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JEE MAIN 2026 · Matrices and Determinants

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 4

Updated
Single correct+4 / -1Answer matches NTA’s final key

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

Solution

Solution

Find adj(A). The cofactors of AA give

adj⁡(A)=[153−6−105α48−4α3α−2].\operatorname{adj}(A) = \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix}.

Add the given matrix.

B=[163−6−100480α−2].B = \begin{bmatrix} 16 & 3 & -6 \\ -10 & 0 & 4 \\ 8 & 0 & \alpha - 2 \end{bmatrix}.

Expanding along the second column, only the entry 33 survives:

det⁡(B)=−3 [(−10)(α−2)−4⋅8]=30α−60+96=30α+36.\det(B) = -3\,\big[(-10)(\alpha - 2) - 4 \cdot 8\big] = 30\alpha - 60 + 96 = 30\alpha + 36.

30α+36=66⇒α=130\alpha + 36 = 66 \Rightarrow \alpha = 1.

Finish. det⁡(A)=α(15)−1(10)+2(8)=15α+6=21\det(A) = \alpha(15) - 1(10) + 2(8) = 15\alpha + 6 = 21, and for a 3×33 \times 3 matrix det⁡(adj⁡A)=(det⁡A)2=441\det(\operatorname{adj} A) = (\det A)^2 = 441.

Answer: 441.

Answer checked against the official answer key. The solution is our own working.

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