If $A =\begin{pmatrix}1 & 2 & 0 \\ 0 & 1 & 2 \\ 2 & 0 & 1\end{pmatrix}$ , then adj A
A. $\begin{pmatrix}1 & 4 & -2 \\ -2 & 1 & 4 \\ 4 & -2 & 1\end{pmatrix}$
B.$\begin{pmatrix}1 & -2 & 4 \\ 4 & 1 & -2 \\ -2 & 4 & 1\end{pmatrix}$
C.$\begin{pmatrix}1 & 2 & 4 \\ -4 & 1 & 2 \\ -4 & -2 & 1\end{pmatrix}$
D. None of these

Hint:
We are given a matrix A and need to find its adjoint. Recall that adj A is the transpose of the cofactor matrix of A.

Complete step-by-step answer:
Given:
$A =\begin{pmatrix}1 & 2 & 0 \\ 0 & 1 & 2 \\ 2 & 0 & 1\end{pmatrix}$
To find the $adj A$ we need to find the minor of each element of A and then, the cofactor matrix.
Now, we find the minor of each element of A.
$M_{11} = \begin{vmatrix}1 & 2 \\ 0 & 1\end{vmatrix} = 1\times1 - (2\times0) = 1$
$M_{12} = \begin{vmatrix}0 & 2 \\ 2 & 1\end{vmatrix} = 0\times0 - (2\times2) = -4$
$M_{13} = \begin{vmatrix}0 & 1 \\ 2 & 0\end{vmatrix} = 0\times0 - (1\times2) = -2$
$M_{21} = \begin{vmatrix}2 & 0 \\ 0 & 1\end{vmatrix} = 2\times1 - (0\times0) = 2$
$M_{22} = \begin{vmatrix}1 & 0 \\ 2 & 1\end{vmatrix} = 1\times1 - (0\times2) = 1$
$M_{23} = \begin{vmatrix}1 & 2 \\ 2 & 0\end{vmatrix} = 1\times0 - (2\times2) = -4$
$M_{31} = \begin{vmatrix}2 & 0 \\ 1 & 2\end{vmatrix} = 2\times2 - (0\times1) = 4$
$M_{32} = \begin{vmatrix}1 & 0 \\ 0 & 2\end{vmatrix} = 1\times2 - (0\times0) = 2$
$M_{33} = \begin{vmatrix}1 & 2 \\ 0 & 1\end{vmatrix} = 1\times1 - (2\times0) = 1$
To find the co-factor for each element
$C_{11} = (-1)^{1+1}M_{11} = (-1)^{2}\times(1) = 1$
$C_{12} = (-1)^{1+2}M_{12} = (-1)^{3}\times(-4) = (-1)\times(-4) = 4$
$C_{13} = (-1)^{1+3}M_{13} = (-1)^{4}\times(-2) = 1\times (-2) = -2$
$C_{21} = (-1)^{2+1}M_{21} = (-1)^{3}\times(2) = (-1)\times2 = -2$
$C_{22} = (-1)^{2+2}M_{22} = (-1)^{4}\times(1) = 1$
$C_{23} = (-1)^{2+3}M_{23} = (-1)^{5}\times(-4) = (-1)\times(-4) = 4$
$C_{31} = (-1)^{3+1}M_{31} = (-1)^{4}\times(4) = 4$
$C_{32} = (-1)^{3+2}M_{32} = (-1)^{5}\times(2) = (-1)\times2 = -2$
$C_{33} = (-1)^{3+3}M_{33} = (-1)^{6}\times(1) = 1$
Therefore, the co-factor matrix of A is,$\begin{pmatrix}1 & 4 & -2 \\ -2 & 1 & 4 \\ 4 & -2 & 1\end{pmatrix}$
Now, adj A = Transpose of co-factor matrix of A = $\begin{pmatrix}1 & 4 & -2 \\ -2 & 1 & 4 \\ 4 & -2 & 1\end{pmatrix}^{T} = \begin{pmatrix}1 & -2 & 4 \\ 4 & 1 & -2 \\ -2 & 4 & 1\end{pmatrix}$

Therefore, the answer is Option B.

Note: While finding the co-factor of the matrix be careful with the signs as one change in the sign will lead to the wrong answer. Also, keep in mind to take the transpose of the co-factor matrix to find the adjoint, most of the time, we will have the co-factor matrix in the options, so always check whether you have taken the transpose or not.