Q. 495.0( 1 Vote )

# If AB = BA for any two square matrices, prove by mathematical induction that (AB)n = An Bn.

By principle of mathematical induction we say that if a statement P(n) is true for n = 1 and if we assume P(k) to be true for some random natural number k and usnig it if we prove P(k+1) to be true we can say that P(n) is true for all natural numbers.

We are given to prove that (AB)n = AnBn

Let P(n) be the statement : (AB)n = AnBn

Clearly, P(1): (AB)1 = A1B1

P(1) : AB = AB

P(1) is true

Let P(k) be true.

(AB)k = AkBk …(1)

Let’s take P(k+1) now:

(AB)k+1 = (AB)k(AB)

(AB)k+1 = AkBk(AB)

NOTE: As we know that Matrix multiplication is not commutative. So we can’t write directly that AkBk(AB) = Ak+1Bk+1

But we are given that AB = BA

(AB)k+1 = AkBk(AB)

(AB)k+1 = AkBk-1(BAB)

As AB = BA

(AB)k+1 = AkBk-1(ABB)

(AB)k+1 = AkBk-1(AB2)

(AB)k+1 = AkBk-2(BAB2)

(AB)k+1 = AkBk-2(ABB2)

(AB)k+1 = AkBk-2(AB3)

We observe that one power of B is decreasing while other is increasing. After certain repetitions decreasing power of B will become I

And at last step:

(AB)k+1 = AkI(ABk+1)

(AB)k+1 = AkABk+1

(AB)k+1 = Ak+1Bk+1

Thus P(k+1) is true when P(k) is true.

(AB)n = An Bn n N when AB = BA.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Triangular Matrices & operations on matrices58 mins
Determinants of Matrices of different order59 mins
Types of Matrices & Properties51 mins
Determining a determinant63 mins