Page 365 - 35Linear Algebra
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365


                          (b) The rows of M span (kerL) ⊥
                          (c) First we put M in RREF:

                                                                                   
                                                1 2     1   3        1    2    1     3
                                                2 1 −1      2        0 −3 −3       −4
                                                                                   
                                      M    =                   ∼                   
                                                1 0     0 −1         0 −2 −1       −4
                                                                                   
                                                4 1 −1      0        0 −7 −5 −12
                                                1 0 −1                1 0 0     −1
                                                            1                   
                                                             3
                                                0 1     1             0 1 0
                                                            4                  8
                                                                                    .
                                                                    
                                                             3                   3 
                                           ∼                4   ∼              4
                                              0 0                   0 0 1 − 
                                                             3                    3
                                                        1 − 
                                                0 0     2 −  8        0 0 0       0
                                                             3
                              Hence
                                                                 8     4
                                              ker L = span{v 1 − v 2 + v 3 + v 4 }
                                                                 3     3
                              and
                                  imL = span{v 1 + 2v 2 + v 3 + 4v 4 , 2v 1 + v 2 + v 4 , v 1 − v 2 − v 4 } .
                              Thus dim ker L = 1 and dim imL = 3 so
                                           dim ker L + dim imL = 1 + 3 = 4 = dim V .

                     17.  (a)
                                                     
                                                      5 = 4a − 2b + c
                                                     
                                                        2 = a − b + c
                                                     
                                                      0 = a + b + c
                                                     
                                                     
                                                        3 = 4a + 2b + c .
                      (b,c,d)
                                                                                        
                                   4 −2 1 5            1    1    1 0          1 0     1 −1
                                   1 −1 1 2            0 −6 −3 5              0 1     0    1
                                                                                        
                                                 ∼                   ∼                  
                                  1   1 1 0         0 −2      0 2        0 0 −3      11  
                                   4   2 1 3           0 −2 −3 3              0 0 −3       3
                              The system has no solutions because c = −1 and c = −  11  is impossible.
                                                                                  3
                          (e) Let
                                                                         
                                                     4 −2 1                 5
                                                     1 −1 1                 2
                                                                         
                                              M =              and V =      .
                                                     1    1 1               0
                                                                         
                                                     4    2 1               3
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