Sure. This was just a side note. I would expect that a "notation" A*B/B == A to be always true (except for division by zero of course), since division the inverse process of multiplication, like for scalar values (a*b/b == a or A.*B./B .== A).
Am Montag, 14. April 2014 00:38:57 UTC+2 schrieb andrew cooke: > > > > On Sunday, 13 April 2014 19:18:31 UTC-3, [email protected] wrote: >> >> (Also note that division does not look like the inverse operation of >> multiplication, since e.g. A=[1 2 3]; B=[1, 2, 4]; A*B/B == A returns >> false.) >> > > what would you expect to happen here? you're taking a scalar product and > asking "this was one of the vectors; what is the other?". that's not well > defined. you've lost information. > > on the other hand: > > julia> A = [1 2;3 4] > 2x2 Array{Int64,2}: > 1 2 > 3 4 > > julia> B = [2 3;4 5] > 2x2 Array{Int64,2}: > 2 3 > 4 5 > > julia> A*B/B > 2x2 Array{Float64,2}: > 1.0 2.0 > 3.0 4.0 > > andrew > >
