begin{aligned} & A(alpha, beta)=left[begin{ar | vedclass

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(B) આપેલ છે,$A(\alpha, \beta) = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \ -\sin \alpha & \cos \alpha & 0 \ 0 & 0 & e^\beta \end{bmatrix}$.પ્રથ

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$A(-\alpha, -\beta)$

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(B) આપેલ છે,$A(\alpha, \beta) = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \ -\sin \alpha & \cos \alpha & 0 \ 0 & 0 & e^\beta \end{bmatrix}$.પ્રથમ,આપણે નિશ્ચાયક શોધીએ $|A(\alpha, \beta)| = e^\beta (\cos^2 \alpha + \sin^2 \alpha) = e^\beta$.આગળ,આપણે સહઅવયવ શ્રેણિક $C$ શોધીએ:$C_{11} = e^\beta \cos \alpha, C_{12} = e^\beta \sin \alpha, C_{13} = 0$$C_{21} = -e^\beta \sin \alpha, C_{22} = e^\beta \cos \alpha, C_{23} = 0$$C_{31} = 0, C_{32} = 0, C_{33} = \cos^2 \alpha + \sin^2 \alpha = 1$તેથી,$	ext{adj}(A(\alpha, \beta)) = C^T = \begin{bmatrix} e^\beta \cos \alpha & -e^\beta \sin \alpha & 0 \ e^\beta \sin \alpha & e^\beta \cos \alpha & 0 \ 0 & 0 & 1 \end{bmatrix}$.પછી,$[A(\alpha, \beta)]^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{e^\beta} \begin{bmatrix} e^\beta \cos \alpha & -e^\beta \sin \alpha & 0 \ e^\beta \sin \alpha & e^\beta \cos \alpha & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & e^{-\beta} \end{bmatrix}$.આને મૂળ શ્રેણિક $A(\alpha, \beta)$ સાથે સરખાવતા,આપણે જોઈ શકીએ છીએ કે $\cos \alpha = \cos(-\alpha)$,$-\sin \alpha = \sin(-\alpha)$,અને $e^{-\beta}$ એ ત્રીજા વિકર્ણ ઘટકને અનુરૂપ છે. તેથી,પરિણામ $A(-\alpha, -\beta)$ છે.

$\begin{aligned} & A(\alpha, \beta)=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^\beta\end{array}\right] \\ & \Rightarrow[A(\alpha, \beta)]^{-1}=\end{aligned}$

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