# Some theorem about block matrix determinants with symmetric inner matrices?

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

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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
Semi-relevant: en.wikipedia.org/wiki/Determinant#Block_matrices –  user2468 Mar 5 '12 at 2:12
and slide 7 of this: rscosan.com/documents/RCTM08_rcostas.pdf –  user2468 Mar 5 '12 at 2:15
–  user2468 Mar 5 '12 at 3:02
Since the matrices $S_{2,2,2}$ and $I_{2,2}$ commute it follows that $det(A)=det(I_{2,2}I_{2,2}-S_{2,2,1}S_{2,2,2})$.
...why do the matrices $S_{2,2,2}$ and $I_{2,2}$ need to commute? –  hhh Mar 5 '12 at 2:34
@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