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(D) The adjoint of a matrix $A$,denoted by $	ext{adj } A$,is the transpose of the cofactor matrix $C$,where $C_{ij} = (-1)^{i+j} M_{ij}$.Given $A = \

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$5$

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(D) The adjoint of a matrix $A$,denoted by $	ext{adj } A$,is the transpose of the cofactor matrix $C$,where $C_{ij} = (-1)^{i+j} M_{ij}$.Given $A = \begin{bmatrix} 1 & 0 & 2 \ -1 & 1 & -2 \ 0 & 2 & 1 \end{bmatrix}$.To find $x$,which is the element at position $(1, 2)$ of $	ext{adj } A$,we calculate the cofactor of the element at position $(2, 1)$ of matrix $A$:$x = C_{21} = (-1)^{2+1} \begin{vmatrix} 0 & 2 \ 2 & 1 \end{vmatrix} = -1(0 - 4) = 4$.To find $y$,which is the element at position $(3, 3)$ of $	ext{adj } A$,we calculate the cofactor of the element at position $(3, 3)$ of matrix $A$:$y = C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 0 \ -1 & 1 \end{vmatrix} = 1(1 - 0) = 1$.Therefore,$x + y = 4 + 1 = 5$.

If $A = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix}$ and $\text{adj } A = \begin{bmatrix} 5 & x & -2 \\ 1 & 1 & 0 \\ -2 & -2 & y \end{bmatrix}$,then the value of $x+y$ is:

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