Mathematics - Matrices Question with Solution | TestHub

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

And $adj\left(A\right)=B,\left|A\right|=2$

$\Rightarrow |adj(A\left)\right|=\left|B\right|$

We know that $\left|adjA\right|={\left|A\right|}^{n-1}$

$\Rightarrow 2^2=(8-3\alpha )-3(4-3\alpha )+\alpha (-\alpha )$

$\Rightarrow {\alpha }^2-6\alpha +8=0$

$\Rightarrow (\alpha -4)(\alpha -2)=0$

$\alpha =4,2$ but $\alpha >2$ so $\boxed{\alpha =4}$

Now $\left[\begin{array}{ccc}\alpha & -2\alpha & \alpha \end{array}\right]B\left[\begin{array}{c}\alpha \\ -2\alpha \\ \alpha \end{array}\right]= \left[\begin{array}{ccc}4& -8& 4\end{array}\right]\left[\begin{array}{ccc}1& 3& 4\\ 1& 2& 3\\ 4& 4& 4\end{array}\right]\left[\begin{array}{c}4\\ -8\\ 4\end{array}\right]$

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

Hence this is the correct option.