A=0−tanαtanα0,B(α)=cosα−sinαsinαcosα

(A):(I+A)=1001+1−tanαtanα 0=1−tanαtanα1(I−A)=1tanα−tanα 1(I−A)−1=11+tan2α1−tanαtanα 1(I+A)(I−A)−1=cos2α1−tanαtanα 1×1−tanαtanα 1=cos2α1−tan2α−tanα−tanαtanα+tanα−tan2α+1=cos2αcos2αcos2α−sin2αcos2αsin2αcos2αcos2αcos2α=cos2α−sin2αsin2αcos2α=B(2α)(B):I−A=1tanα−tanα 1

A=0−tanαtanα0,B(α)=cosα−sinαsinαcosα

(A):(I+A)=1001+1−tanαtanα 0=1−tanαtanα1(I−A)=1tanα−tanα 1(I−A)−1=11+tan2α1−tanαtanα 1(I+A)(I−A)−1=cos2α1−tanαtanα 1×1−tanαtanα 1=cos2α1−tan2α−tanα−tanαtanα+tanα−tan2α+1=cos2αcos2αcos2α−sin2αcos2αsin2αcos2αcos2αcos2α=cos2α−sin2αsin2αcos2α=B(2α)(B):I−A=1tanα−tanα 1

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$A = [\begin{matrix} 0 & - \tan α \\ \tan α & 0 \end{matrix}], B (α) = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

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$(I + A) {(I - A)}^{- 1}$

$(I - A) {(I + A)}^{- 1}$

$B {(α)}^{2}$

$B {(α)}^{- 2}$

$A - B (α)$

$B (2 α)$

$B (- 2 α)$

$A B (- α)$

answer is B, C.

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Detailed Solution

(A):

$\begin{array}{l} (I + A) = [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}] + [\begin{array}{l} 1 & - \tan α \\ \tan α & 0 \end{array}] = [\begin{array}{l} 1 & - \tan α \\ \tan α & 1 \end{array}] \\ (I - A) = [\begin{array}{l} 1 & \tan α \\ - \tan α & 1 \end{array}] \\ {(I - A)}^{- 1} = \frac{1}{1 + \tan^{2} α} [\begin{array}{l} 1 & - \tan α \\ \tan α & 1 \end{array}] \\ (I + A) {(I - A)}^{- 1} = \cos^{2} α [\begin{array}{l} 1 & - \tan α \\ \tan α & 1 \end{array}] \times [\begin{array}{l} 1 & - \tan α \\ \tan α & 1 \end{array}] \\ = \cos^{2} α [\begin{array}{l} 1 - \tan^{2} α & - \tan α - \tan α \\ \tan α + \tan α & - \tan^{2} α + 1 \end{array}] \\ = \cos^{2} α [\begin{array}{l} \frac{\cos 2 α}{\cos^{2} α} & - \frac{\sin 2 α}{\cos^{2} α} \\ \frac{\sin 2 α}{\cos^{2} α} & \frac{\cos 2 α}{\cos^{2} α} \end{array}] \\ = [\begin{array}{l} \cos 2 α & - \sin 2 α \\ \sin 2 α & \cos 2 α \end{array}] \\ = B (2 α) \\ (B) : \\ I - A = [\begin{array}{l} 1 & \tan α \\ - \tan α & 1 \end{array}] \end{array}$