Path: news.mathworks.com!not-for-mail From: "b b" <bogey4@gmail.com> Newsgroups: comp.soft-sys.matlab Subject: Re: Bogey Date: Sun, 26 Oct 2008 18:00:06 +0000 (UTC) Organization: university of Guelph Lines: 28 Message-ID: <ge2b76$msj$1@fred.mathworks.com> References: <gdvjlp$kij$1@fred.mathworks.com> <gdvoje$gnn$1@fred.mathworks.com> <gdvq28$482$1@fred.mathworks.com> <ge0231$lu$1@fred.mathworks.com> Reply-To: "b b" <bogey4@gmail.com> NNTP-Posting-Host: webapp-02-blr.mathworks.com Content-Type: text/plain; charset="ISO-8859-1" Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1225044006 23443 172.30.248.37 (26 Oct 2008 18:00:06 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Sun, 26 Oct 2008 18:00:06 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 944946 Xref: news.mathworks.com comp.soft-sys.matlab:497318 Brilliant! The first line ( tot = X*ksi*Y; ) gave me the correct answer. I'm glad to now know an elegant and straight forward solution to this. As for computational speed, it seems that the equation that you suggested is faster: tot = zeros(2,3); tic; for n = 1: 4; for i = 1: 4; tot = tot+ ksi(n,i)*X(:,n)*Y(i,:); end; end; toc Elapsed time is 0.000118 seconds. tic; X*ksi*Y; toc Elapsed time is 0.000032 seconds. Thanks again! Bruce > In that case try this: > > tot = X*ksi*Y; > > It should give the correct answer. Whether it is faster you will have to determine for yourself. You can experiment with parentheses: (X*ksi)*Y as compared with X*(ksi*Y). > > Roger Stafford >