Q45 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Mathematics
If $A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$, then $(\operatorname{Adj} A)^{-1}=$
- A$A+I$
- B$A-I$
- Correct answerCA
- D$\operatorname{Adj}\left(A^{-1}\right)$
Solution
We have,
$$
\begin{aligned}
& A=\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right] \\
& \therefore|A|=\cos ^2 \alpha+\sin ^2 \alpha=1
\end{aligned}
$$
Now, $|\operatorname{adj} A|=|A|^{n-1}=|A|^{3-1}=|A|^2=(1)^2=1$
Also, $\operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A=|A|^{3-2} \cdot A=|A| A=A$
Now, $(\operatorname{adj} A)^{-1}=\frac{\operatorname{adj}(\operatorname{adj} A)}{|\operatorname{adj} A|} \quad\left[\because A^{-1}=\frac{\operatorname{adj} A}{|A|}\right]$
$$
=A
$$
Hence, option (c) correct.
Hence, the correct answer is option: 'C'.