InterviewSolution
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InterviewSolution
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Let `A ,B`be two matrices such that they commute. Show that for any positiveinteger `n ,`(i) `A B^n=B^n A`(ii) `(A B)^n=A^nB^n` |
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Answer» Let `P(n):(AB)^(n)=A^(n)B^(n)` `therefore P(1):(AB)^(1)=A^(1)B^(1)rArrAB=AB` So P(1) is true Now, `P(k):(AB)^(k)rArrA^(k)B^(k) ,K in N` So, P(K) is true, whenever `P(k+1)` is true `therefore p(k+1:AB)^(k+1)=A^(k+1)B^(k+1)` `rArr AB^(k),AB^(1)` `rArr A^(k)B^(k).BArArrA^(k)B^(k+1)A` `rArr A^(k).A.B^(k+1)rArrA^(k+1)B^(k+1)` `rArr (A.B)^(k+1)=A^(k+1)B^(k+1)` So, P(k+1) is true for all `n in N` , whenever P(k) is true. By mathematical induction (AB)=`A^(n)B^(n)` is true for all `n in N`. |
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