
Re: solving matrices of non liner third order type
Posted:
May 23, 2013 12:54 PM


"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]; > > > 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
hi mr. torsten i used fsolve and found the solution. unfortunately we dont have any initial guess for value .. initially i took 5 as guess i got set of values after solving equations. when i changed the initial guess as 50 i got different set of values as answer.. please suggest a methodme how to proceed!!!

