# Problem in implementing Echelon Form and Solve System of Linear Equations

2 views (last 30 days)
Muhammad Usman on 23 Oct 2022
Answered: Nikhil on 26 Oct 2022
Here is my code:
A = [3 2 1;-4 5 7;1 0 -9];
b = [1;4;5];
% Solving System of Linear Equations by using built-in command for inverse
% method
x1 = A\b;
disp(x1);
-0.3636 1.3434 -0.5960
[m,n]=size(A);
x=zeros(n,1);
% Solving System of Linear Equations by using Echelon Form Method
% Matrix reduced to Echelon Form
for j=1:min(m,n)
x(j)=b(j)/A(j,j);
A(j,:) = A(j,:)/A(j,j);
for i = j+1:m
A(i,:)= A(i,:)- A(j,:)*A(i,j);
end
end
% Backward Substitution
for k=n:-1:1
if (A(k,k)==0)
error('Matrix is singular!');
end
b(1:k-1)=b(1:k-1)-A(1:k-1,k)*x(k);
end
disp(x)
0.2899 -1.4348 5.0000
But I am getting wrong result in implementing Echelon Form Method. Please help me to figure out the error. Possilbly the error lies in backward substitution, but didn't understand how to get rid of.

Nikhil on 26 Oct 2022
Hi Usman, the following code works for me. I first converted into upper triangular and did backward sub.
A = [3 2 1;-4 5 7;1 0 -9];
B = [1;4;5];
[m,n] = size(A);
x = zeros(m,1);
for i = 1:m-1
factor = A(i+1:m,i)/A(i,i);
% i = 1 => factors = A(2,1)/A(1,1) & A(3,1)/A(1,1) short note % A(2:3,1)/A(1,1)
A(i+1:m,:) = A(i+1:m,:) - factor*A(i,:);
% making non-diagonal elements zero
% in this case -4 = -4 - (-4/3)*3
% for first iteration A(2,1) AND A(3,1) become zero
% second iteration A(3,2) becomes zero
B(i+1:m,:) = B(i+1:m,:) - factor*B(i,:);
end
% now we have upper triangular matrix
% so z value = b(3)/a(3,3)
% and keep doing backward sub
x(m,:) = B(m,:)/A(m,m);
for i = m-1:-1:1
x(i,:) = (B(i,:) - A(i,i+1:m)*x(i+1:m,:))/A(i,i);
end
x

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