Page 54 - 35Linear Algebra
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54                                                                 Systems of Linear Equations


                            Example 24 (Undoing A in Ax = b slowly, for A = M = ...)

                                                                                        
                                                         0 1 1      x                        7
                                                         2 0 0      y    =                   4
                                                                                        
                                                         0 0 1      z                        4
                                                                                   
                                             0 1 0       0 1 1      x            0 1 0       7
                                      ⇔     1 0 0     2 0 0        =      1 0 0    
                                                                    y
                                                                                             4
                                             0 0 1       0 0 1      z            0 0 1       4
                                                                                        
                                                         2 0 0      x                        4
                                      ⇔                 0 1 1        =                  
                                                                                             7
                                                                    y
                                                         0 0 1      z                        4
                                             1  0 0      2 0 0      x            1  0 0      4
                                                                                   
                                             2                                   2
                                      ⇔      0 1 0    0 1 1        =      0 1 0    
                                                                                             7
                                                                    y
                                             0 0 1       0 0 1      z            0 0 1       4
                                                                                        
                                                         1 0 0      x                        2
                                      ⇔                 0 1 1        =                  
                                                                    y
                                                                                             7
                                                         0 0 1      z                        4
                                                                                   
                                             1 0   0     1 0 0      x            1 0    0    2
                                      ⇔     0 1 −1    0 1 1        =      0 1 −1   
                                                                    y
                                                                                             7
                                             0 0   1     0 0 1      z            0 0    1    4
                                                                                       
                                                         1 0 0      x                       2
                                      ⇔                 0 1 0        =                   .
                                                                                            3
                                                                    y
                                                         0 0 1      z                       4
                            This is another way of thinking about Gaussian elimination which feels more
                            like elementary algebra in the sense that you “do something to both sides of
                            an equation” until you have a solution.


                            2.3.2    Recording EROs in (M|I )

                                                      −1 −1
                            Just as we put together 3 2     = 6 −1  to get a single thing to apply to both
                            sides of 6x = 12 to undo 6, we should put together multiple EROs to get
                            a single thing that undoes our matrix. To do this, augment by the identity
                            matrix (not just a single column) and then perform Gaussian elimination.
                            There is no need to write the EROs as systems of equations or as matrices
                            while doing this.




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