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