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x(1)=b(1)/L(1,1);
for k=2:n
x(k)=(b(k)-L(k,1:k-1)*x(1:k-1))/L(k,k);
end
x

Example 8.5
Solve the system of equations: UX = B, where U is an upper triangular matrix.
Solution: The matrix form of the problem becomes:

 U 11 U 12 U 13 L U 1 n   x   b 
   1   1 
 0 U 22 U 23 L U 2 n   x 2   b 2 
 M M O M M   M  =  M  (8.15)
     
 0 0 L U n 1 − U n 1   x n 1  b n 1
−
−
−
−
n 1
n
 0 0 0 L U   x   b 
 nn   n   n 
In this case, the solution of this system can also be directly obtained if we pro-
ceed iteratively, but this time in the backward order x , x , …, x , obtaining:
n
1
n–1
b
x = n
n
U
nn
b ( n−1 − U n−1 n x )
n
x =
n−1
U
n−1 n−1
(8.16)
M
 n 
k ∑
  b − U x  
j
kj
=+1
x =  jk 
k
U
kk
The corresponding script M-file is
U=[ ]; % enter the U matrix
b=[ ]; % enter the B column
n=length(b);
x=zeros(n,1);
x(n)=b(n)/U(n,n);
for k=n-1:-1:1

© 2001 by CRC Press LLC