Let `F(alpha)=[cosalpha-sinalpha0sinalphacosalpha0 0 0 1]` and `G(beta)=[cosbeta0sinbeta0 1 0-sinbeta0cosbeta]` . Show that `[F(alpha)]^(-1)=F(-alpha)` (ii) `[G(beta)]^(-1)=G(-beta)` (iii) `[F(alpha)G(beta)]^(-1)=G(-beta)F(-alpha)` .

Options

A. `(A(alpha))^(-1)=A(-alpha)`
B. `A(alpha)A(beta)=A(alpha+beta)`
C. `B(alpha)B(beta)=B(alpha+beta)`
D. `(B(beta))^(-1)=B(-beta)`

Answer

A, B, C, D

Explanation

`A(alpha).A(-alpha)=[{:(cosalpha,sinalpha,0),(sinalpha,cosalpha,0),(0,0,1):}][{:(cosalpha,sinalpha,0),(-sinalpha,cosalpha,0),(0,0,1):}]`
`=[{:(1,0,0),(0,1,0),(0,0,1):}]=I`
Also `|A(alpha)|ne0`
`because(A(alpha))^(-1)=A(-alpha)`
`B(beta)B(-beta)=[{:(cosbeta,0,sinbeta),(0,1,0),(-sinbeta,0,cosbeta):}][{:(cosbeta,0,-sinbeta),(0,1,0),(sinbeta,0,cosbeta):}]`
`=[{:(1,0,0),(0,1,0),(0,0,1):}]=I`
And `|B(beta)|ne0`
`implies(B(beta))^(-1)` exist
`(B(beta))^(-1)=B(beta)`
`A(alpha)A(beta)=[{:(cosalpha,-sinalpha,0),(sinalpha,cosalpha,0),(0,0,1):}][{:(cosbeta,-sinbeta,0),(sinbeta,cosbeta,0),(0,0,1):}]`
`=[{:(cos(alpha+beta),-sin(alpha+beta),0),(sin(alpha+beta),cos(alpha+beta),0),(0,0,1):}]=A(alpha+beta)`

`B(alpha)B(beta)=[{:(cosalpha,0,sinalpha),(0,1,0),(-sinalpha,0,cosalpha)][{:(cosbeta,0,sinbeta),(0,1,0),(-sinbeta,0,cosbeta):}]`
`=[{:(cos(alpha+beta),0,sin(alpha+beta)),(0,1,0),(-sin(alhpa+beta),0,cos(alpha+beta)):}]=B(alpha+beta)`