Let B = [ cases 1 & 3 & α 1 & 2 & 3 α & α & 4 cases ], α > 2 be the...

We have been given that B = \left[\begin{cases}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{cases}\right], \alpha > 2

And $adj ( A ) = B, | A | = 2$

$⇒ | adj ( A ) | = | B |$

We know that \left|adjA\right| = {\left|A\right|}^{n - 1}$</p><p>$⇒ {2}^{2} = ( 8 - 3\alpha ) - 3 ( 4 - 3\alpha ) + \alpha ( - \alpha )$</p><p>$⇒ {\alpha}^{2} - 6\alpha + 8 = 0$</p><p>$⇒ ( \alpha - 4 ) ( \alpha - 2 ) = 0$</p><p>$\alpha = 4, 2$but$\alpha > 2$so$\alpha = 4</p><p>Now \left[\begin{cases}\alpha & - 2\alpha & \alpha\end{cases}\right]B\left[$\begin{array}{ll}\alpha \\ - 2\alpha \\ \alpha\end{array}$\right] = \left[\begin{cases}4 & - 8 & 4\end{cases}\right]\left[\begin{cases}1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4\end{cases}\right]\left[$\begin{array}{ll}4 \\ - 8 \\ 4\end{array}$\right]

= \left[\begin{cases}12 & 12 & 8\end{cases}\right]\left[$\begin{array}{ll}4 \\ - 8 \\ 4\end{array}$\right]$= 48 - 96 + 32 = - 16$

Hence this is the correct option.