Trouble Creating a Vector using a for loop with matrices

5 vues (au cours des 30 derniers jours)
Andrew Knight
Andrew Knight le 7 Nov 2019
Commenté : Stephen23 le 7 Nov 2019
I'm having trouble creating a vector of answers with a for loop involving matrices. When I call on w, it gives me different values than if I were to call on a specific part of w.
B1=[1 0; 0 -7.5];
D1=[0 0;0 10];
X1=[0 0;0 -2.5];
B2=[49.8 0;-0.127 -50];
D2=[25 0;0 25];
A2=[25 0;0 25];
B3=[49.8 0;-0.127 -50];
D3=[25 0;0 25];
A3=[25 0;0 25];
B4=[49.8 0;-0.127 -50];
D4=[25 0;0 25];
A4=[25 0;0 25];
B5=[7.5 0;0 1];
A5=[-10 0;0 0];
Y5=[2.5 0;0 0];
o=[0 0;0 0];
%M
M=[B1 D1 X1 o o;A2 B2 D2 o o;o A3 B3 D3 o;o o A4 B4 D4;o o Y5 A5 B5];
Mi=inv(M);
G=[.21;0;0;0;0;0;0;0;0;0.127];
CG=[.1;.1;.1;.1;.1;.1;.1;.1;.1;.1];
%find L
%L=[b1 o o o o;a2 b2 o o o;o a3 b3 o o;o o a4 b4 o;o o y5 a5 b5]
%U=[I -E1 -x1 o o;o I -E2 o o;o o I -E3 o;o o o I -E4;o o o o I]
n=2;
NJ=5;
w=zeros(2*NJ,1);
for j=1
L(1:(j+1),1:(j+1))=B1; %b(1)=B1
U(1:j+1,j+2:j+3)=-inv(L(1:j+1,1:j+1))*D1; %E(1)=D1*inv(-b(1))
U(1:j+1,2*j+3:2*j+4)=-inv(L(1:j+1,1:j+1))*X1; %x(1)=X1*inv(-b(1))
w(1:2)=inv(L(1:j+1,1:j+1))*G(1:2); %ksi(1) = b(1)^-1 * G(1)
end
for j=2:NJ-1
L(2*j-1:2*j,2*j-3:2*j-2)=A2; %a(2)=A2
L(2*j-1:2*j,2*j-1:2*j)=B2+L(2*j-1:2*j,2*j-3:2*j-2)*U(2*(j-1)-1:2*(j-1),2*(j-1)+1:2*(j-1)+2); %b(2)=B2+a(2)*E(1)
U(2*j-1:2*j,2*j+1:2*j+2)=-inv(L(2*j-1:2*j,2*j-1:2*j))*D2; %E(2)=-inv(b(2))*D(2)
w(2*j-1:2*j)=-inv(L(2*j-1:2*j,2*j-1:2*j))*((-G(2*j-1:2*j)+(L(2*j-1:2*j,2*j-3:2*j-2)*w((2*(j-1)-1):(2*(j-1)))))); %ksi(2)=inv(-b(2))*(-G(2)+a(2)*ksi(1))
end
for j=NJ
L(2*j-1:2*j,2*j-5:2*j-4)=Y5; %y(5)=Y5
L(2*j-1:2*j,2*j-3:2*j-2)=A5+L(2*j-1:2*j,2*j-5:2*j-4)*U(2*(j-2)-1:2*(j-2),2*(j-2)+1:2*(j-2)+2); %a(5)=A5+y(5)*E(NJ-2)
L(2*j-1:2*j,2*j-1:2*j)=B5+L(2*j-1:2*j,2*j-3:2*j-2)*U(2*(j-1)-1:2*(j-1),2*(j-1)+1:2*(j-1)+2); %b(5)=B5+a(5)*E(4)
w(2*j-1:2*j)=-inv(L(2*j-1:2*j,2*j-1:2*j))*(-G(2*j-1:2*j)+L(2*j-1:2*j,2*j-3:2*j-2)*w(2*(j-1)-1:2*(j-1))+L(2*j-1:2*j,2*j-5:2*j-4)*w(2*(j-2)-1:2*(j-2)))%-inv(b(5))*(-G(5)+a(5)*w(4)+y*(w(3))
end
%w=inv(L)*G
%C=inv(U)*w
  1 commentaire
Stephen23
Stephen23 le 7 Nov 2019
Note: to make your code more efficient and robust you should probably be using mldivide instead of inv and *. Explicit matrix inversion is rarely required (because there are better methods for solving such systems of equations).

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