**Adjugate Classical Adjoint**

The matrix formed by taking the transpose of the cofactor matrix of a given original matrix. The adjugate of matrix A is often written adj A.

Note: In the past, the term for adjugate used to be adjoint. The name has changed to avoid ambiguity with a different defintition of the term adjoint. Sometimes the adjugate is called the classical adjoint to refer to this older term. Consult a book on linear algebra for more information.

Example: | Find the adjugate of the following matrix: \[{\rm{A}} = \left[ {\begin{array}{*{20}c} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \\\end{array}} \right]\] |

Solution: | 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: \[\left[ {\begin{array}{*{20}{c}}{24}&{ - 12}&{ - 2}\\5&3&{ - 5}\\{ - 4}&2&4\end{array}} \right]\] |

**See also**

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