Q56 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics

If $F(\propto)=\left[\begin{array}{ccc}\cos \propto & -\sin \propto & 0 \\ \sin \propto & \cos \propto & 0 \\ 0 & 0 & 1\end{array}\right]$, where $\propto \in R$, then $[F(\propto)]^{-1}$

  • Correct answer
    A
    $\mathrm{F}(-\propto)$
  • B
    $F(2 \propto)$
  • C
    $F(\propto)$
  • D
    $\mathrm{F}(3 \propto)$

Solution

$$ \begin{aligned} & F(\propto)=\left[\begin{array}{ccc} \cos \propto & -\sin \propto & 0 \\ \sin \propto & \cos \propto & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \therefore|F(\alpha)|=\cos \alpha(\cos \alpha)+\sin \alpha(\sin \alpha)=\cos ^2 \alpha+\sin ^2 \alpha=1 \\ & \therefore \operatorname{adj}[F(\alpha)]=\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right]^{\mathrm{T}} \end{aligned} $$ $$ \begin{aligned} & =\left[\begin{array}{ccc} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \cos (-\alpha) & -\sin (-\alpha) & 0 \\ \sin (-\alpha) & \cos (-\alpha) & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & {[F(\alpha)]^{-1}=\frac{\operatorname{adj}[F(\alpha)]}{|F(\alpha)|}=F(-\alpha)} \end{aligned} $$

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

Relevant Questions