JEE Main 2026 Apr 8 Shift 2, Mathematics Q4: Algebra Of Determinants

Expanding along row $1$: $|A|=\alpha (15)-1(10)+2(8)=15\alpha +6$.

Computing $\mathrm{adj}(A)$ via the cofactor matrix transpose:

$\mathrm{adj}(A)=\left[\begin{array}{c}153-6\\ -105\alpha 4\\ 8-4\alpha 3\alpha -2\end{array}\right]$

Adding the given matrix to $\mathrm{adj}(A)$ neatly cancels the $-5\alpha$ and $4\alpha$ terms:

$B=\left[\begin{array}{c}163-6\\ -1004\\ 80\alpha -2\end{array}\right]$

Expanding along column $2$: $|B|=-3\cdot [-10(\alpha -2)-32]=-3(-10\alpha -12)=30\alpha +36=6(5\alpha +6)$.

Wait — using direct expansion: $|B|=6\cdot |A|/(\text{factor})$. A clean computation gives $|B|=6(5\alpha +6)$. Setting equal to $66$: $5\alpha +6=11\Rightarrow \alpha =1$.

Hence $|A|=15(1)+6=21$, and for a $3\times 3$ matrix $\det (\mathrm{adj}A)=|A{|}^{n-1}=|A{|}^2=441$.