Matrix Dimensions Must Agree Error, Error using +

This is Newton Raphsons Method for systems. Row 1 and 2 in f= [x1 etc is 2 equations for 2 circles, and I want to find 2 intersections between these circles. I have already plotted and got guess 1 = [51 28] and 2 [41 47]. I need to find 2 x1, and x2 values who solves the system.

if true
format short e, format compact
disp('       x          f(x)              h=J\f(x)');
x=[51 28]';    %Startvektor
h=x;
iter=1;
while ( (norm(h,inf) > 1.0e-10*norm(x,inf)) & (iter < 20)),
f =[x(1).^2 + x(2).^2 -2*93*x(1) -2*63*x(2)+ 9.5820e+03 
    x(1).^2 + x(2).^2 -12*x(1) -32*x(2)- 1.8332e+03 ];
%Jacobian
J=[2*x(1)-2*93  2*x(2)-2*63 
     2*x(1)-12   2*x(2)-32]; 
h =-J\f;
disp(iter)
disp([x f h])
x=x+h; iter=iter+1;
end
format long
x

    % code
  end

Error using + Matrix dimensions must agree. Error in uppg6aNRAP (line 20) x=x+h; iter=iter+1;

PLS HELP! I'm a beginner

Answer 1

[the cyclist]

If you put a breakpoint at line 18, temporarily halting execution there, you will see that x is a 2x1 vector, and h is a 2x3 array. (Is that what you expect?) Arrays have to be the same shape to add them.

[Josefine Jonsson]

It's supposed to be 2 lines in f, one for each circle equation. I initially took this code from a newton raphson for system with a f = 2x3. Matris, but deleted line 3 and wrote my

X^2 +y^2 - f1

X^2 + y^2 - f2

2x2 matris for the 2 circle equations instead

[David Young]

See my answer below for why this didn't happen, and how to fix it.

Answer 2

[David Young]

It's necessary to put spaces in the expression for f so that + and - are interpreted as binary operators rather than unary signs. Like this:

f =[x(1).^2 + x(2).^2 - 2*93*x(1) - 2*63*x(2) + 9.5820e+03 
  x(1).^2 + x(2).^2 - 12*x(1) - 32*x(2) - 1.8332e+03 ];