Page 509 - เตรียมสอบครูผู้ช่วยคอมพิวเตอร์_compressed
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=    ( ) 06 −−
                                                      =    –6

                              − 9     3
                    5.3   A  =       
                                − 4  3 
                                       det (A)    =   [ 9(−  )×  3 ] [ 4(−−  )×  ] 3

                                                      =     –27 + 12

                                                      =      –15







                               1    0    3  
                                           
                    5.4    A  =   − 2  5  −1
                                           
                               2    4   −   3 
                              
                                            1   0    3   1   0
                                  det (A)       =      −  2  5  − 1 −  2  5

                                            2   4 −   3  2   4

                                                 3
                                                             1
                                                                                    (
                                                                                           3
                                                    =  [ ×51  ×( − ) ]+ 0(  ×( − ) × )  + 3 ×( − ) × )  − 2 ×5 × ) −
                                                                                4
                                                                 2
                                                                     (
                                                                            2
                                                              4( × (−  ) 1 ×  ) 1 − [ 3(−  )× (−  ) 2 ×  ] 0
                                                    =     –15 + 0 – 24 – 30 + 4 – 0
                                                    =      –65
                               2    1   −   2
                                           
                    5.5     A  =    3  −1  0
                                           
                               2    5    3  
                               
                                            
                                           2    1   −  2  2  1
                              det (A)           =       3 − 1  0  3 − 1
                                           2    5    3  2   5

                                                    =    2( × (−  ) 1 ×  ) 3 +  1 ( × 0×  ) 2 + [ 2(−  )× 3× 5 ] [2×−  (−  ) 1 × (−  ] ) 2

                                                            (5 0 2) (3 3 1)−  ×  ×  −  ×  ×

                                                    =     –6 + 0 – 30 – 4 – 0 – 9

                                                    =      –49
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