Let `F(alpha)=[{:(cosalpha,-sinalpha,0),(sinalpha,cosalpha,0),(0,0,1):}]` and `G(beta)=[{:(cosbeta,0,sinbeta),(0,1,0),(-sinbeta,0,cosbeta):}]`. Show that `[F(alpha).G(beta)]^(-1)=G(-beta).F(-alpha)`.

Let A_(alpha)=[(cosalpha, -sinalpha,0),(sinalpha, cosalpha, 0),(0,0,1)] , then :

If F(alpha)=[(cosalpha, -sinalpha,0),(sinalpha, cosalpha, 0),(0,0,1)] and G(beta)=[(cosbeta, 0, sinbeta),(0, 1, 0),(-sinbeta, 0, cosbeta)], then [F(alpha)G(beta)]^-1 is equal to (A) F(-alpha)G(-beta) (B) G(-beta)F(-alpha0 (C) F(alpha^-1)G(beta^-1) (D) G(beta^-1)F(alpha^-1)

Let F(alpha)=[cos alpha-sin alpha0sin alpha cos alpha0001] and G(beta)=[cos beta0sin beta010-sin beta0cos beta] Show that [F(alpha)]^(-1)=F(-alpha)(ii)[G(beta)]^(-1)=G(-beta)( iii) [F(alpha)G(beta)]^(-1)=G(-beta)F(-alpha)

Inverse of the matrix {:[(cosalpha,-sinalpha,0),(sinalpha,cosalpha ,0),(0,0,1)]:} is

If A=[{:(sinalpha,-cosalpha,0),(cosalpha,sinalpha,0),(0,0,1):}] then A^(-1) is equal to

If A=[[cosalpha, -sinalpha, 0], [sinalpha, cosalpha, 0], [0, 0, 1]] , then (adjA)^(-1)=

If A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))] , then A(alpha, beta)^(-1) is equal to

If F(alpha)=[[cosalpha, -sinalpha, 0], [sinalpha, cosalpha, 0], [0, 0, 1]] , where alphainR , then (F(alpha))^(-1)=