# Diagonal of the inverse of a 6x6 symmetric partitioned matrix

Let
$$M = \begin{bmatrix} A & B \\ B & C \end{bmatrix}$$ in which $A$, $B$ and $C$ are $3 \times 3$ matrices being also symmetric. In fact, they are quite similar, just differing on a single term: $$A = \begin{bmatrix} \sum_x \sum_y A(x,y) x^2 & \sum_x \sum_y A(x,y) xy & \sum_x \sum_y A(x,y) x\\ \sum_x \sum_y A(x,y) xy & \sum_x \sum_y A(x,y) y^2 & \sum_x \sum_y A(x,y) y\\ \sum_x \sum_y A(x,y) x & \sum_x \sum_y A(x,y) y & \sum_x \sum_y A(x,y) \\ \end{bmatrix}$$ $B$ and $C$ are exactly the same as $A$ but replacing $A(x,y)$ by $B(x,y)$ and $C(x,y)$ respectively. If we assume every matrix is non-singular, is there an analytic way to get the diagonal of the inverse of matrix M, that is $M^{-1}_{i,i}$?

Are $A,B,C$ recursively defined? I can't figure out what you mean by this definition... –  Felix Goldberg Jun 26 '13 at 9:12
Perhaps you can explain how you mean the summation over $x$ and $y$. Are the entries of $A$ polynomials in $x$ and $y$, or do $x$ and $y$ run from $1$ to $k$ ? –  Dietrich Burde Jun 28 '13 at 9:37