Is there $$\left|\begin{matrix}M_1 &M_2\\M_3 &M_4\end{matrix}\right|=\left|\begin{matrix}M_1 &M_2^T\\M_3 &M_4^T\end{matrix}\right|,$$ where $M_1,M_2,M_3,M_4$ are all $n$-by-$n$ matrices, $M^T$ stands for the transpose of matrix $M$, $|M|$ stands for the determinant of $M$?
It follows from the question below:
Let $M$ and $A$ be $2n$-by-$2n$ matrices s.t.
$$A=\left[\begin{matrix}0&E\\-E &0\end{matrix}\right],$$ $$M^TAM=A,$$ where $E$ is the identity matrix. Prove that $|M|=1$.
Letting $M=\left[\begin{matrix} M_1&M_2\\M_3&M_4\end{matrix}\right]$, I calculated that $M_1^TM_4-M_3^TM_2=E$, hence it suffices to show the equality above.
I'd be grateful if anybody could provide any hint, method or a complete solution.
Consider $$ M_1 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix}\quad M_2 = M_2^T = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}\quad M_3 = \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}\quad M_4 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} $$ you will get $$ \det\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ \end{pmatrix} = 1, $$ while $ M_4^T = \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix} $ and $$ \det\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ \end{pmatrix} = 0, $$ thus no.