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Tricky puzzle
Consider a graph, not directed, the nodes of which we'll call n_i and the edges r_i. To each r_i associate a value a_i, complex if you like. Now construct a matrix A having elements as follows:
Each diagonal element a_ii is the sum of any a_u such that r_u connects n_i to some other node, zero otherwise. Note: no edge may connect a node to itself.
Each non-diagonal element a_ij is -a_u if an edge r_u connects n_i and n_j (the sum, if the graph isn't simple), zero otherwise.
Obvious note, A will be symmetric. Chose one node n_p and add a non-zero value a to a_pp. Now invert this matrix; using ^ to indicate inversion, we could call it B = (A + P)^ where P has the one element of value a and all others zero.
I say: B_pp = 1/a. Easy to prove?
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Inutil insegnŕ al mus, si piart timp, in plui si infastiděs la bestie.
Hypography Forum PITA...... er, Administrator. 
Last edited by Qfwfq; 02-02-2006 at 06:44 AM..
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