A) 0
B) 1
C) \[\sin \alpha \cos \alpha \]
D) \[\cos 2\alpha \]
Correct Answer: B
Solution :
Let \[A=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]\] The matrix of cofactors of the elements of A, =\[A=\left| \,\begin{matrix} -1 & 2 & 5 \\ 2 & -4 & a-4 \\ 1 & -2 & a+1 \\ \end{matrix}\, \right|\,=\,\left| \,\begin{matrix} 0 & 0 & a+60 \\ 0 & 0 & -a-6 \\ 1 & -2 & a+1 \\ \end{matrix}\, \right|\] \[\therefore \] \[adjA=\]the transpose of matrix of cofactors of A = \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] \[\therefore \] \[A\,adjA=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]\,\,\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\] = \[\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} k & 0 \\ 0 & k \\ \end{matrix} \right]\] (as given) \[a=2,\,A=\left| \,\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & -8 \\ 1 & -2 & 3 \\ \end{matrix}\, \right|\] \[k=1\].
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