MCQ
If $A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$ and $A\,\,adj$$A = \left[ {\begin{array}{*{20}{c}}k&0\\0&k\end{array}} \right],$ then $ k$  is equal to
  • A
    $0$
  • $1$
  • C
    $\sin \alpha \cos \alpha $
  • D
    $\cos 2\alpha $

Answer

Correct option: B.
$1$
b
(b) Let $A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$

The matrix of cofactors of the elements of  $A,$
$A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{-(-\sin \alpha) }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$

$\therefore $ $adjA = $the transpose of matrix of cofactors of  $ A$

$A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{-\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$

$\therefore $ $A\,adjA = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$

= $\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}k&0\\0&k\end{array}} \right]$ (as given)

=>$ K=1.$

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