Given a $4 \times 4$ matrix $S$ over a commutative ring $R$. I want to consider it as a $2\times 2$ matrix over $M\_2(R)$. Lets say $S=\left(\begin{array}{cc} A&B \\\ C&D\end{array}\right)$ with entries $A,B,C,D\in M\_2(R)$. Asssume further, that they all commute. 


So my question is: Can one then compute the determinant stepwise via


$\det(S)=\det(AD-BC)$.


I checked this with maple for the case, where all the sub-matrices lie in the commutative subring $\{\left(\begin{array}{c}a&b\\\ 2b&a \end{array}\right)|a,b\in \mathbb{Z}\}$ of $M\_2(\mathbb{Z})$. And I also would like to consider the more general case, where one considers a $mn\times mn$ matrix over $R$ as a $m\times m$ matrix over $M\_n(R)$.



Iterated calculation of determinants
The answer is **Yes**. This is done in Exercise 120 of my web page [link text](http://www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf). You can replace $4$ and $2$ by numbers $n$ and $m$ with $m$ dividing $n$.


**Later**. The required complement. Let me take the situation of the question, but with $A,B,C,D$ being $m\times m$ matrices (thus $n=2m$). If $A$ has an inverse, the Schur complement formula tells you that
$$\det S=\det A\cdot\det(D-CA^{-1}B).$$
Because $A$ and $C$ commute, this is nothing but
$$\det A\cdot\det(A^{-1}(AD-BC))=\det(AD-BC).$$
Now, what if neither $A$ nor $B,C,D$ admit an inverse ? Just do the following. Replace $\mathcal R$ by $\mathcal R(X)$ (rational fractions). Then apply the formula to
$$\Sigma=\begin{pmatrix} A+XI\_m & B \\\\ C & D \end{pmatrix},$$
remarking that $A+XI\_m$ is non-singular, because its determinant is invertible.


**Even later**. Here are the details when $p:=n/m$ is larger than $2$. The matrix $S$ is $n\times n$, with commuting blocks $A\_{ij}\in M\_m(\mathcal R)$, $1\le i,j\le p$. To avoid confusion, I use the notation ${\rm Det}$ to denote the determinant in the commutative subring $\mathcal R'$ of $M\_m(\mathcal R)$ spanned by the blocks $A\_{ij}$, whereas $\det$ is the determinant of an ordinary matrix.


The proof is an induction over $p$.
As in the case $p=2$, I may assume that $A\_{11}$ is invertible. Schur's formula gives
$$\det S=\det A\_{11}\det S\_{11},$$
where
$$ \qquad S\_{11}:=\begin{pmatrix} A\_{22} & \cdots & A\_{2p} \\\\ \vdots & & \vdots \\\\ A\_{p2} & \cdots & A\_{pp} \end{pmatrix}-\begin{pmatrix} A\_{21} \\\\ \vdots \\\\ A\_{p1} \end{pmatrix} A\_{11}^{-1}\begin{pmatrix} A\_{12} & \cdots & A\_{1p} \end{pmatrix}.$$
Because $A\_{11}$ commutes to every $A\_{ij}$, we find
$$\det S ( \det A\_{11})^{p-2} = \det T ,$$
where
$$ \qquad T= block(A\_{11}A\_{ij}-A\_{i1}A\_{1j})\_{2\le i,j\le p}.$$
From the induction hypothesis, and because the blocks $A\_{11}A\_{ij}-A\_{i1}A\_{1j}$ commute to each other, one has
$$\det T=\det {\rm Det}((A\_{11}A\_{ij}-A\_{i1}A\_{1j}))\_{2\le i,j\le p}.$$
Now, in every commutative ring, we have
$$\det((x\_{11}x\_{ij}-x\_{i1}x\_{1j}))\_{2\le i,j\le p}=x\_{11}^{p-2}\det X.$$
Therefore
$$\det T=\det\left(A\_{11}^{p-2}{\rm Det} S\right)=(\det A\_{11})^{p-2}\det{\rm Det} S.$$
Simplifying by $(\det A\_{11})^{p-2}$, we obtain the expected formula
$$\det S=\det{\rm Det} S.$$


