Building the Matrix from the two matrices

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surendra kumar Aralapura mariyappa
surendra kumar Aralapura mariyappa 2019 年 6 月 5 日
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
  4 件のコメント
Jan
Jan 2019 年 6 月 5 日
@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
surendra kumar Aralapura mariyappa 2019 年 6 月 5 日
編集済み: surendra kumar Aralapura mariyappa 2019 年 6 月 5 日
@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

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採用された回答

Raghunandan V
Raghunandan V 2019 年 6 月 6 日
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;
  1 件のコメント
surendra kumar Aralapura mariyappa
surendra kumar Aralapura mariyappa 2019 年 6 月 6 日
@Raghunandan:Thank you my fellow Indian bro...

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