Let `A ,\ B` be two matrices such that they commute. Show that for any positive integer `n` , `A B^n=B^n A`

Let A,B be two matrices such that they commute.Show that for any positive integer n,AB^(n)=B^(n)A( ii) (AB)^(n)=A^(n)B^(n)

Let A,B be two matrices such that they commute.Show that for any positive integer n,(AB)^(n)=A^(n)B^(n)

Let A,B be two matrices such that they commute,then for any positive integer nAB^(n)=B^(n)A(AB)^(n)A^(n)B^(n) only (i) and (ii) correct both (i) and (ii) correct only (ii) is correct none of (i) and (ii) is correct

Let A and B are two matrices such that AB = BA, then for every n in N

Let A and B any two n xx n matrices such that the following conditions hold : AB= BA and there exist positive integers k and l such that A^(k)=I (the identity matrix) and B^(l)=0 (the zero matrix). Then-

If A and B are two matrices such that rank of A = m and rank of B = n, then

If A and B are square matrices of the same order and A is non-singular,then for a positive integer n,(A^(-1)BA)^(n) is equal to A^(-n)B^(n)A^(n) b.A^(n)B^(n)A^(-n) c.A^(-1)B^(n)A d.n(A^(-1)BA)

Statement 1: Let A ,B be two square matrices of the same order such that A B=B A ,A^m=O ,n dB^n=O for some positive integers m ,n , then there exists a positive integer r such that (A+B)^r=Odot Statement 2: If A B=B At h e n(A+B)^r can be expanded as binomial expansion.