How do I get the matrix p

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LOKESH UDATHA
LOKESH UDATHA on 17 Mar 2019
Commented: Walter Roberson on 18 Mar 2019
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 Comments
Walter Roberson
Walter Roberson on 18 Mar 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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Answers (1)

Sajeer Modavan
Sajeer Modavan on 17 Mar 2019
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 Comment
LOKESH UDATHA
LOKESH UDATHA on 18 Mar 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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