JEE Main 2026 — 5 April, Morning Shift — Question 26

$\text{Let } A = \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \\ \beta_1 & \beta_2 & \beta_3 \\ \gamma_1 & \gamma_2 & \gamma_3 \end{pmatrix}$

$\text{Now } A^\top \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \alpha_1 & \beta_1 & \gamma_1 \\ \alpha_2 & \beta_2 & \gamma_2 \\ \alpha_3 & \beta_3 & \gamma_3 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$

$\Rightarrow \begin{pmatrix} \gamma_1 \\ \gamma_2 \\ \gamma_3 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} \hspace{4cm} \ldots(1)$

$\text{Now } \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \\ \beta_1 & \beta_2 & \beta_3 \\ \gamma_1 & \gamma_2 & \gamma_3 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$

$\Rightarrow \begin{pmatrix} \alpha_3 \\ \beta_3 \\ \gamma_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} \hspace{4cm} \ldots(2)$

$A^\top \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} - A^\top \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$

$\Rightarrow A^\top \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$

$\Rightarrow \begin{pmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \hspace{4cm} \ldots(3)$

$\text{Now } A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} - A \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$

$\Rightarrow A \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$

$\Rightarrow \begin{pmatrix} \alpha_1 \\ \beta_1 \\ \gamma_1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \hspace{4cm} \ldots(4)$

$A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & \beta_2 & 3 \\ 3 & 1 & 1 \end{pmatrix}$

$|A| = 2(\beta_2 - 3) - 1(1 - 9) + 1(1 - 3\beta_2) = 1$

$\Rightarrow 2\beta_2 - 6 + 8 + 1 - 3\beta_2 = 1$

$\Rightarrow \beta_2 = 2$

$\text{So } A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix}$

$A^2 = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 8 & 5 & 6 \\ 13 & 8 & 10 \\ 10 & 6 & 7 \end{pmatrix}$

$A^2 + A = \begin{pmatrix} 10 & 6 & 7 \\ 14 & 10 & 13 \\ 13 & 7 & 8 \end{pmatrix} \Rightarrow \det(A^2 + A) = 8$

$\Rightarrow \text{Det}(\text{adj}(A^2 + A)) = 8^2 = 64$