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 $|A(\alpha, \beta)| = e^\beta(\cos^2 \alpha + \sin^2 \alpha) = e^\beta$.ત્યારબાદ,આપણે સહઅવયવ શ્રેણિક શોધીએ. સહઅવયવો નીચે મુજબ છે:$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$આમ,$	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}$.તેથી,$[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}$.કારણ કે $\cos(-\alpha) = \cos \alpha$ અને $\sin(-\alpha) = -\sin \alpha$,આ શ્રેણિક $A(-\alpha, -\beta)$ ને સમાન છે.

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

Similar Questions

ધારો કે $A$ એ $4 \times 4$ શ્રેણિક છે અને $P$ તેનો એડજોઈન્ટ શ્રેણિક છે. જો $|P|=\left|\frac{A}{2}\right|$ હોય,તો $\left|A^{-1}\right|=$

જો $A = \begin{bmatrix} 4 & 2 \\ 3 & 4 \end{bmatrix}$ હોય,તો $|adj\,A|$ ની કિંમત શોધો.

શ્રેણિક $\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$ નો વ્યસ્ત શ્રેણિક શોધો.

જો $A = \begin{bmatrix} k & 5 & 2 \\ 2 & -k & 5 \\ 5 & 2 & -k \end{bmatrix}$ અને $\det A = 190$ હોય,તો $\operatorname{Adj} A = $

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