Q50 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
For some $\alpha, \beta \in \mathbf{R}$, let $A=\left[\begin{array}{ll}\alpha & 2 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 1 \\ 1 & \beta\end{array}\right]$ be such that $A^{2}-4 A+2 I=B^{2}-3 B+I=O$. Then $\left(\operatorname{det}\left(\operatorname{adj}\left(A^{3}-B^{3}\right)\right)\right)^{2}$ is equal to $\_\_\_\_$.
Correct answer: 225
Solution
From $A^2 - 4A + 2I = O$: characteristic equation is $\lambda^2 - 4\lambda + 2 = 0$.
For $A$: $\alpha + 2 = 4 \Rightarrow \alpha = 2$.
From $B^2 - 3B + I = O$: characteristic equation is $\lambda^2 - 3\lambda + 1 = 0$.
For $B$: $\beta + 1 = 3 \Rightarrow \beta = 2$.
$A^3 = 14A - 8I$, $B^3 = 8B - 3I$.
$A^3 - B^3 = 14A - 8B - 5I = \begin{bmatrix} 15 & 20 \\ 6 & 7 \end{bmatrix}$.
$\det(A^3 - B^3) = 105 - 120 = -15$.
For 2×2 matrix: $\det(\text{adj}(M)) = \det(M)$.
$(\det(\text{adj}(A^3 - B^3)))^2 = 225$.
For $A$: $\alpha + 2 = 4 \Rightarrow \alpha = 2$.
From $B^2 - 3B + I = O$: characteristic equation is $\lambda^2 - 3\lambda + 1 = 0$.
For $B$: $\beta + 1 = 3 \Rightarrow \beta = 2$.
$A^3 = 14A - 8I$, $B^3 = 8B - 3I$.
$A^3 - B^3 = 14A - 8B - 5I = \begin{bmatrix} 15 & 20 \\ 6 & 7 \end{bmatrix}$.
$\det(A^3 - B^3) = 105 - 120 = -15$.
For 2×2 matrix: $\det(\text{adj}(M)) = \det(M)$.
$(\det(\text{adj}(A^3 - B^3)))^2 = 225$.
Hence, the correct answer is 225.