How do I get the matrix p

Hello,
I am trying to solve the a matrix using solve function, not able to figure it out.
clear all
clc
syms x1 x2 M k L dx1 dx2 p
% Displacement of the bar
xg=(x1+x2)/2;
% Rotation of the bar
thi=(x2-x1)/L;
%moment of inertia
J=M*L^2/12;
dxg=(dx1+dx2)/2;
dthi=(dx2-dx1)/L;
% Kinetic energy
KE=0.5*((M*dxg^2)+(J*dthi^2));
% Potenial Engergy
PE=0.5*(2*k*x1^2+3*k*x2^2);
u=[x1 x2];
du=[dx1;dx2];
s=solve(KE==0.5.*transpose(du)*p*du);

3 comentarios

Walter Roberson
Walter Roberson el 17 de Mzo. de 2019
s = solve(EXPRESSION, p)
LOKESH UDATHA
LOKESH UDATHA el 18 de Mzo. de 2019
Thanks a lot, but a small problem
my p is a 2*2 matrix but I am getting sloution as a scalar.
Walter Roberson
Walter Roberson el 18 de Mzo. de 2019
No it isn't . u is 1 x 2 so transpose of u is 2 x 1. You cannot do 2 x 1 * 2 x 2 * 1 x 2 because none of the inner dimensions match. Your u would have to be 2 x 1 to fix that .
There is not enough information to isolate 2 x 2 p.

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Respuestas (1)

Sajeer Modavan
Sajeer Modavan el 17 de Mzo. de 2019

0 votos

Since you are solving with for p, result of 's' equal to 'p'
to confirm this, you can use following code (which is same as yours, but forcefully solving for 'p')
P = solve(KE==0.5.*transpose(du)*p*du,p);

1 comentario

LOKESH UDATHA
LOKESH UDATHA el 18 de Mzo. de 2019
Thanks a lot for replying,
but my p is a 2*2 matrix but I am getting p as a scalar.

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