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)`

Answer

`alpha =(2pi)/3`

Explanation

`because BAB=A^(-1)`
`rArr ABAB= I`
`rArr (AB)^(2) = I`
Now, `AB= [[cos (alpha+2beta),sin (alpha+2beta)],[sin(alpha+2beta),-cos(alpha + 2beta)]]`
and `(AB)^(2) = (AB) (AB) = [[1,0],[0,1]]= I [because (AB)(AB)=I]`
Also, `BA^(4)B=A^(-1)`
or `A^(4) B= B^(-1) A^(-1) =(AB)^(-1) = AB`
or`A^(4) = A " "...(i)`
Now, `A^(2) = [[cos alpha,-sin alpha],[sin alpha,cos alpha]][[cos alpha,-sin alpha],[sin alpha,cos alpha]]`
`=[[cos 2alpha,-sin 2alpha],[sin 2alpha,cos 2alpha]]`
Similarly, `A^(4)=[[cos 4alpha,-sin 4alpha],[sin 4alpha,cos4alpha]]`
Hence, from Eq. (i)
`[[cos 4alpha,-sin 4alpha],[sin 4alpha,cos4alpha]]=[[cos alpha,-sin alpha],[sin alpha,cos alpha]]`
or `4 alpha = 2pi + alpha`
` therefore alpha = (2pi)/3`