If  A = [1 a 0 1]  , then prove An = [1 na 0 1]  where n is any positive integer

We shall prove the result by using mathematical induction.

Step 1:

P(n): If A = [1 a 0 1]  , then prove An = [1 na 0 1] , n ∈ N

Step 2: Prove for n = 1

For n = 1

L.H.S = A1 = A = [1 a 0 1]

R.H.S = [1 a 0 1] = [1 a 0 1]

L.H.S = R.H.S

  ∴ P(n) is true for n = 1

Step 3: Assume P(k) to be true and then prove P(k+1) is true

Assume that P (k) is true

P(k) : If A= [1 a 0 1] , then Ak = [1 k 0 1]

We will have to prove that P(k +1) is true

P(k + 1) : If A= [1 a 0 1] , then Ak+1 = [1 k 0 1]

Taking L.H.S

  Ak+1

  = Ak  . A

  = [1 (k + 1)] [1 a 0 1 ]

  = [(1(1)  +  ka(0))¦(0(1)  +  1(0))       (1(a)  +  ka(1))¦(0(a)  +  1(1))]

  = [1     a +  ka1]

  = R.H.S

Thus P (k + 1) is true

∴ By the principal of mathematical induction , P(n) is true for n ∈ N

Hence, if A = [1 a 0 1]  , then prove An = [1 na 0 1] n ∈ N.