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Let A and B are two matrices such that A...

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

A

`A^(n) B = BA^(n)`

B

`(AB)^(n) = A^(n)B^(n)`

C

`(A+B)^(n) = ""^(n)C_(0) A^(n) + ""^(n)C_(1) A^(n-1) B+...+ ""^(n) C_(n) B^(n) `

D

`A^(2n) - B ^(2n) = (A^(n)-B^(n) ) (A^(n)+B^(n))`

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
`because A^(2) B = A (AB) = A(BA) = (AB)A = (BA)A= BA^(2)`
Similarly, `A^(3) B = BA^(3) `
In general, `A^(n) B = BA^(n) , AA n ge 1 `
and `(A +B) ^(n) = ""^(n) C_(0)A^(n) + ""^(n)C_(1)A^(n-1) B +""^(n)C_(2) A^(n-2) B^(2) +...+""^(n) C_(n) B^(n)`
Also, `(A^(n) - B^(n)) (A^(n) + B^(n) ) = A^(n) A^(n) + A^(n) B^(n) - B^(n) A^(n) - B^(n) B^(n) `
`= A^(2n)-B^(2n) [ because AB = BA]`
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