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2.1 Row Echelon Form and Rank                                                       43





                                                    Modified Gaussian Elimination
                                       Suppose that U is the augmented matrix associated with the system
                                       after i − 1 elimination steps have been completed. To execute the i th
                                       step, proceed as follows:


                                       •   Moving from left to right in U , locate the first column that contains
                                           a nonzero entry on or below the i th  position—say it is U ∗j .

                                       •   The pivotal position for the i th  step is the (i, j) -position.

                                       •   If necessary, interchange the i th  row with a lower row to bring a
                                           nonzero number into the (i, j) -position, and then annihilate all en-
                                           tries below this pivot.

                                       •   If row U i∗ as well as all rows in U below U i∗ consist entirely of
                                           zeros, then the elimination process is completed.


                                        Illustrated below is the result of applying this modified version of Gaussian
                                    elimination to the matrix given in (2.1.1).
                   Example 2.1.1
                                    Problem: Apply modified Gaussian elimination to the following matrix and
                                    circle the pivot positions:
                                                                  12133
                                                                              
                                                                                .
                                                                 24044 
                                                                  12355
                                                           A = 
                                                                  24047
                                    Solution:

                                                 1
                                                                        1
                                                                                        
                                                     2133                   2   1    3   3
                                                2   4044            0    0     −2    −2 
                                                                                -2
                                                 1   2355               0   0   2    2   2
                                                               −→                       
                                                 2   4047               0   0  −2  −2    1
                                                     2   1    3   3             2    1   3    3
                                                 1                        1                 
                                                0   0      −2   −2       0   0     −2    −2 
                                                         -2
                                                                                    -2
                                                                                                .
                                                 0   0   0    0      −→    0  0    0   0
                                                                                              3
                                                                  0
                                           −→ 
                                                 0   0   0    0   3          0  0    0   0    0
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