JEE Main 2023 — Mathematics Algebra

We have been given that $B=[\begin{matrix}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{matrix}],\alpha >2$B=[11α​32α​α34​],α>2

And $adj(A)=B,|A|=2$adj(A)=B,∣A∣=2

$$\Rightarrow |adj(A)|=|B|$$⇒∣adj(A)∣=∣B∣

We know that $|adjA|={|A|}^{n-1}$∣adjA∣=∣A∣n−1

$$\Rightarrow {2}^{2}=(8-3\alpha )-3(4-3\alpha )+\alpha (-\alpha )$$⇒22=(8−3α)−3(4−3α)+α(−α)

$$\Rightarrow {\alpha }^{2}-6\alpha +8=0$$⇒α2−6α+8=0

$$\Rightarrow (\alpha -4)(\alpha -2)=0$$⇒(α−4)(α−2)=0

$\alpha =4,2$α=4,2 but $\alpha >2$α>2 so $\alpha =4$α=4

Now $[\begin{matrix}\alpha & -2\alpha & \alpha \end{matrix}]B[\begin{matrix}\alpha \\ -2\alpha \\ \alpha \end{matrix}]=[\begin{matrix}4 & -8 & 4\end{matrix}][\begin{matrix}1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4\end{matrix}][\begin{matrix}4 \\ -8 \\ 4\end{matrix}]$[α​−2α​α​]B[α−2αα​]=[4​−8​4​][114​324​434​][4−84​]

$=[\begin{matrix}12 & 12 & 8\end{matrix}][\begin{matrix}4 \\ -8 \\ 4\end{matrix}]$=[12​12​8​][4−84​]$=48-96+32=-16$=48−96+32=−16

Hence this is the correct option.