I don't understand the condition from FOR row 16 to row 26

syms x y z lambda
dist = x.^2+(y-1).^2+z.^2;       % distance of an abitary point (x,y,z) from the point(0,1,0) 
S = z.^2-(x.*y)-1 == 0;          % surface constraint  z^2 = xy+1    ------>   z^2-xy-1 == 0
L = dist + lambda * lhs(S);      % Langrange function
dL_dx = diff(L,x) == 0;          % derivative of L with respect to x
dL_dy = diff(L,y) == 0;          % derivative of L with respect to y
dL_dz = diff(L,z) == 0;          % derivative of L with respect to z
dL_dlambda = diff(L,lambda) == 0;      % derivative of L with respect to lambda
system = [dL_dx; dL_dy; dL_dz; dL_dlambda];    % build the system of equations
[x_val, y_val, z_val,lambda_val] = solve(system, [x y z lambda], 'Real', true);    % solve the system of equations and display the results 
results_numeric = double([x_val, y_val, z_val]);    % show results in a vector of data type double
[rows columns]=size(results_numeric);         % getting number of rows and columns
fprintf('points on the surface S that are closest to the point (0, 1, 0) are :\n');
fprintf('   x         y         z\n');
fprintf('--------------------------\n');
for i=1:rows
    fprintf('(');
    for j=1:columns
        if j < columns
        fprintf("%.4f , ",results_numeric(i,j));
        else 
        fprintf("%.4f",results_numeric(i,j));
        end
    end
    fprintf(")\n");
end

Answer 1

[David Hill]

The whole nested for-loop is just used to display results_numeric. If you don't care about displaying, you can delete the whole thing.

results_numeric%contains all the information within the matrix