If B, C are n rowed matrices and if A = B + C, BC = CB, C2 = O, then show that for every n ϵ N, An+1 = Bn(B + (n + 1)C).
Given A = B + C, BC = CB and C2 = O.
We need to prove that An+1 = Bn(B + (n + 1)C).
We will prove this result using the principle of mathematical induction.
Step 1: When n = 1, we have An+1 = A1+1
⇒ An+1 = B1(B + (1 + 1)C)
∴ An+1 = B(B + 2C)
For the given equation to be true for n = 1, An+1 must be equal to A2.
It is given that A = B + C and we know A2 = A × A.
⇒ A2 = (B + C)(B + C)
⇒ A2 = B(B + C) + C(B + C)
⇒ A2 = B2 + BC + CB + C2
However, BC = CB and C2 = O.
⇒ A2 = B2 + CB + CB + O
⇒ A2 = B2 + 2CB
∴ A2 = B(B + 2C)
Hence, An+1 = A2and the equation is true for n = 1.
Step 2: Let us assume the equation true for some n = k, where k is a positive integer.
⇒ Ak+1 = Bk(B + (k + 1)C)
To prove the given equation using mathematical induction, we have to show that Ak+2 = Bk+1(B + (k + 2)C).
We know Ak+2 = Ak+1 × A.
⇒ Ak+2 = [Bk(B + (k + 1)C)](B + C)
⇒ Ak+2 = [Bk+1 + (k + 1)BkC)](B + C)
⇒ Ak+2 = Bk+1(B + C) + (k + 1)BkC(B + C)
⇒ Ak+2 = Bk+1(B + C) + (k + 1)BkCB + (k + 1)BkC2
However, BC = CB and C2 = O.
⇒ Ak+2 = Bk+1(B + C) + (k + 1)BkBC + (k + 1)BkO
⇒ Ak+2 = Bk+1(B + C) + (k + 1)Bk+1C + O
⇒ Ak+2 = Bk+1(B + C) + Bk+1[(k + 1)C]
⇒ Ak+2 = Bk+1[(B + C) + (k + 1)C]
⇒ Ak+2 = Bk+1[B + (1 + k + 1)C]
∴ Ak+2 = Bk+1[B + (k + 2)C]
Hence, the equation is true for n = k + 1 under the assumption that it is true for n = k.
Therefore, by the principle of mathematical induction, the equation is true for all positive integer values of n.
Thus, An+1 = Bn(B + (n + 1)C) for every n ϵ N.