જો A = begin{bmatrix} 1 & tan frac{alpha}{2} | vedclass

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(D) આપેલ છે કે,$A = \begin{bmatrix} 1 & 	an \frac{\alpha}{2} \ -	an \frac{\alpha}{2} & 1 \end{bmatrix}$ અને $AB = I$. $AB = I$ હોવાથી,$B = A^{-1}$

Vedclass · Accepted Answer

$\cos^2 \frac{\alpha}{2} \cdot A^T$

Vedclass · Answer

(D) આપેલ છે કે,$A = \begin{bmatrix} 1 & 	an \frac{\alpha}{2} \ -	an \frac{\alpha}{2} & 1 \end{bmatrix}$ અને $AB = I$. $AB = I$ હોવાથી,$B = A^{-1}$ થાય. $A$ નો નિશ્ચાયક $|A| = (1)(1) - (	an \frac{\alpha}{2})(-	an \frac{\alpha}{2}) = 1 + 	an^2 \frac{\alpha}{2} = \sec^2 \frac{\alpha}{2}$ છે. $A$ નો એડજોઈન્ટ $	ext{adj}(A) = \begin{bmatrix} 1 & -	an \frac{\alpha}{2} \ 	an \frac{\alpha}{2} & 1 \end{bmatrix}$ છે. તેથી,$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{\sec^2 \frac{\alpha}{2}} \begin{bmatrix} 1 & -	an \frac{\alpha}{2} \ 	an \frac{\alpha}{2} & 1 \end{bmatrix}$. $\frac{1}{\sec^2 \frac{\alpha}{2}} = \cos^2 \frac{\alpha}{2}$ હોવાથી,$B = \cos^2 \frac{\alpha}{2} \begin{bmatrix} 1 & -	an \frac{\alpha}{2} \ 	an \frac{\alpha}{2} & 1 \end{bmatrix}$ મળે. અહીં $A^T = \begin{bmatrix} 1 & -	an \frac{\alpha}{2} \ 	an \frac{\alpha}{2} & 1 \end{bmatrix}$ છે. તેથી,$B = \cos^2 \frac{\alpha}{2} \cdot A^T$ થાય.

જો $A = \begin{bmatrix} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{bmatrix}$ અને $AB = I$ હોય,તો $B$ બરાબર શું થાય?

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