When you have matrices with special structures and symmetries, Julia can encode that information in the type of the matrix, and in those cases the backslash operator should actually be more efficient than in Matlab because there is no need to “guess” what the best algorithm might be. My understanding is that no such thing exists in Matlab. Having said that, take this with a grain of salt, this is not my area of expertise (neither in Julia nor in Matlab).
The relevant Julia manual section is http://docs.julialang.org/en/latest/manual/linear-algebra/. From: julia-users@googlegroups.com [mailto:julia-users@googlegroups.com] On Behalf Of Art Kuo Sent: Wednesday, September 30, 2015 9:30 AM To: julia-users <julia-users@googlegroups.com> Subject: Re: [julia-users] Re: What's the reason of the Success of Python? > (x=inv(A)*b) than it is in many other languages. Julia has at least this You shouldn't ever do this (in either Julia or Matlab, or any language), it is ill-conditioned for general matrices. I think the Matlab function is linsolve. Perhaps more precise to say "A\b is always faster than inv(A)*b, and more accurate if A is ill-conditioned." I also usually prefer Matlab's backslash or mldivide, A\b, over the more specific linsolve, because backslash chooses the right solver (including linsolve) most of the time, for systems that are sparse, over- or under-determined, etc. Julia also implements similar functionality, but I doubt if it is as optimized as Matlab, which has many years' advantage. As for the accuracy advantage, it is true that inv(A)*b can go wrong, and A\b will go less wrong. But Ill-conditioned A is usually an indicator that neither approach will be great.
