If $|A|=-3$ and $A^{-1}=\left[\begin{array}{ccc}1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} &…

If $|A|=-3$ and $A^{-1}=\left[\begin{array}{ccc}1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1\end{array}\right]$, then $(\operatorname{adj} A)$ is
  1. $\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$
  2. $\left[\begin{array}{ccc}-\frac{1}{3} & 0 & 0 \\ \frac{1}{3} & -\frac{1}{9} & 0 \\ 1 & -\frac{2}{9} & \frac{1}{3}\end{array}\right]$
  3. $\left[\begin{array}{ccc}3 & 0 & 0 \\ -3 & 1 & 0 \\ 9 & 2 & -3\end{array}\right]$
  4. $\left[\begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ -\frac{1}{3} & \frac{1}{9} & 0 \\ -1 & \frac{2}{9} & -\frac{1}{3}\end{array}\right]$

Solution

$\begin{aligned} & \because A^{-1}=\frac{\operatorname{adj}(A)}{|A|} \\ & \Rightarrow \operatorname{adj}(A)=|A| A^{-1}=\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]\end{aligned}$

Asked in: MHT CET 2022 (10 Aug Shift 1)

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