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

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(B) दिया गया है, $A(\alpha, \beta) = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \ -\sin \alpha & \cos \alpha & 0 \ 0 & 0 & e^\beta \end{bmatrix}$

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

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(B) दिया गया है, $A(\alpha, \beta) = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \ -\sin \alpha & \cos \alpha & 0 \ 0 & 0 & e^\beta \end{bmatrix}$.सबसे पहले, हम सारणिक ज्ञात करते हैं $|A(\alpha, \beta)| = e^\beta(\cos^2 \alpha + \sin^2 \alpha) = e^\beta$.इसके बाद, हम सहखंडज आव्यूह ज्ञात करते हैं। सहखंड इस प्रकार हैं:$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$अतः, $	ext{adj}(A(\alpha, \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}$.इसलिए, $[A(\alpha, \beta)]^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{e^\beta} \begin{bmatrix} e^\beta \cos \alpha & -e^\beta \sin \alpha & 0 \ e^\beta \sin \alpha & e^\beta \cos \alpha & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & e^{-\beta} \end{bmatrix}$.चूंकि $\cos(-\alpha) = \cos \alpha$ और $\sin(-\alpha) = -\sin \alpha$, यह आव्यूह $A(-\alpha, -\beta)$ के बराबर है।

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

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