If $A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1\end{array} ight]$, then $(\operatorname{Adj} A)^{-1}=$

Question

If $A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1\end{array}ight]$, then $(\operatorname{Adj} A)^{-1}=$

Aicharya · Accepted Answer

We have, $ \begin{aligned} & A=\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ 0 & 0 & 1 \end{array}ight] \ & 	herefore|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, $(\operatorname{adj} A)^{-1}=\frac{\operatorname{adj}(\operatorname{adj} A)}{|\operatorname{adj} A|} \quad\left[\because A^{-1}=\frac{\operatorname{adj} A}{|A|}ight]$ $ =A $ Hence, option (c) correct

If $A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 &…

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