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(B) Given the matrix $A = \begin{bmatrix} 1 & 2 & 0 \ 0 & 1 & 2 \ 2 & 0 & 1 \end{bmatrix}$.To find the adjoint of $A$,we calculate the cofactor matr

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$\begin{bmatrix} 1 & -2 & 4 \ 4 & 1 & -2 \ -2 & 4 & 1 \end{bmatrix}$

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(B) Given the matrix $A = \begin{bmatrix} 1 & 2 & 0 \ 0 & 1 & 2 \ 2 & 0 & 1 \end{bmatrix}$.To find the adjoint of $A$,we calculate the cofactor matrix $C = [C_{ij}]$ and then take its transpose.The cofactors are calculated as follows:$C_{11} = +\begin{vmatrix} 1 & 2 \ 0 & 1 \end{vmatrix} = 1 - 0 = 1$$C_{12} = -\begin{vmatrix} 0 & 2 \ 2 & 1 \end{vmatrix} = -(0 - 4) = 4$$C_{13} = +\begin{vmatrix} 0 & 1 \ 2 & 0 \end{vmatrix} = 0 - 2 = -2$$C_{21} = -\begin{vmatrix} 2 & 0 \ 0 & 1 \end{vmatrix} = -(2 - 0) = -2$$C_{22} = +\begin{vmatrix} 1 & 0 \ 2 & 1 \end{vmatrix} = 1 - 0 = 1$$C_{23} = -\begin{vmatrix} 1 & 2 \ 2 & 0 \end{vmatrix} = -(0 - 4) = 4$$C_{31} = +\begin{vmatrix} 2 & 0 \ 1 & 2 \end{vmatrix} = 4 - 0 = 4$$C_{32} = -\begin{vmatrix} 1 & 0 \ 0 & 2 \end{vmatrix} = -(2 - 0) = -2$$C_{33} = +\begin{vmatrix} 1 & 2 \ 0 & 1 \end{vmatrix} = 1 - 0 = 1$The cofactor matrix is $C = \begin{bmatrix} 1 & 4 & -2 \ -2 & 1 & 4 \ 4 & -2 & 1 \end{bmatrix}$.The adjoint of $A$ is the transpose of the cofactor matrix:$adj(A) = C^T = \begin{bmatrix} 1 & -2 & 4 \ 4 & 1 & -2 \ -2 & 4 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 2 & 0 & 1 \end{bmatrix}$,then find $adj(A)$.

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