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//example 4.1 (pg 149)
//2x1 + 3x2 - x3 = 5
//-2x2 - x3 = -7
//-5x3 = -15
A = [2 3 -1;0 -2 -1;0 0 -5]
b = [5 -7 -15]'
a=[A b]
[nA,mA]=size(A)
n=nA
//Backward substitution
x(3) = a(n,n+1)/a(n,n);
for i = n-1:-1:1
sumk=0;
for k=i+1:n
sumk=sumk+a(i,k)*x(k);
end
x(i)=(a(i,n+1)-sumk)/a(i,i);
end
disp(x)
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