Let `F(a)=[(cos a,-sin a,0),(sin a, cos a, 0),(0,0,1)] and G(B)=[(cos B,0,sin B),(0,1,0),(-sin B,0,cos B)] " Show that " [F(a).G(B)]^(-1)=G(-B).F(-a)`.

Explanation

We have
`F(a).F(-a)=[(cos a, -sin a,0),(sin a, cos a, 0),(0,0,1)] [(cos(-a), -sin (-a),0),(sin (-a), cos (-a),0),(0,0,1)]`
`=[(cosa ,-sin a,0),(sin a,cos a, 0),(0,0,1)][(cos a,sin a,0),(-sin a,cos a,0),(0,0,1)]`
`[(cos^(2)a+sin^(2)a, 0,0),(0, sin^(2)a+cos^(2)a,0),(0,0,1)] [(1,0,0),(0,1,0),(0,0,1)]=I`
`Thus, F(a).F(-a)=I rArr{F(a)}^(-1)=F(-a)`.
Similarly, `G(B). G(-B)=I rArr{G(B)}^(-1)=G(-B)`.
`:." "{F(a).G(B)}^(-1)={G(B)}^(-1).{F(a)}^(-1) ` [by reversal law]
`=G(-B).F(-a)`.
Hence,`{F(a).G(B)}^(-1)= G(-B). F(-a)`.