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.
Written on March 17, 2017, 2 p.m.