100% working code for Steepest Descent iterative algo for finding solution of Ax=b

%%100_working code for Steepest Descent iterative algo for finding

%%solution of Ax=b, where A is symmetric Positive definite

%%written by vikram bhayyA on 18-6-26 ~14:37-14:42

%%general.

clc

clear all

A=input('enter a Symmetric Positive Definite matrix:');

%A=[4,1,1;1,3,-1;1,-1,3]; %has to be symmetric Positve definite

[m,n]=size(A);

n

b=input('enter the output vector of size n,1:');

%b=[6,3 3]';

%x0=zeros(n,1);

x0=[1 -1 0]';

%x0=[0 0 0]' ; % initial guess

%initialize

r0=b-A*x0; %residual

if norm(r0)<0.01

display('solution of Ax=b is')

x0

else

%p0=r0; %search direction

k=1;

while(k)

alpha0=(r0'*r0)/(r0'*A*r0); %step size

x1=x0+alpha0*r0; %update x1

r1=b-(A*x1); % update r1

if norm(r1)<0.01

k=0;

else

k=k+1;

end

r0=r1;

x0=x1;

end

display('solution of Ax=b is')

x1

end

In the command window one can see this

enter a Symmetric Positive Definite matrix:[1 0 0;0 2 0; 0 0 3]

n =

     3

enter the output vector of size n,1:[1 4 12]'

x1 =

    1.0000
    1.9980
    3.9973

To crosscheck, if the solution is correct:
>> A*x1

ans =

    1.0000
    3.9960
   11.9919