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Answer:

Hence proved   $[G(\beta)]^{-1}=G(-\beta)$

Hint:

Here, we use basic concept of determinant and inverse of matrix

$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$

Given:

$F(\alpha)=\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right], G(\beta)=\left[\begin{array}{ccc} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{array}\right]$

Solution:

Let’s find  $|G(\beta)|$

$|G(\beta)|=\cos ^{2} \beta+\sin ^{2} \beta=1$

Cofactor of A

\begin{aligned} &C_{11}=\cos \beta, C_{21}=0, C_{31}=\sin \beta \\ &C_{12}=0, C_{22}=1, C_{32}=0 \\ &C_{13}=\sin \beta, C_{23}=0, C_{33}=\cos \beta \end{aligned}

$\operatorname{Adj}(G(\beta))=\left[\begin{array}{ccc} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{array}\right]^{T}$

\begin{aligned} &\operatorname{Adj}(G(\beta))=\left[\begin{array}{ccc} \cos \beta & 0 & -\sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{array}\right] \\\\ &{[G(\beta)]^{-1}=\frac{1}{|G(\beta)|} \times \operatorname{Adj}(G(\beta))} \end{aligned}

\begin{aligned} &{[G(\beta)]^{-1}=\frac{1}{1}\left[\begin{array}{ccc} \cos \beta & 0 & -\sin \beta \\ 0 & 1 & 1 \\ \sin \beta & 0 & \cos \beta \end{array}\right]} \\\\ &G(-\beta)=\left[\begin{array}{ccc} \cos (-\beta) & 0 & \sin (-\beta) \\ 0 & 1 & 0 \\ \sin (-\beta) & 0 & \cos (-\beta) \end{array}\right]=\left[\begin{array}{ccc} \cos \beta & 0 & -\sin \beta \\ 0 & 1 & 0 \\ \sin \beta & 0 & \cos \beta \end{array}\right] \end{aligned}

Hence

$[G(\beta)]^{-1}=G(-\beta)$

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