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SOLVING A SYSTEM OF LINEAR EQUATIONS  87
                 Without partial pivoting, the Gauss elimination procedure will give us a
                 quite good result.
                             14.6                    14.6
                          110  = 3.981071705534969e+014 10  + 1 = 3.981071705534979e+014
                 [A 1 b 1 ] =
                          0   6.018928294465030e+014    6.018928294465030e+014

                          1 3.981071705534969e+014 3.981071705534979e+014
                       →
                          0         1                  1
                        backward
                                    1
                        substitution
                       −−−−−−−−→ x =
                                    1
                 But, if we exchange the first row with the second row having the larger
                 element in the first column according to the strategy of partial pivoting, the
                 Gauss elimination procedure will give us a rather surprisingly bad result
                 as follows:
                    row       15                     15                      
                   swapping  110  = 1.000000000000000e+015 10 + 1 = 1.000000000000001e+015
                                 15
                                                                   15
                 −−−−−−−→  01 − 10 · 10 −14.6      1 + 10 −14.6  − (1 + 10 ) · 10 −14.6  
                   forward
                  elimination  =−1.5118864315095819   =−1.5118864315095821
                      backward
                     substitution     0.7500000000000000
                    −−−−−−−−→ x =
                                  1.0000000000000002
                    One might be happy to have the scaled partial pivoting scheme
                 [Eq. (2.2.20)], which does not switch the rows in this case, since the
                 relative magnitude (dominancy) of a 11 in the first row is greater than that
                                                            15
                 of a 21 in the second row, that is, 10 −14.6 /1 > 1/10 .
                                    15                15
                                  10     1          10 + 1
              (c) A 3 x = b 3  with A 3 =  −14.6  , b 3 =  −14.6        (E2.2.3)
                                   1   10          1 + 10
                 With any pivoting scheme, we don’t need to switch the rows, since the
                 relative magnitude as well as the absolute magnitude of a 11 in the first row
                 is greater than those of a 21 in the second row. Thus, the Gauss elimination
                 procedure will go as follows:
                         forward
                        elimination     1  1.000000000000000e-015  1.000000000000001e+000
                       −−−−−−−→
                                 0  1.511886431509582e-015  1.332267629550188e-015
                            backward
                           substitution    1.000000000000000
                           −−−−−−−−→ x =
                                        0.811955724875121
            (cf) Note that the coefficient matrix, A 3 is the same as would be obtained by applying
               the full pivoting scheme for A 2 to have the largest pivot element. This example
               implies that the Gauss elimination with full pivoting scheme may produce a worse
               result than would be obtained with scaled partial pivoting scheme. As a matter of
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