Building the Matrix from the two matrices

Hey, I am stuck in building the Matrix for my project "Thermal Network''. The problem is how to buid the Matrix from the two matrices of same dimensions.

Details goes here;

LA = [ 0 77 0 0; 77 0 15 0; 0 15 0 0; 0 0 0 0];

LB = [ 0 0 0 25; 0 0 0 28; 0 0 0 25; 25 28 25 0];

Now I need to build the matrix L like;

L= [LA(2,1)+LB(1,4) (-LA(1,2)) 0 (-(LB(1,4));

(-LA(2,1)) LA(2,1)+LA(2,3)+LB(2,4) (-LA(2,3)) (-LB(2,4));

0 (-LA(3,2)) LA(3,2)+LB(3,4) (-LB(3,4));

(-LB(4,1)) (-LB(4,2)) (-LB(4,3)) 0];

or else like this also fine:

L= [LA(1,2)+LB(4,1) (-LA(2,1)) 0 (-(LB(4,1));

(-LA(1,2)) LA(1,2)+LA(3,2)+LB(4,2) (-LA(3,2)) (-LB(4,2));

0 (-LA(2,3)) LA(2,3)+LB(4,3) (-LB(4,3));

(-LB(1,4)) (-LB(2,4)) (-LB(3,4)) 0];

May be it would be possible using for-loop. It is just sample model, If i change the martices dimensions, Matrix should be built automatically, that's why for-loop would be better option!

If you have any suggestions and your answers are most weclomed and appreciated as well

[Jan]

What's wrong with this code?

L= [LA(1,2)+LB(4,1), -LA(2,1),                0,               -LB(4,1); ...
    -LA(1,2),        LA(1,2)+LA(3,2)+LB(4,2), -LA(3,2),        -LB(4,2); ...
    0,               -LA(2,3)),               LA(2,3)+LB(4,3), -LB(4,3); ...
    -LB(1,4),        -LB(2,4),                -LB(3,4),        0];

Looks nice, so why do you want a loop?

[surendra kumar Aralapura mariyappa]

It is just a simple model, and I need the martix L should be built autmatically, if I change the dimenisions of matrices LA and LB

[Jan]

@surendra: You provide two definitions, in one the first element is LA(2,1)+LB(1,4), in the other it is LA(1,2)+LB(4,1). I cannot guess which one is wanted or what you want, if the dimensions of LA and LB are changed. Without giving a general definition how the output L is created, we cannot suggest the corresponding code.

[surendra kumar Aralapura mariyappa]

@Jan, I understood your concern, Here I am giving more information for the clarity

the dimensions of LA and LB are 4*4, and the matrices are like this:

LA = [ 0 77 0 0;

77 0 15 0;

0 15 0 0;

0 0 0 0];

LB = [ 0 0 0 25;

0 0 0 28;

0 0 0 25;

25 28 25 0];

Now I am changing the matrices size; 5*5

kindly see the similarity between the values in the matrices;

LA = [ 0 77 0 0 0;

77 0 15 0 0;

0 15 0 18 0;

0 0 18 0 0;

0 0 0 0 0];

LB = [ 0 0 0 0 25;

0 0 0 0 28;

0 0 0 0 25;

0 0 0 0 20;

25 28 25 20 0 ];

L = L= [LA(1,2)+LB(1,5) (-LA(1,2)) 0 0 (-(LB(1,5));

(-LA(2,1)) LA(2,1)+LA(2,3)+LB(2,5) (-LA(2,3)) 0 ( -LB(2,5));

0 (-LA(3,2)) LA(3,2)+LA(3,4)+LB(3,5) (-LA(3,4)) (-LB(3,5));

0 0 -(LA(4,3)) LA(4,3)+LB(4,5) (-LB(4,5));

(-LB(5,1)) (-LB(5,2)) (-LB(5,3)) (-LB(5,4)) 0];

Hope this explanation will be fine;

Same for 50*50 dimensions also

Answer 1

[Raghunandan V]

Hi,

Please check the answer

LA = [     0    77      0         0; 
           77    0      15        0;
           0    15      0         0; 
           0     0      0         0];
LB = [     0     0      0      25; 
           0     0      0      28; 
           0     0      0      25;
          25    28     25      0];
[m,n ] = size(LB);
L = -LB;
L(1: m-1, 1:n-1) = -LA(1: m-1, 1:n-1);
for k = 1: m
    L(k,k) = -(sum(L(k,:)));
end
L(m,m) = 0;

[surendra kumar Aralapura mariyappa]

@Raghunandan:Thank you my fellow Indian bro...