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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)
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"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
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