Originally Posted by
aerzo Can anyone tell me why the summation's upper bounds 4 and 3 are used in the proof below? Thanks.
Let A = [aij], B = [bij], C = [cij], AB = D = [dij], BC = E = [eij], (AB)C = F = [fij], and A(BC) = G = [gij]. To show that fij = gij for all i, j:
[tex]
4 4 3
fij = Σ dikckj = Σ (Σ airbrk) ckj
k=1 k=1 r=1
and
3 3 4
gij = Σ airerj = Σ air (Σ brkckj)
r=1 r=1 k=1
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