Page 324 - 35Linear Algebra
P. 324

324                                                                       Sample First Midterm


                                  Use your result to solve the system


                                                      x   +   y   −   z   + 2w     = 7
                                                  
                                                  
                                                  
                                                  
                                                      x   + 3y + 2z + 2w           = 6
                                                  
                                                   −x − 3y − 4z + 6w              = 12
                                                  
                                                  
                                                              4y + 7z − 2w = −7
                                                  
                            Solutions
                               1. As an additional exercise, write out the row operations above the ∼ signs
                                  below.

                                                                                       3   11  
                                         1   3 0 4           1    3 0     4         1 0
                                                                                            5   5
                                                                                                   .
                                       1 −2 1 1  ∼  0 −5 1 −3  ∼  0 1 −                1 5  3 
                                                                                 
                                                                            
                                                      
                                      
                                                           
                                                                                                5
                                         2   1 1 5           0 −5 1 −3              0 0     0   0
                                  Solution set is
                                                               11
                                                                     3        
                                                    x         5         − 5         
                                                       y
                                                        =   3  + µ    1    : µ ∈ R  .
                                                                5          5
                                                       z        0          1
                                                                                    
                                                                                              11
                                                                                            
                                                                                              5
                                                                        3
                                  Geometrically this represents a line in R through the point   3  running
                                                                                              5
                                                                                              0
                                                         3  
                                                        −
                                                          5
                                                           .
                                                         1 
                                  parallel to the vector   5
                                                          1
                                               11                                 3  
                                                5                                  −  5
                                  The vector   3   is a particular solution and    1    is a homogeneous
                                                5                                    5
                                                0                                    1
                                  solution.
                                  As a double check note that
                                                      11
                                                                                  3     
                                       1    3 0                4          1    3 0       −         0
                                                      5                                    5
                                       1 −2 1          3  =    1   and    1 −2 1           1   =   0   .
                                                                                        
                                                       5                                   5
                                       2    1 1        0       5          2    1 1         1       0
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