InterviewSolution
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InterviewSolution
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If `A=[[cos alpha, -sin alpha] , [sin alpha, cos alpha]], B=[[cos2beta, sin 2beta] , [sin 2 beta, -cos2beta]]` where `0 lt beta lt pi/2` then prove that `BAB=A^(-1)` Also find the least positive value of `alpha` for which `BA^4B= A^(-1)` |
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Answer» Correct Answer - `alpha=(2pi)/3` `BAB=A^(-1)` `implies ABAB=I` `implies (AB)^(2)=I` Now, `AB=[(cos alpha,-sin alpha),(sin alpha,cos alpha)][(cos 2 beta,sin 2 beta),(sin 2 beta,-cos 2 beta)]` `=[(cos(alpha+2 beta),sin (alpha+2beta)),(sin (alpha+2beta),- cos (alpha +2beta))]` and `(AB)^(2)=[(cos (alpha+2beta),sin (alpha+2beta)),(sin (alpha+2beta),-cos (alpha+2beta))]xx[(cos(alpha+2beta),sin (alpha+2beta)),(sin (alpha+2beta),-cos (alpha+2beta))]` `=[(cos^(2)(alpha+2beta)+sin^(2) (alpha+2beta),0),(0,cos^(2) (alpha+2beta)+sin^(2) (alpha+2beta))]` `=[(1,0),(0,1)]` `=I` `BA^(4)B=A^(-1)` or `A^(4)B=B^(-1)A^(-1)=(AB)^(-1)=AB` or `A^(4)=A` (1) Now, `A^(2)=[(cos alpha,-sin alpha),(sin alpha,cos alpha)][(cos alpha,-sin alpha),(sin alpha,cos alpha)]` `=[(cos 2 alpha,-sin 2 alpha),(sin 2 alpha,cos 2 alpha)]` and `A^(4)=[(cos 4 alpha,-sin 4 alpha),(sin 4 alpha,cos 4 alpha)]` Hence, from Eq. (1), `[(cos 4 alpha,-sin 4 alpha),(sin 4 alpha,cos 4 alpha)]=[(cos alpha,-sin alpha),(sin alpha,cos alpha)]` or `4alpha=2pi+alpha` or `alpha=(2pi)/3` |
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