Hello,
The issue with your code is not because there is a 0 in row 1 and column 2 of A. Key corrections in your code is as follows:
- Forward Elimination Corrected: Properly calculates the factor (k) for each element to be eliminated, ensuring correct elimination of elements below the pivot.
- Back Substitution Implemented: Solves for variables from the bottom up after converting (A) to an upper triangular form, ensuring accurate calculation of solution vector (x).
clc; clear all; close all;
A = [2 0 1 ; -2 4 1 ;-1 -1 3];
b = [8; 0; 2];
sz = size(A,1);
disp ([A b]);
% Forward elimination
for i = 1:sz-1
for j = i+1:sz
k = A(j,i)/A(i,i);
A(j,:) = A(j,:) - k * A(i,:);
b(j) = b(j) - k * b(i);
end
end
% Back substitution
x = zeros(sz,1); % Initialize solution vector
for i = sz:-1:1
x(i) = (b(i) - A(i,i+1:sz)*x(i+1:sz))/A(i,i);
end
disp(x);
