Page 49 - FISIKA MATEMATIKA I PERSAMAAN LINEAR DAN PENERAPAN DALAM RANGKAIAN LISTRIK_Neat
P. 49

11    −3      0       1     30
                                               [−3     6     −1] [   ] = [ 5 ]
                                                                    2
                                                 0    −1      3       3    −25


                            Selanjutnya,  akan  menyelesaikan  persamaan  linear  diatas


                                    dengan menggunakan metode crammer


                                    11    −3      0
                            |  | = [−3     6     −1 ] = 11 |   6    −1  | – (-3) | −3   −1 | +
                                     0    −1      3           −1     3            0      3
                                        3    6
                                    0  |        |
                                        0 −1
                                       = 11 (19) – ((-3)(-9)) + 0(0) = 209-27+0 = 160


                                      30     −3     0
                            |   | = [ 5       6    −1] = 30 |    6    −1  | – (-3) |  5    −1  | +
                               1
                                     −25 −1         3           −1     3           −25      3
                                          5      6
                                    0  |            |
                                        −25 −1
                                       = 30 (17) – ((-3)(-10)) + 0(145) = 480

                                     11      30      0
                            |   | = [−3      5      −1] = 11 |    5     −1  | – 30 | −3 −1    | +
                               2
                                      0     −25      3          −25      3           0     3
                                        −3      5
                                    0  |            |
                                         0    −25
                                       = 11 (-10) – 30(-9) + 0(75) = 160


                                     11     −3     30
                            |   | = [−3      6      5  ] = 11 |   6      5  | – (-3)
                               3
                                      0     −1 −25               −1    −25
                                     −3       5           −3      6
                                    |            | + 30  |          |
                                      0     −25            0     −1
                                       = 11 (-145) – ((-3)(75)) + 30(3) = -1280







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