Question
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
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Answer
b
(b) Let $A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$
(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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