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2.6 Matrices  183




                                  DEFINITION 10       Let A be a square zero–one matrix and let r be a positive integer. The rth Boolean power of
                                                      A is the Boolean product of r factors of A. The rth Boolean product of A is denoted by A .
                                                                                                                                [r]
                                                      Hence

                                                         A [r]  = A   A   A   ···   A.

                                                                       r times
                                                      (This is well defined because the Boolean product of matrices is associative.) We also define
                                                      A [0]  to be I n .


                                                            ⎡         ⎤
                                                              0  0   1
                                      EXAMPLE 9      Let A =  ⎣ 1  0  0 ⎦ . Find A [n]  for all positive integers n.
                                                              1  1   0


                                                     Solution: We find that
                                                                       ⎡        ⎤
                                                                        1   1  0
                                                        A [2]  = A   A =  ⎣ 0  0  1 ⎦  .
                                                                        1   0  1

                                                     We also find that

                                                                         ⎡        ⎤                       ⎡        ⎤
                                                                          1   0  1                         1   1  1
                                                        A [3]  = A [2]    A =  ⎣ 1  1  0 ⎦  ,  A [4]  = A [3]    A =  ⎣ 1  0  1 ⎦ .
                                                                          1   1  1                         1   1  1

                                                    Additional computation shows that

                                                              ⎡         ⎤
                                                                1  1   1
                                                        A [5]  =  ⎣ 1  1  1 ⎦  .
                                                                1  1   1

                                                     The reader can now see that A [n]  = A [5]  for all positive integers n with n ≥ 5.  ▲

                                 Exercises



                                            ⎡           ⎤
                                             1  1   1  3                                         −1   0  5   6
                                                                                         b) A =                 ,
                                             2  0   4  6 .                                      −4   −35 −2
                                                        ⎦
                                   1. Let A = ⎣
                                             1  1   3  7
                                                                                                −3    9 −3   4
                                     a) What size is A?                                     B =   0  −2  −12    .
                                     b) What is the third column of A?
                                     c) What is the second row of A?                   3. Find AB if

                                     d) What is the element of A in the (3, 2)th position?  a) A =  2  1  , B =  0  4  .
                                               t
                                     e) What is A ?                                              32         1  3
                                                                                               ⎡      ⎤
                                   2. Find A + B, where                                          1  −1
                                           ⎡            ⎤                                                     3 −2   −1
                                              1    0   4                                 b) A = ⎣ 0  1 ⎦ , B =  1  0  2  .
                                     a) A = ⎣ −1   2   2 ⎦ ,                                     2   3
                                              0  −2  −3                                        ⎡  4  −3 ⎤

                                           ⎡            ⎤                                      ⎢ 3             −1    3 2   −2
                                             −1    3   5                                 c) A = ⎣  0  −1⎥        0  −1  4  −3  .
                                                                                                        ⎦ , B =
                                                                                                     −2
                                        B = ⎣ 2    2  −3 ⎦ .                                     −1    5
                                              2  −3    0
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