Re: solution to undetermined linear equation

• To: mathgroup at smc.vnet.net
• Subject: [mg83976] Re: solution to undetermined linear equation
• From: dh <dh at metrohm.ch>
• Date: Thu, 6 Dec 2007 02:37:59 -0500 (EST)
• References: <fj64qu\$fbo\$1@smc.vnet.net>

```
Hi Shame,

M becomes a 4 x 13 matrix with and additional row of {1,1,1...} and your

M.g == rhs, where g is the searched solution and rhs the vector

{0,0,..,0,13}. You have now 4 equations and 13 variables. Therefore,

there maybe many solutions. You can now either use LinearSolve, this

returns one of the many possible solution. The rest you may get from the

NullSpace. Or you can use Solve that gives you 4 g's as a function of

the remaining g's

hope this helps, Daniel

shama shahbaz wrote:

> hi

>

>   my basic problem is i have a matrix  13*3  of values

>

>   -0.2028            0.12778    -0.09129

>   0.1278             -0.09129    0.07089

>   -0.09129           0.07089     -0.05731

>   0.07089            -0.05731     0.04902

>   -0.05731             0.04902     -0.04188

>   0.04902             0.04188        0.03747

>

>   .............

>   ............

>

>

>   i want to find wieght factors which when applied to above values would result in a value close to 0

>

>   such as

>

>   -0.2028  g1 +  0.1278  g2 +   -0.09129 g3 +0.07089 g4+........................g13=min

>    0.12778 g1+-0.09129g2 + 0.07089 g3  +   -0.05731g4+.........................g13=min

>    -0.09129g1+0.07089g2+   -0.05731g3  +    0.04902 g4+........................g13=min

>

>

>   howeverthe values of g's should be subjected to the constraint

>

>   g1+g2+g3+.........................g13=13 and all the wieghts must be nonzero

>

>   Is it possible

>

>   regards,

>

>

```

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