I have a square matrix of ones and zeroes and I am trying to change all the ones to a user entered number. I am only successfully changing the first value in the matrix.

for i = matrix_A(1,:)

if i == 1

matrix_A(i) = user_input

end

end

Answer 1

[KSSV]

A = rand(3) ; 
A(A<0.5) = 0 ;
A(A>=0.5) = 1 ; 

idx = find(A==1) ; 
for i = 1:length(idx)
    [r,c] = ind2sub(size(A),idx(i)) ; 
    fprintf('Enter value for (%d,%d) position\n',r,c)   
    val  = input('Enter value:') ;
    A(idx) = val ;
end

If you have a fixed single value:

A = rand(3) ; 
A(A<0.5) = 0 ;
A(A>=0.5) = 1 ; 

idx = find(A==1) ; 
val  = input('Enter value:') ;
A(idx) = val ;

Answer 2

[DGM]

You can do this with logical indexing.

For example:

% test array
A = randi([0 1],5,5)

A = 5×5

1 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0 1

newvalue = 12; % value to substitute
A(A==1) = newvalue

A = 5×5

12 0 0 0 12 12 12 0 0 0 0 0 12 0 0 0 0 12 0 12 0 0 12 0 12

As to why your loop isn't working

for i = matrix_A(1,:) % i is always either 0 or 1
    if i == 1 % if it's 1
        matrix_A(i) = user_input; % assign a value to matrix_A(1)
    end
end

You could instead do something like this

for col = 1:size(matrix_A,2)
    if matrix_A(1,col) == 1
        matrix_A(1,col) = user_input;
    end
end

but as indicated above, it's entirely unnecessary to use loops.