Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: solving matrices of non liner third order type
Replies: 12   Last Post: May 24, 2013 2:30 AM

 Messages: [ Previous | Next ]
 Hari Kishore Posts: 7 Registered: 4/10/13
Re: solving matrices of non liner third order type
Posted: May 15, 2013 6:44 AM

"Torsten" wrote in message <kmvm1t\$d83\$1@newscl01ah.mathworks.com>...
> "Hari Kishore " <harikishoreguptha@gmail.com> wrote in message <kmvibo\$3rj\$1@newscl01ah.mathworks.com>...
> > "Torsten" wrote in message <kmve6l\$mq3\$1@newscl01ah.mathworks.com>...
> > > "Hari Kishore " <harikishoreguptha@gmail.com> wrote in message <kmvcrf\$jbj\$1@newscl01ah.mathworks.com>...
> > > > hiii..
> > > > i tried to solve this matrices but i couldnot because it ends up with non linear third order equations please help me out in this issue and give a program to solve this....
> > > > your help is highly appriciated!!
> > > > syms bcf0 bcf1 bcf2 bcf3 bcf4 bcf5 q0 q1 q2
> > > >
> > > >
> > > > amatrix=[ 12.0, 0, 9.0, 12.0, 0, 9.0; 12.0, 4.39*10^(-4), 9.0, 12.0, 0, 9.0; 12.0, 0, 9.0, 12.0, 0, 9.0; 12.0, 0, 9.0, 12.3, 0, 9.01; 12.0, 0, 9.0, 12.0, 0.00703, 9.0; 12.0, 0, 9.0, 12.0, 0, 9.0];
> > > >
> > > >
> > > > acoeff=[ bcf0;bcf1;bcf2;bcf3;bcf4;bcf5];
> > > >
> > > >

mr.torsten
actually there is no need of any guess for bcf.. bcfs's are to be find int terns of q0 q1 q2 fromfirst equation and substitute in second equation.. so final equtions would be.. 3 equations in 3 variables..
> > > > bmatrix =[ 47.3, 42.3, 19.4; -1.63*10^(-19), 4.53*10^(-21), -1.03*10^(-15);1.69, 3.52, 1.47;23.6, 21.1, 9.7;-2.45*10^(-19), 6.79*10^(-21), -5.17*10^(-16);0.844, 1.76, 0.737];
> > > >
> > > > bcoeff=[q0^2;q1^2;q2^2];
> > > >
> > > > cmatrix =[ 16.4, -0.0774, 0.193; -0.0774, 0.724, -0.00177; 0.193, -0.00177, 0.0645];
> > > >
> > > > ccoeff=[q0;q1;q2];
> > > > dmatrix =[ 1.65, 0.0571, 0.00251, -1.65, -0.127, -0.00528;2.45*10^(-18), -4.26*10^(-19), 7.37*10^(-20), 7.09*10^(-17), 4.26*10^(-19), 4.58*10^(-19);0.00302, 8.52*10^(-4), 6.98*10^(-5), -0.0733, -9.76*10^(-4), -6.98*10^(-5)];
> > > >
> > > > dcoeff=[bcf0*q0;bcf1*q1;bcf2*q2; bcf3*q0;bcf4*q1;bcf0*q2];
> > > >
> > > > (amatrix)*(acoeff)=bmatrix*(bcoeff);
> > > > %from first statement find bcf values and use it in second equation
> > > > (cmatrix)*(ccoeff)=dmatrix*(dcoeff);
> > > > %from second equation find the q0 q1 q2 value

> > >
> > > I don't see an easier way than using MATLAB's FSOLVE for your system of nonlinear equations.
> > >
> > > Best wishes
> > > Torsten.

> > Hi. mr.torsten..
> > please can you give the code for that operation?

>
> x0 = [-5; -5; -5; -5; -5; -5; -5; -5; -5]; % Make a starting guess at the solution
> options = optimoptions('fsolve','Display','iter'); % Option to display output
> [x,fval] = fsolve(@myfun,x0,options) % Call solver
>
> function F = myfun(x)
> bcf0=x(1);
> bcf1=x(2);
> bcf2=x(3);
> bcf3=x(4);
> bcf4=x(5);
> bcf5=x(6);
> q0=x(7);
> q1=x(8);
> q2=x(9);
>
> amatrix=[ 12.0, 0, 9.0, 12.0, 0, 9.0; 12.0, 4.39*10^(-4), 9.0, 12.0, 0, 9.0; 12.0, 0, 9.0, 12.0, 0, 9.0; 12.0, 0, 9.0, 12.3, 0, 9.01; 12.0, 0, 9.0, 12.0, 0.00703, 9.0; 12.0, 0, 9.0, 12.0, 0, 9.0];
> acoeff=[ bcf0;bcf1;bcf2;bcf3;bcf4;bcf5];
> bmatrix =[ 47.3, 42.3, 19.4; -1.63*10^(-19), 4.53*10^(-21), -1.03*10^(-15);1.69, 3.52, 1.47;23.6, 21.1, 9.7;-2.45*10^(-19), 6.79*10^(-21), -5.17*10^(-16);0.844, 1.76, 0.737];
> bcoeff=[q0^2;q1^2;q2^2];
> cmatrix =[ 16.4, -0.0774, 0.193; -0.0774, 0.724, -0.00177; 0.193, -0.00177, 0.0645];
> ccoeff=[q0;q1;q2];
> dmatrix =[ 1.65, 0.0571, 0.00251, -1.65, -0.127, -0.00528;2.45*10^(-18), -4.26*10^(-19), 7.37*10^(-20), 7.09*10^(-17), 4.26*10^(-19), 4.58*10^(-19);0.00302, 8.52*10^(-4), 6.98*10^(-5), -0.0733, -9.76*10^(-4), -6.98*10^(-5)];
> dcoeff=[bcf0*q0;bcf1*q1;bcf2*q2; bcf3*q0;bcf4*q1;bcf0*q2];
>
> F=[amatrix*acoeff-bmatrix*bcoeff; cmatrix*ccoeff-dmatrix*dcoeff];
>
> By the way: bcf0=bcf1=bcf2=bcf3=bcf4=bcf5=q0=q1=q2=0 is a solution which
> might complicate the solution process.
> Furthermore, scaling of your equations might be necessary because of the
> difference in magnitude of the matrix coefficients.
>
> Best wishes
> Torsten.

Date Subject Author
5/15/13 Hari Kishore
5/15/13 Torsten
5/15/13 Hari Kishore
5/15/13 Torsten
5/15/13 Hari Kishore
5/15/13 Torsten
5/15/13 Torsten
5/15/13 Hari Kishore
5/15/13 Torsten
5/15/13 Torsten
5/15/13 Hari Kishore
5/23/13 Hari Kishore
5/24/13 Torsten