Page 372 - 35Linear Algebra
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                            Any value of µ will give a solution of the system, and any system can be written
                            in this form for some value of µ. Since there are multiple solutions, we can
                            also express them as a set:
                                                                          
                                                   x 1      2        −3         
                                                     x 2    =   1   + µ   0   µ ∈ R  .
                                                                                
                                                     x 3     0         1
                            Worked Examples of Gaussian Elimination
                            Let us consider that we are given two systems of equations that give rise to
                            the following two (augmented) matrices:
                                                                                
                                                  2  5  2  0    2        5  2     9
                                                 1 1   1  0    1      0  5    10 
                                                  1 4   1  0    1        0  3     6
                            and we want to find the solution to those systems. We will do so by doing
                            Gaussian elimination.
                               For the first matrix we have
                                                                                    
                                               2  5  2  0    2          1  1  1  0    1
                                                                  ∼
                                              1  1  1  0    1   R 1 ↔R 2   2  5  2  0  2 
                                               1  4  1  0    1          1  4  1  0    1
                                                                                      
                                                                        1  1  1  0    1
                                                          R 2 −2R 1 ;R 3 −R 1
                                                               ∼       0  3  0  0    0 
                                                                        0  3  0  0    0
                                                                                      
                                                                        1  1  1  0    1
                                                                   1
                                                                   3  R 2
                                                                   ∼   0  1  0  0    0 
                                                                        0  3  0  0    0
                                                                                      
                                                                        1  0  1  0    1
                                                          R 1 −R 2 ;R 3 −3R 2   0
                                                               ∼           1  0  0    0 
                                                                        0  0  0  0    0
                            1. We begin by interchanging the first two rows in order to get a 1 in the
                               upper-left hand corner and avoiding dealing with fractions.
                            2. Next we subtract row 1 from row 3 and twice from row 2 to get zeros in the
                               left-most column.
                            3. Then we scale row 2 to have a 1 in the eventual pivot.
                            4. Finally we subtract row 2 from row 1 and three times from row 2 to get it
                               into Reduced Row Echelon Form.
                            Therefore we can write x = 1 − λ, y = 0, z = λ and w = µ, or in vector form
                                                                        
                                                     x      1        −1        0
                                                                               0
                                                     y
                                                            0
                                                                 0      
                                                      =     + λ      + µ     .
                                                    z     0     1       0 
                                                     w      0         0        1
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