If A and B are square matrices of the same order such that `A B = B A`, then proveby induction that `A B^n=B^n A`. Further, prove that `(A B)^n=A^n B^n`for all `n in N`.

Answer

N/a

Explanation

First part
Given that , AB=BA .. (1)
To prove `AB^(n) =B^(n) A`
for n=1
`AB^(1)=B^(1)A`
`implies Ab=BA`
which si given ,
therefore , given statement is true for n=1.
Let given statement is true for n=K.
`therefore AB^(k)=B^(k)A.. .(2)`
for n=K+1,
`AB^(K+1)=A(b^(k)B)`
`=(AB^(k) )B`
(from the associative law of multiplication )
`=(B^(k) A)B `[from eq. (2)]
`=B^(k)(AB)`
(from the associative law of multiplication )
`=B^(k) (BA) `[from eq. (1)]
`=(B^(k)B)A`
(from the associative law of multiplication )
`=B^(K+1)A`
`implies ` Given statement is also true for n=K+1.
therefore , from the principle of mathematical inducation , `AB^(n)=B^(n)A` is true for all natural number n.
SEcond part
to Prove `(AB)^(n) =A^(n) B^(n)`
for n=1,
`(AB)^(1)=A^(1) B^(1) `
implies AB=AB
which is true
`therefore ` GIven statement is true for n=1.
Let given statement is true for n=k.
`therefore (AB)^(k)=A^(k)B^(k).. .(3)`
for n=K+1,
`(AB)^(K+1) =(AB)^(k)(AB) `
` =(A^(k)B^(k))(AB)` [from eq . (3)]
`=A^(k)(B^(k)A)B`
(from the associative law of multiplication )
`=A ^(k) (AB^(k))B `[from eq. (2) ]
`=(A^(k)A)(B^(k)B)`
(form the associative law of multiplation )
`=A^(k+1)B^(K+1)`
`implies ` Given statements is also true for n=K+1.
thereofore , from the principle of mathematical inducation ,`(AB)^(n)=A^(n) B^(n)` is true for all natural numbers n.
Hence proved .