Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Solving a Overdeterminated System - More Equantions than unknows Date: Sat, 16 Apr 2011 17:51:05 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 47 Message-ID: <iocku9$nnp$1@fred.mathworks.com> References: <ioaic9$avp$1@fred.mathworks.com> <ioaulj$fek$1@speranza.aioe.org> <ioc9ep$d1f$1@fred.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: www-03-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1302976265 24313 172.30.248.48 (16 Apr 2011 17:51:05 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Sat, 16 Apr 2011 17:51:05 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:722246 "Carlos Junior" <carlosjunior@gmail.com> wrote in message <ioc9ep$d1f$1@fred.mathworks.com>... > Hi Nasser, > Thanks by the attention, but I did not understand your solution... I tried now to apply that at MatLab and it did not work ! > > My necessity is to solve the system below (to find H) : > > IMPORTANT : H is a 3x3 matrix. Thus, I have nine unknows. > > H = [ h11 h12 h13 ; h21 h22 h23 ; h31 h32 h33 ] > > [296 199 1]' = H * [0.1 0.1 1]' > [367 210 1]' = H * [0.2 0.1 1]' > [273 267 1]' = H * [0.1 0.2 1]' > [346 281 1]' = H * [0.2 0.2 1]' > > How should I use your proposition A\b ? > > Thanks, > > Carlos > carlosjunior@gmail.com - - - - - - - - - I think what Nasser has indicated to you is that you are trying to "solve" the matrix equation H*A = b where H is a 3 x 3 matrix of unknowns with A = [0.1 0.2 0.1 0.2; 0.1 0.1 0.2 0.2; 1 1 1 1]; and b = [296 367 273 346; 199 210 267 281; 1 1 1 1]; As you have stated this involves 12 equations but only 9 unknowns so it cannot actually be solved. However matlab's 'slash' operator is designed to give the least squares approximation to it: H = b/A Notice that it is analogous to doing a right multiply by the inverse of A, though A's size is 3 x 4 and cannot have an actual inverse. Similar statements can be made about the 'backslash' operator. Notice also that each of the three rows of H represent an independent set of 4 equations and 3 unknowns. You could have applied least squares analysis to each set separately. Roger Stafford