If I have an array of 4 dimensions say A=complex(​rand(2,2,2​,2),rand(2​,2,2,2)). If I need to calculate the inverse of this matrix, as defined beow , how should I do it?

I need to compute B(x,y,z,w) such that if I multiply the terms of A and B , I should get an identity matrix I(a,b,c,d):

       I=zeros(a,b,c,d);
           for a=1:2
             for b=1:2
               for c=1:2
                  for d=1:2
                      for i=1:2
                            for j=1:2
                               I(a,b,c,d)=A(a,b,i,j)*B(i,j,c,d)+I(a,b,c,d);
                            end
                      end
                  end
               end
            end
        end

The I(a,b,c,d)= 1 only when a=b=c=d

[Matt J]

It is puzzling that you are organizing your data in 4D form, when you appear to want to do simple 2D matrix algebra with it. Are you sure it wouldn't be better just to reshape your data into 2D form

A=reshape(A,4,4)

and keep it that way?

[vishav Rattu]

But if I reshape it, will I get the above B matrix that is required? That is, will B satisfy the above condition, that is I(a,b,c,d)=1 only if a=b=c=d? If I am able to get such a B, then I don't need to keep it in a 4d form.

Answer 1

[Matt J]

[a,b,c,d]=ndgrid(1:2);

I=reshape(a==b & b==c & c==d, 4,4);

A=reshape(A,4,4);

B=reshape( A\I , 2,2,2,2);

[vishav Rattu]

This does satisfiy the above required condition. Thanks!!!

Answer 2

[KSSV]

Are you looking for something like this?

A = rand(2,2,2,2) ;
B = zeros(2,2,2,2) ;

for i = 1:2
    for j = 1:2
        B(:,:,i,j) = inv(A(:,:,i,j)) ;
    end
end
% check 
for i = 1:2
    for j = 1:2
        A(:,:,i,j)*B(:,:,i,j) 
    end
end

[vishav Rattu]

This does not satisfy the above condition that I mentioned about I(a,b,c,d)