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

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 Hari Kishore Posts: 7 Registered: 4/10/13
solving matrices of non liner third order type
Posted: May 15, 2013 3:18 AM
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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

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

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