Find $X$ for$\ A^T(BA^{-1}-I)^TX=B^T $ If:
$$A =\left( \begin{array}{cc} 1 & -1 & 0\\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{array} \right)$$ and $$B =\left( \begin{array}{cc} 2 & 1 & 3\\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{array} \right)$$
I got: $\ X=((BA^{-1}-I)^T)^{-1}(A^T)^{-1}B^T $ However I think it's quite complicated to solve this equation in this way and I feel that there should be some way of simplifying it to some extent.
If you transpose the equation first you get $$X^T(BA^{-1}-I)A=B\quad \Rightarrow\quad X^T(B-A)=B\quad \Rightarrow\quad X^T=B(B-A)^{-1}$$ which is not too hard to calculate. And from $X^T$, finding $X$ is trivial.