Q28 / 75 · JEE Main 2026 (05 April Shift 1) · Mathematics
Let $A$ be a $3 \times 3$ matrix such that $A^T \begin{bmatrix}1\\0\\1\end{bmatrix} = \begin{bmatrix}5\\2\\2\end{bmatrix}$, $A^T \begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}3\\1\\1\end{bmatrix}$, $A \begin{bmatrix}1\\0\\1\end{bmatrix} = \begin{bmatrix}3\\4\\4\end{bmatrix}$ and $A \begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}1\\3\\1\end{bmatrix}$. If $\det(A) = 1$, then $\det(\operatorname{adj}(A^2 + A))$ is equal to:
- A$16$
- B$25$
- C$49$
- Correct answerD$64$
Solution
Let $C_1, C_2, C_3$ be the columns of $A$ and $R_1, R_2, R_3$ be the rows of $A$.
From the given equations, we have:
$A \begin{bmatrix}0\\0\\1\end{bmatrix} = C_3 = \begin{bmatrix}1\\3\\1\end{bmatrix}$
$A \begin{bmatrix}1\\0\\1\end{bmatrix} = C_1 + C_3 = \begin{bmatrix}3\\4\\4\end{bmatrix} \Rightarrow C_1 = \begin{bmatrix}2\\1\\3\end{bmatrix}$
Similarly, for the transpose $A^T$, the columns correspond to the rows of $A$:
$A^T \begin{bmatrix}0\\0\\1\end{bmatrix} = R_3^T = \begin{bmatrix}3\\1\\1\end{bmatrix} \Rightarrow R_3 = \begin{bmatrix}3 & 1 & 1\end{bmatrix}$
$A^T \begin{bmatrix}1\\0\\1\end{bmatrix} = R_1^T + R_3^T = \begin{bmatrix}5\\2\\2\end{bmatrix} \Rightarrow R_1^T = \begin{bmatrix}2\\1\\1\end{bmatrix} \Rightarrow R_1 = \begin{bmatrix}2 & 1 & 1\end{bmatrix}$
Using the elements of $C_1, C_3, R_1,$ and $R_3$, we can construct the matrix $A$ with an unknown central element $b$:
$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & b & 3 \\ 3 & 1 & 1 \end{bmatrix}$
We are given that $\det(A) = 1$. Expanding along the first row:
$\det(A) = 2(b - 3) - 1(1 - 9) + 1(1 - 3b) = 1$
$2b - 6 + 8 + 1 - 3b = 1$
$3 - b = 1 \Rightarrow b = 2$
Thus, the matrix $A$ is:
$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$
We need to find $\det(\operatorname{adj}(A^2 + A))$. Using the properties of determinants and adjoints:
$\det(\operatorname{adj}(A^2 + A)) = (\det(A^2 + A))^2 = (\det(A) \cdot \det(A + I))^2$
First, find $A + I$:
$A + I = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & 3 \\ 3 & 1 & 2 \end{bmatrix}$
Now, calculate $\det(A + I)$:
$\det(A + I) = 3(6 - 3) - 1(2 - 9) + 1(1 - 9)$
$\det(A + I) = 3(3) - 1(-7) + 1(-8) = 9 + 7 - 8 = 8$
Since $\det(A) = 1$, we have:
$\det(A^2 + A) = 1 \times 8 = 8$
Finally, for a $3 \times 3$ matrix:
$\det(\operatorname{adj}(A^2 + A)) = 8^{3-1} = 8^2 = 64$
Answer: $64$
From the given equations, we have:
$A \begin{bmatrix}0\\0\\1\end{bmatrix} = C_3 = \begin{bmatrix}1\\3\\1\end{bmatrix}$
$A \begin{bmatrix}1\\0\\1\end{bmatrix} = C_1 + C_3 = \begin{bmatrix}3\\4\\4\end{bmatrix} \Rightarrow C_1 = \begin{bmatrix}2\\1\\3\end{bmatrix}$
Similarly, for the transpose $A^T$, the columns correspond to the rows of $A$:
$A^T \begin{bmatrix}0\\0\\1\end{bmatrix} = R_3^T = \begin{bmatrix}3\\1\\1\end{bmatrix} \Rightarrow R_3 = \begin{bmatrix}3 & 1 & 1\end{bmatrix}$
$A^T \begin{bmatrix}1\\0\\1\end{bmatrix} = R_1^T + R_3^T = \begin{bmatrix}5\\2\\2\end{bmatrix} \Rightarrow R_1^T = \begin{bmatrix}2\\1\\1\end{bmatrix} \Rightarrow R_1 = \begin{bmatrix}2 & 1 & 1\end{bmatrix}$
Using the elements of $C_1, C_3, R_1,$ and $R_3$, we can construct the matrix $A$ with an unknown central element $b$:
$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & b & 3 \\ 3 & 1 & 1 \end{bmatrix}$
We are given that $\det(A) = 1$. Expanding along the first row:
$\det(A) = 2(b - 3) - 1(1 - 9) + 1(1 - 3b) = 1$
$2b - 6 + 8 + 1 - 3b = 1$
$3 - b = 1 \Rightarrow b = 2$
Thus, the matrix $A$ is:
$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$
We need to find $\det(\operatorname{adj}(A^2 + A))$. Using the properties of determinants and adjoints:
$\det(\operatorname{adj}(A^2 + A)) = (\det(A^2 + A))^2 = (\det(A) \cdot \det(A + I))^2$
First, find $A + I$:
$A + I = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & 3 \\ 3 & 1 & 2 \end{bmatrix}$
Now, calculate $\det(A + I)$:
$\det(A + I) = 3(6 - 3) - 1(2 - 9) + 1(1 - 9)$
$\det(A + I) = 3(3) - 1(-7) + 1(-8) = 9 + 7 - 8 = 8$
Since $\det(A) = 1$, we have:
$\det(A^2 + A) = 1 \times 8 = 8$
Finally, for a $3 \times 3$ matrix:
$\det(\operatorname{adj}(A^2 + A)) = 8^{3-1} = 8^2 = 64$
Answer: $64$
Hence, the correct answer is option: 'D'.