Q. If $A(\alpha,\beta) = \ \begin{bmatrix} \cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & e^{\beta} \end{bmatrix}$then $A(\alpha,\beta)^{- 1}$ is equal to
A) $A( - \alpha, - \beta)$
B) $A( - \alpha,\beta)$
C) $A(\alpha, - \beta)$
D) $A(\alpha,\beta)$
Explanation
(a)
We have,
$A(\alpha,\beta)^{- 1} = \frac{1}{e^{\beta}}\begin{bmatrix}
e^{\beta}\cos\alpha & - e^{\beta}\sin\alpha & 0 \\
e^{\beta}\sin\alpha & e^{\beta}\cos\alpha & 0 \\
0 & 0 & 1
\end{bmatrix}$
$= A( - \alpha, - \beta)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.