change one column of matrix

I have A

A=[1,0,4,0,5,0;0,1,0,3,-1,0;0,4,0,0,0,8];
B=[1.2,-1,-2.1,-1.3,-1.4,1;-2.1,1.3,-1,-2,-3,-1.8;1.9,-1.2,3.1,2.7,2.5,-0.5];
index=2;

I want to add nonzero element in A to corresponding B element.

A(index,:)=B(index,A(index, :)~=0)+A(index, :);

result should be

A=[1,0,4,0,5,0;   0, 1+1.3,0,3-2,-1-3,0;     0,4,0,0,0,8];

Answer 1

[Walter Roberson]

mask = A(index, :) ~= 0;
A(index, mask) = A(index, mask) + B(index, mask) ;

[NA]

if index=[2;3;4]

A=[1,0,4,0,5,0;0,1,0,3,-1,0;0,4,0,0,0,8; 1,0,4,0,5,0;0,1,0,3,-1,0;0,4,0,0,0,8];
B=[1.2,-1,-2.1,-1.3,-1.4,1;-2.1,1.3,-1,-2,-3,-1.8;1.9,-1.2,3.1,2.7,2.5,-0.5 ;1.2,-1,-2.1,-1.3,-1.4,1;-2.1,1.3,-1,-2,-3,-1.8;1.9,-1.2,3.1,2.7,2.5,-0.5];
index=[2;3;4];

mask = A(index, :) ~= 0;
A(index, mask) = A(index, mask) + B(index, mask) ;

I have this error

The logical indices in position 2 contain a true value outside of the
array bounds.

[NA]

I use this one but it makes program slow, if the size of A is big. Is there any better way?

mask = A(index, :) ~= 0;
for i=1:size(mask,1)
A(index(i), mask(i,:)) = A(index(i), mask(i,:)) + B(index(i), mask(i,:)) ;
end

[Walter Roberson]

Andrei's solution looks good.

Answer 2

[Andrei Bobrov]

A(index,:) = A(index,:) + B(index,:).*(A(index,:) ~= 0);