Let \[A = \left[ {\begin{array}{*{20}{c}}1&0&0\\5&2&0\\{ - 1}&6&1\end{array}} \right]\] then the adjoint of \[{\rm{A}}\] is
A. \[\left[ {\begin{array}{*{20}{c}}2&{ - 5}&{32}\\0&1&{ - 6}\\0&0&2\end{array}} \right]\]
В. \[\left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\{ - 5}&{ - 2}&0\\1&{ - 6}&1\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\{ - 5}&{ - 2}&0\\1&{ - 6}&{ - 1}\end{array}} \right]\]
D. None of these

Hint:
We must first find the adjoint of a matrix by taking the cofactor of each element in the matrix and then taking the matrix's transpose. The adjoint of a matrix \[A = {(a_{ij})_{n \times n}}\] is defined as the transpose of the matrix \[{(A_{ij})_{n \times n}}\] where \[(A_{ij})\] is the element's cofactor \[(a_{ij})\]

Formula Used:
Cofactor can be found using formula:
\[{C_{ij}} = {( - 1)^{i + j}}{M_{ij}}\]

Complete Step by Step Solution:
We have been provided in the question the \[3 \times 3\] matrix ‘A’
The \[3 \times 3\] matrix of A is
\[A = \left[ {\begin{array}{*{20}{c}}1&0&0\\5&2&0\\{ - 1}&6&1\end{array}} \right]\]
First we have to determine the cofactor of the matrix
We have been already known that the cofactor matrix is made up of all the cofactors of the provided matrix, which are determined using the formula \[{C_{ij}} = {( - 1)^{i + j}}{M_{ij}}\] where minor is \[{M_{ij}}\]
\[{C_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}2&0\\6&1\end{array}} \right| = 2\]
\[{C_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}5&0\\{ - 1}&1\end{array}} \right| = - 5\]
\[{C_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}5&2\\{ - 1}&6\end{array}} \right| = 32\]
\[{C_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{*{20}{c}}0&0\\6&1\end{array}} \right| = 0\]
\[{C_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{*{20}{c}}1&0\\{ - 1}&1\end{array}} \right| = 1\]
\[{C_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{*{20}{c}}1&0\\{ - 1}&6\end{array}} \right| = - 6\]
\[{C_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{*{20}{c}}0&0\\2&0\end{array}} \right| = 0\]
\[{C_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{*{20}{c}}1&0\\5&0\end{array}} \right| = 0\]
\[{C_{33}} = {( - 1)^{3 - 3}}\left| {\begin{array}{*{20}{l}}1&0\\5&2\end{array}} \right| = 2\]
The cofactor matrix is \[\left[ {\begin{array}{*{20}{c}}2&{ - 5}&{32}\\0&1&{ - 6}\\0&0&2\end{array}} \right]\]
\[ \Rightarrow {\mathop{\rm adj}\nolimits} (A) = {\left[ {\begin{array}{*{20}{c}}2&{ - 5}&{32}\\0&1&{ - 6}\\0&0&2\end{array}} \right]^T}\]
Now, we have to determine the transpose of co-factor matrix
Here, rows becomes columns and columns becomes rows, we have
The matrix’s transpose is
\[ = \left[ {\begin{array}{*{20}{c}}2&0&0\\{ - 5}&1&0\\{32}&{ - 6}&2\end{array}} \right]\]
Therefore, the adjoint of \[{\rm{A}}\] is \[\left[ {\begin{array}{*{20}{c}}2&0&0\\{ - 5}&1&0\\{32}&{ - 6}&2\end{array}} \right]\]
Hence, the option D is correct

Note: In this type of question, we must first get the cofactor of each element, and then use transpose to find the adjoint of the provided matrix. Students must be mindful of signs when calculating cofactors and transposing the matrix. When the adjoint matrix is evaluated, the value must be the transpose of the obtained value. Another consideration is that anytime the adjoint matrix is assessed, we must ensure that the cofactor and minor of that matrix have the correct sign.