JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Determinants

$B$ is given as the cofactor matrix of $A$. Matching individual cofactors of $A=\left[\begin{array}{ccc}\beta & \alpha & 3\\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2\alpha \end{array}\right]$ against the entries of $B$:
Cofactor of $a_{11}=\beta$: $\alpha (2\alpha )-\beta (\alpha )=2{\alpha }^2-\alpha \beta =3\alpha ⟹\beta =2\alpha -3$.
Cofactor of $a_{13}=3$: $\alpha (\alpha )-\alpha (-\beta )={\alpha }^2+\alpha \beta =3\alpha ⟹\alpha +\beta =3$.
Solving together: $\alpha +(2\alpha -3)=3⟹\alpha =2,\beta =1$ (consistent with the remaining cofactor entries).
So $A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 2& 1\\ -1& 2& 4\end{array}\right]$, and

Since $B$ is the cofactor matrix, $\text{adj}(A)=B^T$, so $\det (B)=\det (\text{adj }A)=[\det A{]}^2$, and