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Question: Answered & Verified by Expert
If \(M(\alpha)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\); \(M(\beta)=\left[\begin{array}{ccc}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta\end{array}\right]\) then \([M(\alpha) M(\beta)]^{-1}\) is equal to -
MathematicsApplication of DerivativesBITSATBITSAT 2009
Options:
  • A \(\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
  • B \(\mathrm{M}(-\alpha) \mathrm{M}(-\beta)\)
  • C \(\mathrm{M}(-\beta) \mathrm{M}(-\alpha)\)
  • D \(-\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
Solution:
2690 Upvotes Verified Answer
The correct answer is: \(\mathrm{M}(-\beta) \mathrm{M}(-\alpha)\)
\(\begin{aligned}
& \left.[\mathrm{M}(\alpha) \mathrm{M}(\beta)]^{-1}=\mathrm{M}(\beta)^{-1} \mathrm{M}(\alpha)^{-1}\right] \\
& \operatorname{Now~M(\alpha )^{-1}}=\left[\begin{array}{ccc}
\cos \alpha & \sin \alpha & 0 \\
-\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\cos (-\alpha) & -\sin (-\alpha) & 0 \\
\sin (-\alpha) & \cos (-\alpha) & 0 \\
0 & 0 & 1
\end{array}\right]=\mathrm{M}(-\alpha) \\
& M(\beta)^{-1}=\left[\begin{array}{ccc}
\cos \beta & 0 & -\sin \beta \\
0 & 1 & 0 \\
\sin \beta & 0 & \cos \beta
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\cos (-\beta) & 0 & \sin (-\beta) \\
0 & 1 & 0 \\
-\sin (-\beta) & 0 & \cos (-\beta)
\end{array}\right]=\mathrm{M}(-\beta) \\
& {[\mathrm{M}(\alpha) \mathrm{M}(\beta)]^{-1}=\mathrm{M}(-\beta) \mathrm{M}(-\alpha)} \\
&
\end{aligned}\)

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