InterviewSolution
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InterviewSolution
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If `A=[[costheta,isintheta],[isintheta,costheta]],`then prove by principal of mathematical induction that`A^n=[[cosntheta, i sinn theta],[isin n theta, cos n theta]]`for all n`in` `NN`. |
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Answer» Here, it is given that, `A = [[cos theta, isintheta],[isintheta,cos theta]]`. We have to prove, `A^n = [[cos ntheta, isin ntheta],[isin ntheta,cos ntheta]]`. Now, for `n = 1`. `A^1 = [[cos 1theta, isin 1theta],[isin 1theta,cos 1theta]]`. `=>A = [[cos theta, isintheta],[isintheta,cos theta]]`, which is true. So, given equation is true for `n = 1`. Now, let given equation is true for `n = k`. Then, `A^k = [[cos ktheta, isin ktheta],[isin ktheta,cos ktheta]]`->(1).Now, we have to prove given equation is true for `n = k+1`. We have to prove, `A^(k+1) = [[cos (k+1)theta, isin (k+1)theta],[isin (k+1)theta,cos (k+1)theta]]` Now, `L.H.S. = A^(k+1) = A^k*A = [[cos ktheta, isin ktheta],[isin ktheta,cos ktheta]] [[cos theta, isintheta],[isintheta,cos theta]]` `= [[cos kthetacostheta + i^2sin kthetasin theta, icoskthetasin theta + isinkthetacos theta],[isin ktheta cos theta+ i cosktheta sin theta,i^2sin ktheta sin theta+cos k theta cos theta]] ` `= [[cos kthetacostheta - sin kthetasin theta, i(coskthetasin theta + sinkthetacos theta)],[i(sin ktheta cos theta+ cosktheta sin theta),-sin kthetasin theta+cos k theta cos theta]] ` `=[[cos (k+1)theta, isin (k+1)theta],[isin (k+1)theta,cos (k+1)theta]] = R.H.S.` Thus, our equation is true for `n = k+1`. `:.A^n = [[cos ntheta, isin ntheta],[isin ntheta,cos ntheta]]`. |
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