Given A = [ cc 0 & - & 0 ] and B ( )= [ cc & - & ], then ( I + A )( I - A )^ -1 is equal to

Given A = [ cc 0 & - & 0 ] and B ( )= [ cc & - & ], then ( I + A …

MCQ

Given $A =\left[\begin{array}{cc}0 & -\tan \alpha \\ \tan \alpha & 0\end{array}\right]$ and $B (\alpha)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$, then $( I + A )( I - A )^{-1}$ is equal to

A
$B (\alpha)$
B
$B (-\alpha)$
✓
$B (2 \alpha)$
D
$B (-2 \alpha)$

✓

Answer

Correct option: C.

$B (2 \alpha)$

(C) $I+A=\left[\begin{array}{cc}1 & -\tan \alpha \\ \tan \alpha & 1\end{array}\right]$
$I-A=\left[\begin{array}{cc}1 & \tan \alpha \\ -\tan \alpha & 1\end{array}\right]$
$\therefore| I - A |=1+\tan ^2 \alpha=\sec ^2 \alpha \neq 0$
$\operatorname{adj}(I-A)=\left[\begin{array}{cc}1 & \tan \alpha \\ -\tan \alpha & 1\end{array}\right]^{\top}=\left[\begin{array}{cc}1 & -\tan \alpha \\ \tan \alpha & 1\end{array}\right]$
$\Rightarrow( I - A )^{-1}=\frac{1}{| I - A |}[\operatorname{adj}( I - A )]$
$=\left[\begin{array}{cc}\cos ^2 \alpha & -\sin \alpha \cdot \cos \alpha \\ \sin \alpha \cdot \cos \alpha & \cos ^2 \alpha\end{array}\right]$
$\therefore(I+A)(I-A)^{-1}$
$\begin{array}{l}=\left[\begin{array}{cc}1 & -\tan \alpha \\ \tan \alpha & 1\end{array}\right]\left[\begin{array}{cc}\cos ^2 \alpha & -\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha & \cos ^2 \alpha\end{array}\right] \\ =\left[\begin{array}{cc}\cos ^2 \alpha-\sin ^2 \alpha & -2 \sin \alpha \cos \alpha \\ 2 \sin \alpha \cos \alpha & \cos ^2 \alpha-\sin ^2 \alpha\end{array}\right]\end{array}$
$\therefore \quad(I+A)(I-A)^{-1}=\left[\begin{array}{cc}\cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha\end{array}\right]$ ...(i)
$B (\alpha)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
$\therefore \quad B (2 \alpha)=\left[\begin{array}{cc}\cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha\end{array}\right]$ ...(ii)
$\therefore \quad(I+A)(I-A)^{-1}=B(2 \alpha) \ldots[$ From (i) and (ii) $]$