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If $B=\left[\begin{array}{lll}1 & \alpha & 2 \\ 1 & 2 & 2 \\ 2 & 3 & 3\end{array}\right]$ is the adjoint of a $3 \times 3$ matrix $\mathrm{A}$ and $|\mathrm{A}|=5$, then $\alpha$ is equal to
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27
Using $|\operatorname{Adj} A|=|A|^{n-1}$
But $\mathrm{B}=\operatorname{Adj}(\mathrm{A})$...['Given]
$\therefore \quad|\mathrm{B}|=|\mathrm{A}|^2$
$\begin{aligned} & \Rightarrow\left|\begin{array}{lll}1 & \alpha & 2 \\ 1 & 2 & 2 \\ 2 & 3 & 3\end{array}\right|=|\mathrm{A}|^2 \\ & \Rightarrow \alpha-2=5^2 \\ & \Rightarrow \alpha-2=25 \\ & \Rightarrow \alpha=27\end{aligned}$
But $\mathrm{B}=\operatorname{Adj}(\mathrm{A})$...['Given]
$\therefore \quad|\mathrm{B}|=|\mathrm{A}|^2$
$\begin{aligned} & \Rightarrow\left|\begin{array}{lll}1 & \alpha & 2 \\ 1 & 2 & 2 \\ 2 & 3 & 3\end{array}\right|=|\mathrm{A}|^2 \\ & \Rightarrow \alpha-2=5^2 \\ & \Rightarrow \alpha-2=25 \\ & \Rightarrow \alpha=27\end{aligned}$
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