Proving the cross product matrix tranformation identity with an alternative solution
Solution 1:
Using the scalar-triple product: we know that $$ A \cdot (B \times C) = \det[A \ \ B\ \ C], $$ where $A \cdot B = A^TB$ denotes a dot-product (and ${}^T$ denotes a transpose). With that, we can deduce the entries of $(MS) \times (MT)$.
Let $e_1,e_2,e_3$ denote the $x,y,z$ unit vectors. Using the above and the rule $\det(PQ) = \det(P)\det(Q)$, we see that for all $i$,
$$
\begin{align}
(Me_i) \cdot [(MS) \times (MT)] & = \det[Me_i \ \ MS\ \ MT] \\
& = \det(M) \det[e_i\ \ S\ \ T] \\
& = \det(M) (e_i \cdot (S\times T)) \\
& = \det(M) (S\times T)_i.
\end{align}
$$
That is, we have
$$
\begin{align}
M^T ((MS) \times (MT)) & = \pmatrix{(Me_1)^T\\ (Me_2)^T\\ (Me_3)^T}((MS) \times (MT)) \\
& = \det(M)\pmatrix{(S \times T)_1\\ (S \times T)_2 \\ (S \times T)_3} \\
& = \det(M) (S \times T).
\end{align}
$$
Solving for $S \times T$, we have
$$
M^T ((MS) \times (MT)) = \det(M) (S \times T) \implies
(MS) \times (MT) = \det(M)M^{-T} (S \times T),
$$
which was what we wanted.