I am asked to show the determinant of adjoint of n by n matrix is $n$ or $0$ or $1$. I know that $\det(adj A) =[ \det (A) ]^{(n-1)}$. I also know inverse of A in terms of $\det (A)$. Any help will be appreciated.
[Math] To find determinant of adjoint of a square matrix
linear algebra
Best Answer
First, the relation you say you know is true and easy to prove: we indeed have that
$$\color{red}{(**)}\;\;\;A\cdot\text{Adj}\,A=|A|\cdot I\implies |A|\cdot|\text{Adj}\,A|=|A|^n$$
so if $\;A\;$ is regular then we indeed have $\;|\text{Adj}\,A|=|A|^{n-1}\;$ . Now, this equals $\;1,\,n\;$ iff $\;|A|^{n-1}=1,\,n\;$ , so $\;|A|\;$ must then be an $\;(n-1)\,-$ th root of $\;1\;$ or of $\;n\;$.
On the other hand, if $\;A\;$ is singular then $\;|A|=0\;$ , so from $\;(**)\;$ above we get $\;A\cdot\text{adj}\,A=0\;$ and this means $\;\text{Adj}\,A\;$ is singular, too nand thus $\;|\text{Adj}\,A|=0\;$