If `A =[[cos alpha,-sin alpha],[sinalpha, cos alpha]] , B= [[cos 2 beta, sin 2 beta],[sin beta ,-cos beta]]` where
`0 lt beta lt pi/2`, then prove that `BAB=A^(-1)`. Alsp, find the least
value of `alpha` of which `BA^(4) B= A^(-1)`

Answer

`alpha=(2pi)/3`

Explanation

`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`