Backward dan Forward Substitution

Source code backward dan forward substitution

//Forward Substitution Ly=B
    y(1,1) = b(1)/l(1,1);
    for i = 2:n
        y(i,1)=(b(i)-l(i,1:i-1)*y(1:i-1,1))./l(i,i);
    end 

//Backward substitution Ux=y
    x(n,1) = y(n)/u(n,n);
    for i = n-1:-1:1
        x(i,1)=(y(i)-u(i,i+1:n)*x(i+1:n,1))./u(i,i);
    end 

    if x==0 then
        printf('Trivial atau penyelesaian nol\n')
    end

Sumber: https://www.youtube.com/watch?v=4mOFl6VvH-0

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About Izzuddin Fathin Azhar

A young man with awesome spirit.