Page 325 - 35Linear Algebra
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325


                      2.  (a) The augmented matrix

                                                     1    0 −1     2 −1
                                                                         
                                                     1    1    1 −1     2
                                                                         
                                                                         
                                                                         
                                                   0 −1 −2        3 −3 
                                                     5    2 −1     4    1
                              encodes the system of equations.
                          (b) Again, write out the row operations as an additional exercise.
                              The above augmented matrix is row equivalent to
                                                                                     
                                       1    0 −1      2 −1          1 0 −1       2 −1
                                       0    1    2 −3      3        0 1     2 −3      3
                                                                                     
                                                                                     
                                     0 −1 −2                     0 0      0    0
                                                             ∼                       
                                                      3 −3                           0 
                                       0    2    4 −6      6        0 0     0    0    0
                              which is in reduced row echelon form.
                          (c) Solution set is

                                                                               
                                              −1           1          −2
                                                                                     
                                                                                     
                                                                                     
                                              3        −2         3 
                                       X =        + µ 1      + µ 2      : µ 1 , µ 2 ∈ R  .
                                              0         1         0 
                                                                                     
                                                                                     
                                               0           0           1
                                                                                     
                                                  
                                                −1
                                                  3
                                                  
                          (d) The vector X 0 =      is a particular solution and the vectors
                                                  0
                                                  
                                                  0
                                                                         
                                                         1               −2
                                                       −2               3 
                                                 Y 1 =      and Y 2 =    
                                                         1                 0
                                                                         
                                                         0                 1
                              are homogeneous solutions. They obey
                                               MX = V ,       MY 1 = 0 = MY 2 .
                              where
                                                                               
                                                 1    0 −1      2               −1
                                                 1    1    1 −1                  2
                                                                               
                                         M =                       and V =       .
                                                 0 −1 −2        3               −3
                                                                               
                                                 5    2 −1      4                1
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