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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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Matrix $B$ is not symmetric. –  Gerry Myerson Mar 5 '12 at 1:57
    
@GerryMyerson: yes but the matrices inside are, I am trying to break this puzzle into parts and try to find some elegant way to solve this (without the bruteforce and actuallying doing everything one-by-one). –  hhh Mar 5 '12 at 2:00
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Semi-relevant: en.wikipedia.org/wiki/Determinant#Block_matrices –  user2468 Mar 5 '12 at 2:12
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and slide 7 of this: rscosan.com/documents/RCTM08_rcostas.pdf –  user2468 Mar 5 '12 at 2:15
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Also this mth.kcl.ac.uk/~jrs/gazette/blocks.pdf –  user2468 Mar 5 '12 at 3:02

1 Answer 1

up vote 2 down vote accepted

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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...why do the matrices $S_{2,2,2}$ and $I_{2,2}$ need to commute? –  hhh Mar 5 '12 at 2:34
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@hhh $\det(\begin{matrix}A&B\\C&D\end{matrix}) = \det(D)\det(A-BD^{-1}C)$ $= \det(A-BD^{-1}C)\det(D) = \det(AD - BD^{-1}CD)$. First equality is a property of determinants. Second equality by commutativity of scalars, and third equality is an identity of determinants under multiplication. If $C, D$ commute then $CD = DC$ and $BD^{-1}CD = BD^{-1}DC = BC.$ Hence $\det(\cdot) = \det(AD-BC)$. In your case, $A = D = I_{2,2}$ and $B = S_{2,2,1}$ and $C = S_{2,2,2}$. –  user2468 Mar 5 '12 at 2:57
    
I wish I could somehow copy-paste this LaTex to a separate answer to make it more visually-pleasing, @mods anyway? Also, I would like to move the references to an answer... –  hhh Mar 5 '12 at 3:42

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