I different framing of Kevin's answer could be that it solves your systems of equations: three systems of one equation in three variables
julia> x=a/b 3x3 Array{Float64,2}: 0.0519481 0.0649351 0.0779221 0.103896 0.12987 0.155844 0.155844 0.194805 0.233766 julia> x*b 3-element Array{Float64,1}: 1.0 2.0 3.0 The solutions are not unique and Julia chooses those with minimum norm. 2014-02-05 Kevin Squire <kevin.squ...@gmail.com>: > > Hello Fil, > > Generally, division is like multiplying by the inverse of the divisor. For vectors, the inverse is actually a pseudoinverse or generalized inverse. In this case, > > a / b = a * pinv(b) > > where > > julia> pinv(b) > 1x3 Array{Float64,2}: > 0.0519481 0.0649351 0.0779221 > > julia> pinv(b) * b > 1-element Array{Float64,1}: > 1.0 > > julia> a * pinv(b) # the outer product of a and pinv(b) > 3x3 Array{Float64,2}: > 0.0519481 0.0649351 0.0779221 > 0.103896 0.12987 0.155844 > 0.155844 0.194805 0.233766 > > Cheers! > Kevin > > > On Tue, Feb 4, 2014 at 9:19 PM, Fil Mackay <f...@vertigotechnology.com> wrote: >> >> Can anyone explain what's going on here? :) >> >> julia> a = [1,2,3] >> 3-element Array{Int64,1}: >> 1 >> 2 >> 3 >> >> julia> b = [4,5,6] >> 3-element Array{Int64,1}: >> 4 >> 5 >> 6 >> >> julia> a / b >> 3x3 Array{Float64,2}: >> 0.0519481 0.0649351 0.0779221 >> 0.103896 0.12987 0.155844 >> 0.155844 0.194805 0.233766 >> >> I get the principle of how two vectors (3x) lead to a matrix (3x3) result - but the actual values are a mystery? >> > -- Med venlig hilsen Andreas Noack Jensen