Example 5.  Use  PA = LU  factorization with pivoting to solve the linear system  .  

Solution 5.

Remark.  The  LU  factorization part does not depend on  B.  To emphasize this, we only enter the matrix  A  at this time.

Execute the  LUfactor  subroutine.  Note that the dimension of the system must be used, in this case n = 3.

Now print out the matrices  L,  U,  L.U,  L.U and  P.A.  (Remember we stored a copy of  A  in A0).

Now enter the matrix  B.  

Solve the system  AX = B for  X.

We are done.

Aside.  For curiosity, we can check to see the "in between" steps that were made in the solution.  First, we solved    for    using forward substitution.  This is just for fun

Second, we solved   for    using back substitution.

Finally, we can verify that the residual is  zero, i. e. use matrix multiplication to see if    AX - B  =  0.

Aside.  Suppose a new linear system  AX = B  is to be solved, where  A  is the same we used above, but  B  is a new vector.
Then we enter  B  and only invoke the  SolveLU  subroutine.

Solve the new system  AX = B for  X.

(c) John H. Mathews 2004