question_answer 1)

Let  show that (aI + bA)n = anI + nan?1 bA, where I is the identity matrix of order 2 and

Answer:

We shall prove the result by mathematical induction.       For n = 1       L.H.S. = (aI + bA)? = aI + bA       R.H.S. = anI + nan?1bA                    given result is true for n =1.       Let us suppose that given statement is true for n = k.                  ?.(1)       Consider for n = k +1       (aI + bA)k+1 = (aI + bA)k. (aI + bA)       = (akI + kak?1 bA) (aI + bA)                         [Using (1)]       = ak+1I + akbIA + akba AI             + kb2 ak?1A2                         which is true for n = k + 1.       Hence by P.M.I., given statement is true for