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(D) To find the adjoint of matrix $A$,we first find the matrix of cofactors $C_{ij}$.Given $A = \begin{bmatrix} 1 & 0 & 0 \ 5 & 2 & 0 \ -1 & 6 & 1 \

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None of these

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(D) To find the adjoint of matrix $A$,we first find the matrix of cofactors $C_{ij}$.Given $A = \begin{bmatrix} 1 & 0 & 0 \ 5 & 2 & 0 \ -1 & 6 & 1 \end{bmatrix}$.The cofactors are:$C_{11} = +\begin{vmatrix} 2 & 0 \ 6 & 1 \end{vmatrix} = 2 - 0 = 2$$C_{12} = -\begin{vmatrix} 5 & 0 \ -1 & 1 \end{vmatrix} = -(5 - 0) = -5$$C_{13} = +\begin{vmatrix} 5 & 2 \ -1 & 6 \end{vmatrix} = 30 - (-2) = 32$$C_{21} = -\begin{vmatrix} 0 & 0 \ 6 & 1 \end{vmatrix} = -(0 - 0) = 0$$C_{22} = +\begin{vmatrix} 1 & 0 \ -1 & 1 \end{vmatrix} = 1 - 0 = 1$$C_{23} = -\begin{vmatrix} 1 & 0 \ -1 & 6 \end{vmatrix} = -(6 - 0) = -6$$C_{31} = +\begin{vmatrix} 0 & 0 \ 2 & 0 \end{vmatrix} = 0 - 0 = 0$$C_{32} = -\begin{vmatrix} 1 & 0 \ 5 & 0 \end{vmatrix} = -(0 - 0) = 0$$C_{33} = +\begin{vmatrix} 1 & 0 \ 5 & 2 \end{vmatrix} = 2 - 0 = 2$Thus,the cofactor matrix is $C = \begin{bmatrix} 2 & -5 & 32 \ 0 & 1 & -6 \ 0 & 0 & 2 \end{bmatrix}$.The adjoint of $A$ is the transpose of the cofactor matrix:$adj(A) = C^T = \begin{bmatrix} 2 & 0 & 0 \ -5 & 1 & 0 \ 32 & -6 & 2 \end{bmatrix}$.Comparing this with the given options,none of them match. Therefore,the correct option is $(D)$.

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 5 & 2 & 0 \\ -1 & 6 & 1 \end{bmatrix}$,then the adjoint of $A$ is

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