Page 104 - Applied Numerical Methods Using MATLAB
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DECOMPOSITION (FACTORIZATION)  93
            The usage of a permutation matrix is exemplified by

                                                             
                        001         a 11  a 12  a 13  a 31  a 32  a 33
                  PA =    100     a 21  a 22  a 23    =    a 11  a 12  a 13    (2.4.2)
                        010         a 31  a 32  a 33  a 21  a 22  a 23

            which denotes switching the first and third rows followed by switching the second
            and third rows. An interesting and useful property of the permutation matrix is
            that its transpose agrees with its inverse.
                                    T            T    −1
                                  P P = I,     P = P                     (2.4.3)

            To take a close look at the LU decomposition, we consider a 3 × 3 nonsingular
            matrix:

                                                      
                                  1   0   0
               a 11  a 12  a 13                u 11  u 12  u 13
              a 21  a 22  a 23    =   l 21  1  0   0  u 22  u 23  
               a 31  a 32  a 33  l 31  l 32  1  0   0   u 33
                                                                     
               a 11  a 12  a 13   u 11       u 12            u 13
               a 21  a 22  a 23  l 21 u 11  l 21 u 12 + u 22  l 21 u 13 + u 23
                            =                                         (2.4.4)
               a 31  a 32  a 33  l 31 u 11  l 31 u 12 + l 32 u 22  l 31 u 13 + l 32 u 23 + u 33
            First, equating the first rows of both sides yields

                                  u 1n = a 1n ,  n = 1, 2, 3            (2.4.5a)

            Then, equating the second rows of both sides yields

                      a 21 = l 21 u 11 ,  a 22 = l 21 u 12 + u 22 ,a 23 = l 21 u 13 + u 23
            from which we can get

                l 21 = a 21 /u 11 ,  u 22 = a 21 − l 21 u 12 ,  u 23 = a 23 − l 21 u 13  (2.4.5b)

            Now, equating the third rows of both sides yields

               a 31 = l 31 u 11 ,  a 32 = l 31 u 12 + l 32 u 22 ,  a 33 = l 31 u 13 + u 32 u 23 + u 33

            from which we can get

             l 31 = a 31 /u 11 ,  l 32 = (a 32 − l 31 u 12 )/u 22 ,  u 33 = (a 33 − l 31 u 13 ) − l 32 u 23
                                                                        (2.4.5c)
              In order to put these formulas in one framework to generalize them for matri-
            ces having dimension greater than 3, we split this procedure into two steps
            and write the intermediate lower/upper triangular matrices into one matrix for
            compactness as
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