then \[A{{(\alpha ,\beta )}^{-1}}\]is equal to
A) \[A(-\alpha ,-\beta )\]
B) \[A(-\alpha ,\beta )\]
C) \[A(\alpha ,-\beta )\]
D) \[A(\alpha ,\beta )\]
Correct Answer: A
Solution :
[a] we have, \[A{{(\alpha ,\beta )}^{-1}}=\frac{1}{{{e}^{\beta }}}\left[ \begin{matrix} {{e}^{\beta }}\cos \alpha & -{{e}^{\beta }}\sin \alpha & 0 \\ {{e}^{\beta }}\sin \alpha & {{e}^{\beta }}\cos \alpha & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] \[=A(-\alpha ,-\beta )\]
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