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I could do this problem with bruteforce but I think there must be some elegant theorem that helps to calculate the determinant with the block matrix (here having symmetric matrices inside) such as:

$B=\begin{pmatrix}1 & 1 & 4 & 5 \\ 1 & 1 & 5 & 4 \\ 2 & 4 & 1 & 1 \\ 4 & 2 & 1 & 1 \\\ \end{pmatrix}=\begin{pmatrix}I_{2,2} & S_{2,2,1} \\ S_{2,2,2} & I_{2,2}\end{pmatrix}$

Actually, look this one

$B= \begin{pmatrix}1 & 1 & 2 & 4 \\ 1 & 1 & 4 & 2 \\ 2 & 4 & 1 & 1 \\ 4 & 2 & 1 & 1 \\ \end{pmatrix}+ \begin{pmatrix} 0 & 0 & 2 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\\ \end{pmatrix}$

and now I am thinking how I could use this one to speed up the calculation...determinant over this-kind-of-matrix-sum?

The problem is booringly stated as with Gaus method but I am interested to find some trick to calculate the determinante. My first idea was to do $4-3$ -row-minus and $1-2$ -row-minus (so getting some ones away but there must be some theorem to simplify the monotonous Gaussian elimination and determinant finding).

Page 741 here.

References by J.D. for further research

  1. http://en.wikipedia.org/wiki/Determinant#Block_matrices

  2. http://rscosan.com/documents/RCTM08_rcostas.pdf

  3. http://mth.kcl.ac.uk/~jrs/gazette/blocks.pdf

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    @J.D.: I gathered the references to the question (trying to keep things tidy) but you could easily make an answer with the details in comments, perhaps clarifying things.2012-03-05

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Since the matrices $S_{2,2,2}$ and $I_{2,2}$ commute it follows that $det(A)=det(I_{2,2}I_{2,2}-S_{2,2,1}S_{2,2,2})$.

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    I wish I could somehow copy-paste this LaTex to a separate a$n$swer to make it more visually-pleasing, @mods anyway? Also, I would like to move the references to an answer...2012-03-05