If A and B are square matrices of the same order such that AB = BA, th

1.	If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further, prove that (AB)n = AnBn for all n ∈ N.
Answer» (i) Let P(n) be the statement P(n) ⇒ Abⁿ = BⁿA P(1) ⇒ AB = BA ∴ P( 1) is true Let P(k) be true P(k) ⇒ AB^k = B^kA P(k +1) ⇒ AB^k+1 = A(B^kB) = (AB^k)B ∵ matrix multiplication is associative. ⇒ (B^kA) B ⇒ B^k (AB) = B^k (BA) = (B^kB) A = B^k+1 A Hence P(k + 1) is true. ∴ P(n) is true for all the values of n ∈ N (ii) Let P(n) be the statement P(n) ⇒ (AB)ⁿ = AⁿBⁿ P(1) = (AB)¹ = AB ∴ P(1) is true Let P(k) be true P(k) ⇒ (AB)^k = A^kB^kP(k+1) ⇒ (AB)^k+1 = (AB)^kAB = (A^kB^k) AB = A^k (B^kA)B = A^k (AB^k) B (∵ ABⁿ = BⁿA whenever Ab = BA) = A^k+1B^k+1 ∴ P (k+1) is true Hence P(n) is true for all the natures of n ∈ N

If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further, prove that (AB)n = AnBn for all n ∈ N.

Answer»

(i) Let P(n) be the statement

P(n) ⇒ Abⁿ = BⁿA

P(1) ⇒ AB = BA

∴ P( 1) is true

Let P(k) be true

P(k) ⇒ AB^k = B^kA

P(k +1) ⇒ AB^k+1

= A(B^kB) = (AB^k)B

∵ matrix multiplication is associative.

⇒ (B^kA) B ⇒ B^k (AB) = B^k (BA)

= (B^kB) A = B^k+1 A

Hence P(k + 1) is true.

∴ P(n) is true for all the values of n ∈ N

(ii) Let P(n) be the statement

P(n) ⇒ (AB)ⁿ = AⁿBⁿ

P(1) = (AB)¹ = AB

∴ P(1) is true

Let P(k) be true

P(k) ⇒ (AB)^k = A^kB^kP(k+1)

⇒ (AB)^k+1 = (AB)^kAB = (A^kB^k) AB

= A^k (B^kA)B = A^k (AB^k) B

(∵ ABⁿ = BⁿA whenever Ab = BA)

= A^k+1B^k+1

∴ P (k+1) is true

Hence P(n) is true for all the natures of n ∈ N

Discussion

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