Q65 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
If $A = [a_{ij}]_{3\times3}$, where $a_{ij} = \begin{cases} 1, & \text{if } i+j \text{ is even} \\ 0, & \text{if } i+j \text{ is odd} \end{cases}$, then $\text{adj}(A) = \ldots$ ..
- A$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
- Correct answerB$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$
- C$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
- D$\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
Solution
Given $A = [a_{ij}]_{3\times3}$ where $a_{ij} = 1$ if $i+j$ is even, and $0$ if $i+j$ is odd.
Constructing the matrix $A$:
$A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$
The adjoint of $A$ is the transpose of the cofactor matrix $C$. Calculating the cofactors $C_{ij} = (-1)^{i+j} M_{ij}$:
$C_{11} = 1, C_{12} = 0, C_{13} = -1$
$C_{21} = 0, C_{22} = 0, C_{23} = 0$
$C_{31} = -1, C_{32} = 0, C_{33} = 1$
The cofactor matrix is $C = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$.
Taking the transpose of $C$ gives the adjoint matrix:
$\text{adj}(A) = C^T = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$
Answer: $\begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$
Constructing the matrix $A$:
$A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$
The adjoint of $A$ is the transpose of the cofactor matrix $C$. Calculating the cofactors $C_{ij} = (-1)^{i+j} M_{ij}$:
$C_{11} = 1, C_{12} = 0, C_{13} = -1$
$C_{21} = 0, C_{22} = 0, C_{23} = 0$
$C_{31} = -1, C_{32} = 0, C_{33} = 1$
The cofactor matrix is $C = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$.
Taking the transpose of $C$ gives the adjoint matrix:
$\text{adj}(A) = C^T = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$
Answer: $\begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$
Hence, the correct answer is option: 'B'.