MCQ
If $A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$, then find $(\operatorname{adj} A)^{-1}$.
  • $A$
  • B
    $-A$
  • C
    $A^{\prime}$
  • D
    None of these

Answer

Correct option: A.
$A$
(a): We have, $|A|=\cos ^2 \alpha+\sin ^2 \alpha=1 \neq 0$
So, $A^{-1}$ exists.
We know adj $A=|A| A^{-1}$
$
\begin{array}{lll}
\Rightarrow & \text { adj } A=A^{-1} & {[\because|A|=1]} \\
\Rightarrow & (\operatorname{adj} A)^{-1}=\left(A^{-1}\right)^{-1}=A &
\end{array}
$

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