JEE Main 2026 — Mathematics Algebra

The cofactor matrix of $A$A is calculated as follows:

$C_{11} = 15$C11​=15, $C_{12} = -10$C12​=−10, $C_{13} = 8$C13​=8

$C_{21} = 3$C21​=3, $C_{22} = 5\alpha$C22​=5α, $C_{23} = -4\alpha$C23​=−4α

$C_{31} = -6$C31​=−6, $C_{32} = 4$C32​=4, $C_{33} = 3\alpha - 2$C33​=3α−2

The adjoint of $A$A is the transpose of the cofactor matrix:

$$\text{adj}(A) = \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix}$$adj(A)=​15−108​35α−4α​−643α−2​​

The matrix $B$B is given by:

$$B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix} = \begin{bmatrix} 16 & 3 & -6 \\ -10 & 0 & 4 \\ 8 & 0 & \alpha - 2 \end{bmatrix}$$B=​100​0−5α4α​00−2α​​+​15−108​35α−4α​−643α−2​​=​16−108​300​−64α−2​​

Expanding the determinant of $B$B along the second column:

$$\det(B) = -3 \begin{vmatrix} -10 & 4 \\ 8 & \alpha - 2 \end{vmatrix} = -3(-10(\alpha - 2) - 32) = -3(-10\alpha - 12) = 30\alpha + 36$$det(B)=−3​−108​4α−2​​=−3(−10(α−2)−32)=−3(−10α−12)=30α+36

Given $\det(B) = 66$det(B)=66, we get:

$$30\alpha + 36 = 66 \Rightarrow 30\alpha = 30 \Rightarrow \alpha = 1$$30α+36=66⇒30α=30⇒α=1

The determinant of $A$A is:

$$\det(A) = \alpha(15 - 0) - 1(10 - 0) + 2(8 - 0) = 15\alpha + 6$$det(A)=α(15−0)−1(10−0)+2(8−0)=15α+6

Substituting $\alpha = 1$α=1:

$$\det(A) = 15(1) + 6 = 21$$det(A)=15(1)+6=21

Using the property $\det(\text{adj}(A)) = (\det(A))^{n-1}$det(adj(A))=(det(A))n−1 for a $3 \times 3$3×3 matrix:

$$\det(\text{adj}(A)) = (\det(A))^2 = (21)^2 = 441$$det(adj(A))=(det(A))2=(21)2=441

Answer: $441$441