Write a program so that with help of 3 for-end loops we form a matrix C that will be equivalent to C= A*B - B*A

Let 2 matrices (A and B) be given in MatLab.Write a program so that with help of 3 for-end loops we form a matrix C that will be equivalent to C= A*B - B*A

[Jakub Devera]

Has A and B same dimensions? Is A and B two dimensional or three dimensional matrix (3 for loops required)? You could at least except copy - paste the homework assignment give some constraints or example data.

[Bruno Luong]

@Jakub Devera matrix is 2D, and naive multipling two matrices requires 3 nested loops.

[Jan]

This is a homework question. As usual, post, what you have tried so far and ask a specific question concern the remaining Matlab problem.

The actual problem is interesting. You can avoid to calculate A*B and B*A explicitly.

@Jakub Devera: If A*B and B*A is mathematically defined, A and B must be square matrices.

Answer 1

[Srija Kethiri]

Hi Iva,

From my understanding you want to to get a matrix C, where C = A*B - B*A. To get the matrix C it is possible only if A and B both are square matrices with same dimension.

The code for finding C is:

[c,d] = size(A); 
[e,f] = size(B); 
C = zeros(c,c); 
if(c==d && e==f && c==e)    
    for i=1:c 
        for j=1:c 
           sum1 =0; 
           sum2 =0; 
            for p = 1:c 
            sum1 = sum1 + A(i,p)*B(p,j); 
            sum2 = sum2 +B(i,p)*A(p,j);     
            C(i,j)  = sum1-sum2; 
            end 
        end 
    end 
end 

[John D'Errico]

Please don't answer obvious homework assignment questions, when no effort was made or shown. This does not help the student. It only teaches them to post the rest of their homework on Answers, and that hurts the site.

[Jan]

I agree with John.

Your code overwrites C(i,j) in each iteration of the inner loop. One variable would be enough for the accumulation.

Answer 2

[Matt J]

A=rand(4); B=rand(4); %synthetic input data


M={A,B;B,A};
mask=[1 1;1,-1];

for i=1:4
 M{i}=mask(i)*M{i};
end


for i=1:2
 M{i,1}=M{i,1}*M{i,2};
end

C=0;
for i=1:2
 C=C+M{i};
end

C-(A*B-B*A)%Check the result

ans = 4×4

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0