Metodo da substituição retroativa

clear all, clc, close all,
help gauss

%A=[[3 4 -5 1 -10];[0 1 1 -2 -1];[0 0 4 -5 3];[0 0 0 2 2]];N=4;     %EX 2.4
%A=[[3 4 -5 1 -10];[0 0 1 -2 0];[0 0 4 -5 3];[0 0 0 2 2]];N=4;      %EX 2.5
%A=[[3 4 -5 1 -10];[0 0 1 -2 -1];[0 0 4 -5 3];[0 0 0 2 2]];N=4;     %EX 2.6
%A=[[1 3 -2 7 0 -9 6 -1 6.25];[0 4 3 -1 8 6 -7 4 55.08];
%A=[0 0 7 4 2 -4 -8 2 -2.454];[0 0 0 -3 5 9 5 1 51.442];
%A=[0 0 0 0 2 -6 -4 8 0];[0 0 0 0 0 -5 0 3 -0.008];
%A=[0 0 0 0 0 0 -9 5 7.228];[0 0 0 0 0 0 0 6 24];N=8;                %EX 2.7



disp('MATRIZ AUMENTADA DO SISTEMA:')
disp(num2str(A));

N1=N+1;K=N-1;
X(N)=A(N,N1)/A(N,N);
for I=1:K
    L=N-I;
    X(L)=A(L,N1);
    M=L+1;
    for J=M:N
        X(L)=X(L)-A(L,J)*X(J);
    end
    X(L)=X(L)/A(L,L);
end

disp('   '); disp('VETOR SOLUCAO:'); disp(num2str(X)); disp('   ');