# 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)

• Your $O, F$ are not square matrices and do not have determinants. Commented Nov 16, 2021 at 23:54

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.