question_answer 74)

 If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further prove that (AB)n = AnBn for all  

Answer:

. For n = 1.       L.H.S. = ABn = AB? = AB       R.H.S. = BbA = B?A = BA       But AB = BA        L.H.S. = R.H.S.       Let us suppose that given statement is true for n = k.             Now consider for n = k + 1       n = k + 1             [by associatively of matrix multiplication]             = (Bk A) B                      [Using (1)]             = Bk (AB)             = Bk (BA)                    ABk+1 = Bk+1 A,         which is true for n = k + 1.             2nd Part : To prove             (AB)n = An Bn             We shall prove this result also by the principle of Mathematical Induction.       For n = 1.             L.H.S. = (AB)? = AB             R.H.S. = An Bn = A?B? = AB             L.H.S. = R.H.S.        given statement is true for n = 1       Let us suppose that given statement is true for n = k.             Consider for n = k + 1       (AB)k+1 = (AB)k . (AB)1 = AkBk (AB)                         [Using (2)]                               [By associatively of  matrix multiplication]       = Ak (Bk+1A) = AkABk+1                                     which is true for n = k + 1.       Hence by PMI, given statement is true for n = k + 1.