JEE Main 2026 — 21 January, Morning Shift — Question 24

Given $A^2 - 4A + 2I = O$ and $B^2 - 3B + I = O$. From trace: $\operatorname{tr}(A)=4 \Rightarrow \alpha+2=4 \Rightarrow \alpha=2$. $\operatorname{tr}(B)=3 \Rightarrow \beta+1=3 \Rightarrow \beta=2$. Thus $A = \begin{pmatrix} 2 & 2 \\ 1 & 2 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$. Using $A^2 = 4A - 2I$, compute $A^3 = A \cdot A^2 = A(4A-2I) = 4A^2 - 2A = 4(4A-2I) - 2A = 16A - 8I - 2A = 14A - 8I$. $A^3 = 14\begin{pmatrix} 2 & 2 \\ 1 & 2 \end{pmatrix} - 8\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 28 & 28 \\ 14 & 28 \end{pmatrix} - \begin{pmatrix} 8 & 0 \\ 0 & 8 \end{pmatrix} = \begin{pmatrix} 20 & 28 \\ 14 & 20 \end{pmatrix}$. Similarly, $B^2 = 3B - I$, so $B^3 = B \cdot B^2 = B(3B-I) = 3B^2 - B = 3(3B-I) - B = 9B - 3I - B = 8B - 3I$. $B^3 = 8\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} - 3\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 8 & 8 \\ 8 & 16 \end{pmatrix} - \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 5 & 8 \\ 8 & 13 \end{pmatrix}$. $A^3 - B^3 = \begin{pmatrix} 20 & 28 \\ 14 & 20 \end{pmatrix} - \begin{pmatrix} 5 & 8 \\ 8 & 13 \end{pmatrix} = \begin{pmatrix} 15 & 20 \\ 6 & 7 \end{pmatrix}$. $\det(A^3 - B^3) = 15\cdot7 - 20\cdot6 = 105 - 120 = -15$. For any square matrix $M$, $\det(\operatorname{adj}(M)) = (\det M)^{n-1}$. Here $n=2$,

so $\det(\operatorname{adj}(A^3-B^3)) = \det(A^3-B^3) = -15$. Thus $\left(\det(\operatorname{adj}(A^3-B^3))\right)^2 = (-15)^2 = 225$.