Page 235 -
P. 235

202                                 ALGEBRA

                          There exist various methods for the construction of LU-decompositions. In particular
                       if the following conditions hold:

                                                     a 11  a 12

                                          a 11 ≠ 0,           ≠ 0,  ... ,  det A ≠ 0,
                                                    a 21  a 22

                       then the elements of the desired matrices L and U can be calculated by the formulas
                                                        j–1
                                                ⎧
                                                ⎪

                                                ⎨  a ij –         for i ≥ j,
                                            l ij =         l is u sj
                                                        s=1
                                                   0              for i < j,
                                                ⎪
                                                ⎩
                                                 ⎪ 1
                                                 ⎧          i–1
                                                       a ij –           for i < j
                                                 ⎪
                                                 ⎨             l is u sj
                                            u ij =  l ii    s=1
                                                 ⎪ 1                    for i = j,
                                                 ⎪
                                                   0                    for i > j.
                                                 ⎩
                       5.5.2-3. Solutions of a square system with different right-hand sides.
                        ◦
                       1 . One often has to solve a system of linear equations with a given basic matrix A
                                                                                               (1)
                       and different right-hand sides. For instance, consider the systems AX (1)  = B , ... ,
                       AX (m)  = B (m) .These m systems can be regarded as a single matrix equation AX = B,
                       where X and B are matrices of size n × m whose columns coincide with X (j)  and B (j)
                       (j = 1, 2, ... , m).
                          Example 5. Suppose that we have to solve the equation AX = B with the given basic matrix A and
                       different right-hand sides:
                                                                  7
                                            (  1  2  –3  )       ( )         (  10  )
                                         A =   3  –2  1  ,  B  (1)  =  1  ,  B  (2)  =  6  .
                                              –2  1   3           5            –5
                       Using the Gauss-Jordan procedure, we obtain
                         1  2 –3   710       12 –3     7  10       10 –1/2    2  4       10 0   34
                       ⎛               ⎞   ⎛                 ⎞   ⎛                 ⎞   ⎛           ⎞
                       ⎝ 3  –21    16 ⎠ → ⎝ 0 –810    –20 –24 ⎠ → ⎝ 01 –5/4  5/23 ⎠ → ⎝ 01 0    53 ⎠ .
                         –21   3   5 –5      05 –3    19 15        00 13/4   1/32 0      00 1   20
                       Therefore,
                                                          3
                                                                      4
                                                        ( )         ( )
                                                    (1)          (2)
                                                   X  =   5  ,  X  =  3  .
                                                          2           0
                       2 .If B = I,where I is the identity matrix of size n × n, then the solution of the matrix
                        ◦
                                                                   –1
                       equation AX = I coincides with the matrix X = A .
                          Example 6. Find the inverse of the matrix
                                                            2  1 0
                                                          (        )
                                                       A =  –307     .
                                                            –541
                       Let us transform the augmented matrix of the system, using the Gauss-Jordan method. We get
                         ⎛               ⎞   ⎛  1 1/2  0  1/2 00  ⎞
                            21 0 100
                           –30 7 010            0 3/2  7  3/2 10
                         ⎝               ⎠ → ⎝                   ⎠ →
                           –54 –1 001           0 13/2 –1 5/2 01
                                                10 –7/3   0  –1/3 0       10 0 14/47 –1/94 –7/94
                                             ⎛                     ⎞    ⎛                        ⎞
                                                01 14/3   1  2/3  0       01 0 19/47 1/47   7/47 ⎠ .
                                           → ⎝                     ⎠ → ⎝
                                                00 –94/3 –4 –13/3 1       00 1   6/47 13/94 –3/94
   230   231   232   233   234   235   236   237   238   239   240