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11. ⇒ (MHT CET 2021 20th September Morning Shift )

$A (\propto) = [\begin{array}{cc} \cos \propto & \sin \propto \\ - \sin \propto & \cos \propto \end{array}]$ , then ${[A^{2} (\propto)]}^{- 1} =$

A. $A (\propto)$

B. $A^{2} (\propto)$

C. $A (- 2 \propto)$

D. $A (2 \propto)$

Correct answer option is (C)

$\begin{aligned} \begin{aligned} A (α) = [\begin{array}{cc} \cos α & \sin α \\ - \sin α & \cos α \end{array}] \Rightarrow | A (α) | = \cos^{2} α + \sin^{2} α = 1 \\ A^{2} (α) = [\begin{array}{cc} \cos α & \sin α \\ - \sin α & \cos α \end{array}] [\begin{array}{cc} \cos α & \sin α \\ - \sin α & \cos α \end{array}] \\ = [\begin{array}{cc} \cos^{2} α - \sin^{2} α & 2 \sin α \cos α \\ - 2 \sin α \cos α & \cos^{2} α - \sin^{2} α \end{array}] = [\begin{array}{cc} \cos 2 α & \sin 2 α \\ - \sin 2 α & \cos 2 α \end{array}] \\ ∴ adj [A^{2} (α)] = [\begin{array}{cc} \cos 2 α & - \sin 2 α \\ \sin 2 α & \cos 2 α \end{array}] \\ ∴ {[A^{2} (α)]}^{- 1} = [\begin{array}{cc} \cos 2 α & - \sin 2 α \\ \sin 2 α & \cos 2 α \end{array}] . . . . [∵ | A^{2} | = | A |^{2} = 1] \\ ∴ {[A^{2} (α)]}^{- 1} = [\begin{array}{cc} \cos (- 2 α) & \sin (- 2 α) \\ - \sin (- 2 α) & \cos (- 2 α) \end{array}] = A (- 2 α) \end{aligned} \end{aligned}$

⇒Matrices