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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)`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)` |
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Answer» Correct Answer - A::B::C::D `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)` |
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