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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
Let $F(\alpha ) = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }&0\\{\sin \alpha }&{\cos \alpha }&0\\0&0&1\end{array}} \right]$, where $\alpha \in R.$ Then ${[F(\alpha )]^{ - 1}}$ is equal to
  • ✓
    $F( - \alpha )$
  • B
    $F({\alpha ^{ - 1}})$
  • C
    $F(2\alpha )$
  • D
    None of these
✓

Answer

Correct option: A.
$F( - \alpha )$
a
(a) We have

$F(\alpha )\,F( - \alpha )\, = \,\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }&0\\{\sin \alpha }&{\cos \alpha }&0\\0&0&1\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }&0\\{ - \sin \alpha }&{\cos \alpha }&0\\0&0&1\end{array}} \right]$

= $\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = I$

$\therefore $ $F( - \alpha ) = {[F(\alpha )]^{ - 1}}$.

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