First find the cofactor of each element.
\[\begin{array}{*{20}{l}}{{{\rm{A}}_{11}} = \left| {\begin{array}{*{20}{c}}4&5\\0&6\end{array}} \right| = 24}&{{{\rm{A}}_{12}} = - \left| {\begin{array}{*{20}{c}}0&5\\1&6\end{array}} \right| = 5}&{{{\rm{A}}_{13}} = \left| {\begin{array}{*{20}{c}}0&4\\1&0\end{array}} \right| = - 4}\\{}&{}&{}\\{{{\rm{A}}_{21}} = - \left| {\begin{array}{*{20}{c}}2&3\\0&6\end{array}} \right| = - 12}&{{{\rm{A}}_{22}} = \left| {\begin{array}{*{20}{c}}1&3\\1&6\end{array}} \right| = 3}&{{{\rm{A}}_{23}} = - \left| {\begin{array}{*{20}{c}}1&2\\1&0\end{array}} \right| = 2}\\{}&{}&{}\\{{{\rm{A}}_{31}} = \left| {\begin{array}{*{20}{c}}2&3\\4&5\end{array}} \right| = - 2}&{{{\rm{A}}_{32}} = - \left| {\begin{array}{*{20}{c}}1&3\\0&5\end{array}} \right| = - 5}&{{{\rm{A}}_{33}} = \left| {\begin{array}{*{20}{c}}1&2\\0&4\end{array}} \right| = 4}\end{array}\]
As a result the cofactor
matrix of A is
\[\left[ {\begin{array}{*{20}{c}}{24}&5&{ - 4}\\{ - 12}&3&2\\{ - 2}&{ - 5}&4\end{array}} \right]\]
Finally the adjugate of A is the transpose of the cofactor
matrix:
\[\left[ {\begin{array}{*{20}{c}}{24}&{ - 12}&{ - 2}\\5&3&{ - 5}\\{ - 4}&2&4\end{array}} \right]\] |