Determinants of 'block' matrices

Question

I am trying to simplify the determinant of \begin{pmatrix}C&A\\B&0\end{pmatrix}

where $A$ and $B$ are square $m\times m$ and $n\times n$ matrices, and $C$ is some $m\times n$ matrix, $0$ is $n\times m$.

I believe I can say it is equal to $\det(-AB)=-\det(A)\det(B)$ but my argument is very vague, involving a lot of 'hand-waving'. I was thinking there was a way of forming this matrix from the products of upper and lower triangular matrices - and thus the result following - but I can't seem to figure it out?

Answer 1

First step: if you exchange two columns in a matrix, then the determinant gets multiplied by $(-1)$.

Second step: we can therefore put last $m$ columns to the beginning, all this doesn't change the determinant - up to a possible multiplication by $(-1)$ (you can easily find when you need to do it depending on $m$ and $n$).

Now you matrix writes

$$\begin{pmatrix}A&C\\0&B\end{pmatrix}.$$ Can you take it from here?

Answer 2

Notice that

$$\begin{pmatrix}C&A\\B&0\end{pmatrix}=\begin{pmatrix}C&I\\B&0\end{pmatrix}\begin{pmatrix}I&0\\0&A\end{pmatrix}$$ so $$\det\begin{pmatrix}C&A\\B&0\end{pmatrix}=\det\begin{pmatrix}C&I\\B&0\end{pmatrix}\det\begin{pmatrix}I&0\\0&A\end{pmatrix}$$

and we develop the two determinants on the RHS relative to the last columns and to the first columns to get $$\det\begin{pmatrix}C&A\\B&0\end{pmatrix}=(-1)^{m(n+m+1)}\det B\det A$$