Determinant of a matrix with non-square properties Let A be an n x n matrix such that the det(A)=5;
Let E be an m x m matrix such that the det(E)=4;
Let F be an n x m matrix.
Find the det\begin{bmatrix}0&A\\E&F\end{bmatrix}
The answer can be expressed in terms of n and m.
I'm having trouble with this question as why my answer is wrong is throwing me off. Using properties of the determinant, I can rewrite this question as ;
det(0)det(F) - det(A)det(E)
My assumption was that det(0)det(F) equates to 0, but I'm not so sure anymore considering that my answer of -20 is incorrect. Any clues on maybe how to express the F determinant (which is non-square)
 A: By switching the "block-columns", we end up with
$$
\det \pmatrix{0 & A\\ E & F} = (-1)^{m(m+n-1)}\det \pmatrix{A & 0\\F & E}.
$$
On the other hand, using the formula for the determinant of a block upper-triangular matrix (proved here for instance) gives us
$$
\det\pmatrix{A & 0\\F & E} = \det(A) \det(E) = 20.
$$

Regarding the exponent of $(-1)$: to get from $\pmatrix{0 & A\\ E & F}$ to $\pmatrix{A & 0\\F & E}$, we apply a permutation of the columns of the block matrix. The value $(-1)^{m(m+n-1)}$ is the "parity" or "sign" of this permutation. The exponent of $(-1)$ here is any number of transpositions (i.e. column-switches) with which one could achieve this permutation.
So, why $m(m+n-1)$? Here is how one could achieve the desired permutation using exactly $m(m+n-1)$ transpositions. First, we consider the cyclic permutation
$$
[2\ \ 3\ \ \cdots \ \ m \ \ \cdots \ \ (m + n)\ \ 1]
$$
(where I have expressed the permutation in one-line notation). This permutation can be implemented in $(m+n-1)$ switches: first switch the $1$st and $2$nd elements, then the $2$nd and $3$rd, and so forth until you switch $(m + n - 1)$th and $(m + n)$th.
Now, to do what we wanted to do, which was the permutation
$$
[m+1 \ \ \cdots \ \ (m + n - 1)\ \ 1 \ \ \cdots\ \ m],
$$
it suffices to apply the first permutation $m$ times in a row. If we want to break it down into transpositions, we can repeat the above sequence $m$ times. Thus, we can achieve this permutation with $m(m + n - 1)$ transpositions in total.
