Adding Matrices Diagonally for FEM

Hello.

I am trying to make a diagnoal matrix in which the three matrices overlap.

I am attaching a picture which will make things more clear.

I can do it for 2x2, but for 3x3 or more like 12x12. I am having confusion.

Can somebody help me expand the logic of 2x2, that would be very nice

here is the code:

B = [ 1 -1;

-1 1];

C = [ 1 -1;

-1 1];

D = [ 1 -1;

-1 1];

T = zeros(4,4)

d=zeros(2,2,3);

d(:,:,1) = B;

d(:,:,2) = C;

d(:,:,3) = D;

for k = 1:1:3

for i = 1:1:2

for j = 1:1:2

T(i+k-1,j+k-1) = T(i+k-1,j+k-1) + d(i,j,k);

end

end

end

[Chris Dan]

per isakson

[jannat alsaidi]

You can do this,

R=zeros(12,12) W=[1 - 1;-1 1] R(1:2,1:2)=W R(3:4,3:4)=W ... For more help give me the two matrices

[Chris Dan]

jannat alsaidi

The two matrices are

A == [ 1 -1 1;-1 2 -1; 1 -1 1]

B == [ 1 -1 1;-1 2 -1; 1 -1 1]

Answer 1

[jannat alsaidi]

A(:,:,1) = [1 -1 1;-1 2 -1; 1 -1 1];

A(:,:,3) = [1 -1 1;-1 2 -1; 1 -1 1];

A(:,:,5) = [1 -1 1;-1 2 -1; 1 -1 1];

R=zeros(7,7)

for n=1:2:5

i=n+[0 1 2]

j=i

R(i,j)=R(i,j)+A(:,:,n)

end

Answer 2

[jannat alsaidi]

A = [ 1 -1 1;-1 2 -1; 1 -1 1];

%A=B

R=zeros(6,6)

for n=1:3

    i=n+[0 1 2]

    j=i

    R(i,j)=R(i,j)+A

end

R

You can change vector [1 2 3], to choose the elements that you need to add them.

[Chris Dan]

will this work for different matrices?

if the values of A and B are not same?

[jannat alsaidi]

yes, it work for different matrices, try it,

A(:,:,1) = [1 -1 1;-1 2 -1; 1 -1 1];
A(:,:,2) = [2 -1 3; -2 5 -6; 7 8 9];
A(:,:,3) = [-8 2 4; -7 9 5; 2 -5 8];

R=zeros(5,5)
for n=1:3
    i=n+[0 1 2]
    j=i
    R(i,j)=R(i,j)+A(:,:,n)
end
R

if there is a different size of (A) matrix change the vector [0 1 2] to the size that is needed, like size(A)= 6×6 the vector will be [0 1 2 3 4 5].

[Chris Dan]

jannat alsaidi

Hey, I ran ur code.

The answer is not correct

can you please check it

The answer which I should get for

A(:,:,1) = [1 -1 1;-1 2 -1; 1 -1 1];

A(:,:,2) = [1 -1 1;-1 2 -1; 1 -1 1];

A(:,:,3) = [1 -1 1;-1 2 -1; 1 -1 1];

R=zeros(6,6)

for n=1:3

i=n+[0 1 2]

j=i

R(i,j)=R(i,j)+A(:,:,n)

end

is

R =[ 1 -1 1 0 0 0 0;

-1 2 -1 0 0 0 0;

-1 -1 2 -1 -1 0 0;

0 0 -1 2 -1 0 0;

0 0 1 -1 2 -1 -1;

0 0 0 0 -1 2 -1;

0 0 0 0 1 -1 1 ]

BUT I am getting :

1 -1 1 0 0 0

-1 3 -2 1 0 0

1 -2 4 -2 1 0

0 1 -2 3 -1 0

0 0 1 -1 1 0

0 0 0 0 0 0

0 0 0 0 0 0

This is wrong...