Does every invertible matrix A has a matrix B such that A=Adj(B)?

Question

I'm trying to understand if it's always true, always true over $\mathbb C$ or never true.

I know that if $A$ is invertible, than there exists $A^{-1}$. $$A=\frac{1}{det (A^{-1})}Adj(A^{-1})$$ So I have an adjoint matrix multiplied by a scalar, but how do I know if the result is an adjoint by itself?

Answer 1

There is no solution over $\mathbb R$ if $n \ge 3$ is odd and $\det(A) < 0$.

$\det(\text{adj}(B))= \det(\det(B) B^{-1}) = \det(B)^{n-1}$, which can't be negative in this case.

Answer 2

The case $n=1$ is immediate. For the case $n\ge 2$, by user1551's comment, $${\rm adj}({\rm adj}{(A)})=(\det A)^{n-2}A.$$ Also, there is a fact that ${\rm adj}(kA)=k^{n-1}{\rm adj}(A)$ for any scalar $k\in\mathbb{C}$. Thus we have \begin{align} {\rm adj}((\det A)^{-\frac{n-2}{n-1}}{\rm adj}{(A)}) &=(\det A)^{-(n-2)}{\rm adj}({\rm adj}{(A)})\\ &=(\det A)^{-(n-2)}(\det A)^{n-2}A\\ &=A, \end{align} and the result follows by taking $B=(\det A)^{-\frac{n-2}{n-1}}{\rm adj}{(A)}$.