Building up a Matrix using for LOOP and summation

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Abdulrahman Odhah
评论： Abdulrahman Odhah 2020-4-16
Hello fellows,
I am trying to make this equation but I couldn't do the summaiton part , i tried to use the double for loop but still not sure how to do the summation part,
I have the matrix " a " already defined 4x4 also initial l and u matrices are defiend l=eye(N),u=zeros(N,N);
any hints?

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采纳的回答

David Hill
David Hill 2020-4-5
I believe the below is correct.
I=eye(4);
u=zeros(4);
u(1,1:4)=a(1,1:4);
I(1:4,1)=a(1:4,1);
for i=2:4
for j=1:4
u(i,j)=a(i,j)-sum(I(i,1:i-1).*u(1:i-1,j)');
if i<j
I(j,i)=(I(j,i)-sum(I(j,1:i-1).*u(1:i-1,i)'))/u(i,i);
end
end
end
  2 个评论
Abdulrahman Odhah
Abdulrahman Odhah 2020-4-8
I think its the correct one also, I'll try it on some values,
thanks a lot, for your suuport,
Abdulrahman Odhah
Abdulrahman Odhah 2020-4-16
I've applied the code on some known values, this code is supposed to find the lu matrices of a matrix A, using the above expression
what i've got from the code given by you is the following,
I =
6.0000 0 0 0
4.0000 1.0000 0 0
-2.0000 -0.5333 1.0000 0
3.0000 0.8000 0.0845 1.0000
u =
6.0000 4.0000 2.0000 1.0000
-20.0000 -15.0000 -8.0000 -2.0000
-0.6667 2.0000 4.7333 8.9333
1.0563 1.8310 4.0000 -1.1549
I thing that the code representing the mathematical expression has to produce the following
I =
1.0000 0 0 0
0.6667 -0.5000 0.5000 1.0000
-0.3333 1.0000 0 0
0.5000 0 1.0000 0
u=
6.0000 4.0000 2.0000 1.0000
0 3.3333 5.6667 8.3333
0 0 3.0000 0.5000
0 0 0 5.2500
Thanks again for your suuport.

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