Question:
If AB = BA for any two square matrices, prove by mathematical induction that (AB)n = An Bn.
Solution:
Let P(n) : (AB)n = AnBn
So, P(1) : (AB)1 = A1B1
AB = AB
So, P(1) is true.
Let P(n) is true for some k ∈ N
Now,
$(A B)^{k+1}=(A B)^{k}(A B)$ (using (i))
$=A^{k} R^{k}(A B)$
$=A^{k} B^{k-1}(B A) B$
$=A^{k} B^{k-1}(A B) B \quad($ as given $A B=B A)$
$=A^{k} B^{k-1} A B^{2}$
$=A^{k} B^{k-2}(B A) B^{2}$
$=A^{k} B^{k-2} A B B^{2}$
$=A^{k} B^{k-2} A B^{3}$
..........
..........
$=A^{k+1} B^{k+1}$
Hence, P(1) is true and whenever P(k) is true P(k + 1) is true.
Thus, P(n) is true for all n ∈ N.