Back substitution

gistfile1.txt

R x = b  => x = R^-1 b

compute using back substitution: Say R has the form

    [a00 a01 a02]
R = [ 0  a11 a12]
    [ 0   0  a22]
    
then our system Rx = b is

    [a00 a01 a02] [x0]   [b0] 
    [ 0  a11 a12] [x1] = [b1]
    [ 0   0  a22] [x2]   [b2]
    
so:
                    a22*x2 = b2    => x2 = b2 / a22
           a11*x1 + a12*x2 = b1    => x1 = (b1 - a12*x2) / a11
  a00*x0 + a01*x1 + a02*x2 = b0    => x0 = (b0 - a01*x1 - a02*x2) / a00
  
and so forth. That's back substitution (because builds the solution from the
last element to the first).

The equivalent process for lower triangular matrices is called forward substitution
and proceeds in the other direction.