Page 330 - 35Linear Algebra
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330                                                                       Sample First Midterm


                                                                                    
                                                         1   0 0 0       1 1 −1      2
                                                        1   1 0 0     0 2     3   0 
                                                   =                                
                                                       −1 −1 1 0         0 0 −2      8
                                                                                    
                                                         0   2 0 1       0 0     1 −2
                                                                                   
                                                        1    0    0 0     1 1 −1 2
                                                        1    1    0 0     0 2     3 0
                                                                                   
                                                  =                                  .
                                                     −1 −1       1 0   0 0 −2 8    
                                                        0    2 −  1  1    0 0     0 2
                                                                  2
                                  To solve MX = V using M = LU we first solve LW = V whose augmented
                                  matrix reads

                                                                                           
                                               1    0    0   0   7          1 0    0   0   7
                                               1    1    0   0   6          0 1    0   0 −1
                                                                                           
                                                                     ∼                     
                                             −1 −1      1   0  12        0 0    1   0  18  
                                               0    2   − 1  1 −7           0 2 −   1  1 −7
                                                          2                         2
                                                                              
                                                              1 0 0 0       7
                                                              0 1 0 0 −1
                                                                              
                                                         ∼                      ,
                                                              0 0 1 0     18  
                                                              0 0 0 1       4
                                  from which we can read off W. Now we compute X by solving UX = W
                                  with the augmented matrix

                                                                                         
                                                1 1 −1 2       7          1 1 −1 2       7
                                                0 2    3   0 −1           0 2    3   0 −1
                                                                                         
                                                                   ∼                     
                                               0 0 −2 8       18       0 0 −2 0       2  
                                                0 0    0   2   4          0 0    0   1   2
                                                                                         
                                                  1 1 −1 2       7         1 0 0 0       1
                                                  0 2    0   0   2         0 1 0 0       1
                                                                                         
                                             ∼                       ∼                    .
                                                 0 0    1   0 −1         0 0 1 0 −1      
                                                  0 0    0   1   2         0 0 0 1       2
                                  So x = 1, y = 1, z = −1 and w = 2.













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