How to do the calculation of these three different size matrices ?

Answer 1

[Amritesh]

First we have to create and initialise the arrays

C = [100,150]; 
B = [1,1;0.923077,0.615385;1,1;1,1;1,1];
D1 = [0,1,1,0,0,1.923077,3389.76;0,0,1,0,1,2,1793;0,1,1,0,0,1.923077,3389.76;0,0,1,0,1,2,1793];
D2 = [1,1,0,0,0,1.615385,5538.952;1,1,0,0,0,1.615385,5538.952;1,0,0,1,0,2,3601;1,0,0,1,0,2,3601];
A(:,:,1) = D1;
A(:,:,2) = D2;

Now,

F2 in first row of second column of D1 will correspond to A(1,2,1). So,

F2 = B[F2,C1]*C[D1]/A[1,6,D1] will be equivalent to

A(1,2,1) = B(2,1)*C(1)/A(1,6,1)

Similarly, you can calculate for other cells.

Hope this solves your issue.

[Akash Pal]

Yes ,It's only for the permanant value ,but i was trying to solve for the random values so that time I need to again change my all values manually .I want that it will be automatically calculated for the any value ,I dont need to give the everytime matrix position by myself

[Amritesh]

You can use nested loop like below for updating values

for k = 1:2
    for i = 1:4
        for j =1:5
            A(i,j,k) = B(j,k)*C(k)/A(i,6,k);
        end
    end
end

Hope this solves your problem.

[Akash Pal]

But where you have used the value od D1 and D2 matrix I mean the value of A(:,:,1) and the value of A(:,:,2)

[Amritesh]

Here, k = 1 signify D1 and k = 2 signify D2.

Just try dry run of loop using pen and paper.

You can accept the answer if you understood.

[Akash Pal]

Thank you .i understand it