2'(1,2,3)$, $M_3'(-1,-2,0)$, $M_4'(0,0,0)$.

Step 1: $1^\circ$ Standard frame

The translation is $M_1'$, and the columns of the matrix are $M_k'-M_1'$ for $k=2,3,4$: $\vec b=(3,5,-2),\quad \text{columns } (-1,-6,1),\ (1,-5,3),\ (-2,-5,4)$.

\[x'=-x+y-2z+3,\quad y'=-6x-5y-5z+5,\quad z'=x+3y+4z-2,\quad\Delta=50\]

Step 2: $2^\circ$ General points

Each row $(a_{i1},a_{i2},a_{i3},b_i)$ solves a $4\times4$ system with coefficient rows $(x_k,y_k,z_k,1)$. Subtracting the equation of $M_3$ from the others and eliminating (as in the source) gives:

\[\begin{aligned} x'&=-2x-\tfrac95y-\tfrac{29}5z+\tfrac{22}5\\ y'&=-\tfrac{11}3x-\tfrac{10}3y-\tfrac{31}3z+8\\ z'&=-\tfrac73x-\tfrac{28}{15}y-\tfrac{88}{15}z+\tfrac{28}5\end{aligned}\]

$\Delta=\tfrac15\ne0$, so the map is non-degenerate. The source does not check this.

Step 3: Verification of $2^\circ$ (all four points)

For example, for $M_2(-3,2,1)$: \[x'=6-\tfrac{18}5-\tfrac{29}5+\tfrac{22}5=1,\quad y'=11-\tfrac{20}3-\tfrac{31}3+8=2,\quad z'=7-\tfrac{56}{15}-\tfrac{88}{15}+\tfrac{28}5=3.\]