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
