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CHAP.  71                      STATE SPACE ANALYSIS



                  Thus, A  = 0 is the eigenvalues of A with multiplicity 3. By  Eq. (7.66) we have
                                                 eA' = b,I  + b,A + b2A2

                  where b,,  b,, and b2 are determined by setting  A  = 0 in  the following equations [App. A, Eqs.
                  (A. 59) and (A. 60 )I:

                                                 b, + b,A + b2h2 = eA'





                  Thus,




                  Hence,

                                                               t
                                                  eA' = I + tA + -A'
                                                                2
                 which is the same result obtained in  Prob. 7.44(b).

            7.46.  Show that

                                                     eA+B  = eAeB
                 provided A and B commute, that is, AB  = BA.
                     By Eq. (7.53)

















                                                    1      1      1      1
                                       =I+A+B+-A~+-AB+-BA+-B~+                  ...
                                                    2!     2      2     2!
                 and                        eAeB  - eA+B  = L , (AB - BA) +  .
                 Thus, if AB = BA, then
                                                     eA+B = eAeB


           7.47.  Consider the matrix
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