The inverse of [ cc 1 & - & -1 ] is - Vidyadip

MCQ

The inverse of $\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]$ is

A
$\left[\begin{array}{cc}1 & -\sin \alpha \\ -\sin \alpha & -1\end{array}\right]$
B
$-\sec ^2 \alpha\left[\begin{array}{cc}1 & -\sin \alpha \\ \sin \alpha & -1\end{array}\right]$
✓
$\sec ^2 \alpha\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]$
D
$\cos ^2 \alpha\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]$

✓

Answer

Correct option: C.

$\sec ^2 \alpha\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]$

(C) $|A|=\left|\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right|=-1+\sin ^2 \alpha \neq 0$
$\therefore \quad A ^{-1}=\frac{1}{-1+\sin ^2 \alpha}\left[\begin{array}{cc}-1 & -\sin \alpha \\ \sin \alpha & 1\end{array}\right]$ ...[Using Shortcut 2]
$\begin{array}{l}=\frac{1}{\cos ^2 \alpha}\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right] \\ =\sec ^2 \alpha\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]\end{array}$

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