InterviewSolution
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InterviewSolution
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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`. |
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Answer» We are given, `AB = BA` We have to prove, `AB^n = B^nA` When `n = 1`, `AB^1 = B^1A => AB = BA` So, given equation is true for `n =1` Let this equation is true for `n=k`. Then, `AB^k = B^kA` Now, we have to prove it is true for `n = k+1` that is `AB^(k+1) = B^(k+1)A` `L.H.S = AB^(k+1) = AB^k*B =B^kA*B = B^k(AB) = B^k(BA)= B^(k+1)A = R.H.S.` So, given equation is true for `n =k+1` if it is true for `n=k`. `:. AB^n = B^nA` Now, we will prove the second part. `(AB)^n = A^nB^n` When `n = 1`, `L.H.S. = (AB)^1 = AB` `R.H.S. = A^1B^1 = AB` So, given equation is true for `n =1` Let this equation is true for `n=k`. Then, `(AB)^k = A^kB^k` Now, we have to prove it is true for `n = k+1` that is `(AB)^(k+1) = A^(k+1)B^(k+1)` `L.H.S = (AB)^(k+1) = (AB)^(k)AB =(A^kB^k)AB = (A^kA)*(B^kB) = A^(k+1)B^(k+1) = R.H.S.` So, given equation is true for `n =k+1` if it is true for `n=k`. `:. (AB)^n = A^nB^n` |
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