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Solution of Simultaneous Equations with Matrices

               Example 4.17

               For the electric circuit of Figure 4.1, the mesh equations are

                                                    1 Ω        2 Ω        2 Ω


                                                +            9 Ω       9 Ω        4 Ω
                                                −
                                       V = 100 v   I 1        I 2        I 3

                                              Figure 4.1. Circuit for Example 4.17
                                                10I –  9I      =  100
                                                   1    2
                                               – 9I +  20I – 9I =   0                                  (4.56)
                                                        2
                                                  1
                                                             3
                                                    – 9I +  15I =   0
                                                       2
                                                             3
               Use the inverse matrix method to compute the values of the currents I I  and I,,  3 .
                                                                                      2
                                                                                   1
               Solution:
                                                                        1 –
               For this example, the matrix equation is RI =  V  or I =  R V , where

                                                  –
                                               10 9 0           100             I 1
                                          R =  – 9 20 9  ,   V =  0     and   I =  I 2
                                                     –
                                                   9
                                               0 – 15             0             I 3

               The next step is to find R  1 –  . This is found from the relation

                                                           1
                                                   R  1 –  =  ------------ adjR                        (4.57)
                                                         detR
               Therefore, we find the determinant and the adjoint of  . For this example, we find that
                                                                    R

                                                               219 135 81
                                          detR =  975   adjR =,  135 150 90                            (4.58)
                                                               81 90 119

               Then,
                                                                   219 135 81
                                                                1
                                                    1
                                            R  1 –  =  ------------adjR =  ---------  135 150 90
                                                  detR         975
                                                                    81 90 119
               and





               Numerical Analysis Using MATLAB® and Excel®, Third Edition                             4−27
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