If M(α)= [ cos α -sin α 0 sin α cos α 0 0 0 1 ] ; M(β)= [ cos β 0 sin β 0 1 0 -sin β 0 cos β ] then [M(α) M(β)]^-1 is equal to -

Correct answer: (C) M(-β) M(-α). .[M(α) M(β)]^-1=M(β)^-1 M(α)^-1 ] operatornameNow~M(α )^-1= [ cos α sin α 0 -sin α cos α 0 0 0 1 ] = [ cos (-α) -sin (-α) 0 sin (-α) cos (-α) 0 0 0 1 ]=M(-α) M(β)^-1= [ cos β 0 -sin β 0 1 0 sin β 0 cos β ] = [ cos (-β) 0 sin (-β) 0 1 0 -sin (-β) 0 cos (-β) ]=M(-β) [M(α) M(β)]^-1=M(-β) M(-α)

Q64 / 125 · BITSAT 2009 · Mathematics

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 -

A
\(\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
B
\(\mathrm{M}(-\alpha) \mathrm{M}(-\beta)\)
Correct answer
C
\(\mathrm{M}(-\beta) \mathrm{M}(-\alpha)\)
D
\(-\mathrm{M}(\beta) \mathrm{M}(\alpha)\)

Solution

\(\begin{aligned} & \left.[\mathrm{M}(\alpha) \mathrm{M}(\beta)]^{-1}=\mathrm{M}(\beta)^{-1} \mathrm{M}(\alpha)^{-1}\right] \\ & \operatorname{Now~M(\alpha )^{-1}}=\left[\begin{array}{ccc} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{ccc} \cos (-\alpha) & -\sin (-\alpha) & 0 \\ \sin (-\alpha) & \cos (-\alpha) & 0 \\ 0 & 0 & 1 \end{array}\right]=\mathrm{M}(-\alpha) \\ & M(\beta)^{-1}=\left[\begin{array}{ccc} \cos \beta & 0 & -\sin \beta \\ 0 & 1 & 0 \\ \sin \beta & 0 & \cos \beta \end{array}\right] \\ & =\left[\begin{array}{ccc} \cos (-\beta) & 0 & \sin (-\beta) \\ 0 & 1 & 0 \\ -\sin (-\beta) & 0 & \cos (-\beta) \end{array}\right]=\mathrm{M}(-\beta) \\ & {[\mathrm{M}(\alpha) \mathrm{M}(\beta)]^{-1}=\mathrm{M}(-\beta) \mathrm{M}(-\alpha)} \\ & \end{aligned}\)

Hence, the correct answer is option: 'C'.

Solution

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