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JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 4

Let A=[α12230045]A=\begin{bmatrix}\alpha&1&2\\2&3&0\\0&4&5\end{bmatrix} and B=[1000−5α004α−2α]+adj⁡(A)B=\begin{bmatrix}1&0&0\\0&-5\alpha&0\\0&4\alpha&-2\alpha\end{bmatrix}+\operatorname{adj}(A). If det⁡(B)=66\det(B)=66, then det⁡(adj⁡(A))\det(\operatorname{adj}(A)) equals :

Choose one option
Answer and worked solution

Official answer: C

Adjugate of A (in terms of α):
adj⁡A=[153−6−105α48−4α3α−2],det⁡A=15α+6.\operatorname{adj}A=\begin{bmatrix}15&3&-6\\-10&5\alpha&4\\8&-4\alpha&3\alpha-2\end{bmatrix},\qquad\det A=15\alpha+6.

B. Adding the given matrix:
B=[163−6−100480α−2],det⁡B=−3(−10(α−2)−32)=30α+36.B=\begin{bmatrix}16&3&-6\\-10&0&4\\8&0&\alpha-2\end{bmatrix},\qquad\det B=-3\left(-10(\alpha-2)-32\right)=30\alpha+36.

Solve. 30α+36=6630\alpha+36=66 gives α=1\alpha=1, so det⁡A=21\det A=21.

Answer. det⁡(adj⁡A)=(det⁡A)2=441\det(\operatorname{adj}A)=(\det A)^2=441.

(C)

Topics:Adjoint and inverseDeterminants and properties
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