If F(x)=cosx-sinx0sinxcosx0001 and G(x)=cosx0sinx010-sinx0cosx then F(x)G(x)-1=

F(x)=cos x-sin x0sin xcos x0001adj F(x)=cos xsin x0-sin xcos x0001|F(x)|=1F(x)-1=F(-x)G(x)=cos x0sin x010-sin x0cos xadj G(x)=cos x0-sin x010sin x0cos x =G(-x)|G(x)|=1Gx-1=G(-x)F(x) G(x)-1=G(x)-1 F(x)-1= G(-x) F(-x)

If F(x)=cosx-sinx0sinxcosx0001 and G(x)=cosx0sinx010-sinx0cosx then F(x)G(x)-1=

F(x)=cos x-sin x0sin xcos x0001adj F(x)=cos xsin x0-sin xcos x0001|F(x)|=1F(x)-1=F(-x)G(x)=cos x0sin x010-sin x0cos xadj G(x)=cos x0-sin x010sin x0cos x =G(-x)|G(x)|=1Gx-1=G(-x)F(x) G(x)-1=G(x)-1 F(x)-1= G(-x) F(-x)

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If F(x) $= [\begin{matrix} cosx & - sinx & 0 \\ sinx & cosx & 0 \\ 0 & 0 & 1 \end{matrix}] and G (x) = [\begin{matrix} cosx & 0 & sinx \\ 0 & 1 & 0 \\ - sinx & 0 & cosx \end{matrix}] then {[F (x) G (x)]}^{- 1} =$

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G(–x) F(–x)

F(-x)G(-x)

${\{F (x^{- 1})\}}^{} {\{G (x^{- 1})\}}^{}$

$G (x^{- 1}) F (x^{- 1})$

answer is A.

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Detailed Solution

$\begin{array}{rcl} F (x) & = & [\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}] \\ ad j F (x) & = & [\begin{matrix} \cos x & \sin x & 0 \\ - \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}] \\ | F (x) | & = & 1 \\ {(F (x))}^{- 1} & = & F (- x) \end{array}$

$\begin{array}{rcl} G (x) & = & [\begin{matrix} \cos x & 0 & \sin x \\ 0 & 1 & 0 \\ - \sin x & 0 & \cos x \end{matrix}] \\ ad j G (x) & = & [\begin{matrix} \cos x & 0 & - \sin x \\ 0 & 1 & 0 \\ \sin x & 0 & \cos x \end{matrix}] \\ = & G (- x) \\ | G (x) | & = & 1 \\ {(G (x))}^{- 1} & = & G (- x) \\ {(F (x) G (x))}^{- 1} & = & {(G (x))}^{- 1} {(F (x))}^{- 1} \\ = & G (- x) F (- x) \end{array}$