MHTCET 19 Apr 2026 Shift 1 Paper

Concept:For a matrix $A$, $\operatorname{adj}(A)$ is the transpose of the cofactor matrix of $A$.Explanation:For $A = [a_{ij}]_{3 \times 3}$, we have $a_{ij} = 1$ when $i+j$ is even, and $a_{ij} = 0$ when $i+j$ is odd.So the elements with both $i$ and $j$ odd or both even are $1$, and the rest are $0$.Thus, $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$.Now compute the cofactor matrix of $A$.Cofactor of each element gives $\begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$.Since $\operatorname{adj}(A)$ is the transpose of the cofactor matrix, and this cofactor matrix is symmetric, we get$\operatorname{adj}(A) = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$.Answer:Hence, $\operatorname{adj}(A) = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$, which is Option B.