Talk:Value of Adjugate of Determinant

29 October 2019
The Theorem statement:


 * Expected in the statement are $n \times n$ matrices $A$ and $\adj {A}$.


 * A determinant value $D^*$ cannot be called an adjugate matrix.


 * Notation det(A) crosses language barriers. Notation D depends on the English sentence defining it.


 * But this is an English language website, I don't understand what you mean. --prime mover (talk) 04:09, 29 October 2019 (EDT)


 * If $D=0$ (1x1 matrix), then $D^{n-1}$ is $0^0$. What is true: $\det \paren { \adj {A} } \det \paren {A} = \paren{ \det \paren {A} }^n$.

The proof:


 * One tool is the adjugate identity: $\adj {A} \, A = A \, \adj {A} = \det \paren {A} I$, valid for any square matrix $A$.


 * Another tool is the determinant product theorem $\det \paren {EF} = \det \paren {E} \det \paren {F}$.


 * Equalities $\det \paren { c X } = \det \paren { \paren {c I} X} = \det \paren {cI} \det \paren {X} = c^n \det \paren {X}$ apply.