If $B=\left[\begin{array}{ccc}3 & \alpha & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right]$ is the adjoint…

If $B=\left[\begin{array}{ccc}3 & \alpha & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right]$ is the adjoint of $a$ $3 \times 3$ matrix $\mathrm{A}$ and $|\mathrm{A}|=4$, then $\alpha$ is equal to8
  1. $1$
  2. $0$
  3. $-1$
  4. $-2$

Solution

$\begin{aligned} & \quad \text { Using }|\operatorname{adj} A|=|A|^{\mathrm{n}-1} \\ & \quad \text { But } \mathrm{B}=\operatorname{Adj}(\mathrm{A}) ...[Given]\\ & \therefore \quad|\mathrm{B}|=|\mathrm{A}|^2 \\ & \Rightarrow\left|\begin{array}{ccc} 3 & \alpha & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right|=|\mathrm{A}|^2 \\ & \Rightarrow 24-4 \alpha-4=4^2 \\ & \Rightarrow 20-4 \alpha=16 \\ & \Rightarrow \alpha=1 \end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 2)

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