prove: $\left(A^*\right)^*=\left|A\right|^{n-2}A$ Suppose square matrix $A$ with order-n, and $A^*$ is it's adjugate matrix, when $n>2$, prove: $\left(A^*\right)^*=\left|A\right|^{n-2}A$
Proof:



*

*when $A$ is invertible, $A^{-1}=\frac{1}{|A|}A^*$, so $A^*=|A|A^{-1}$, and then


\begin{align}
 \left(A^*\right)^*
  = \left|\left(|A|A^{-1}\right)\right|
         \left(|A|A^{-1}\right)^{-1}
  = \left|A\right|^n\left|A\right|^{-1}A|A|^{-1}
  = \left|A\right|^{n-2}A
\end{align}


*

*how about the case $A$ is not invertible?


$|A|=0$, then seems we should show $\left(A^*\right)^*=0$
 A: You don't need to prove anything different for the singular case.
You can do your calculations in the field of rational functions in $n^2$ indeterminates $x_{ij}$ $(i,j=1,2,\dots,n)$ about the matrix $X$ having those indeterminates as coefficients, which is clearly invertible. Thus the relation
$$
\operatorname{Adj}(\operatorname{Adj}(X))=(\det X)^{n-2}X
$$
holds true with the same proof as above; but this expression lives in the ring of polynomials in those indeterminates, so substituting $a_{ij}$ for $x_{ij}$ gives a true expression.
Note that the relation holds also for $n=2$, for
$$
\operatorname{Adj}(\operatorname{Adj}(A))=A
$$
for $2\times2$ matrices, but only if we set $0^0=1$.
For $1\times1$ matrices, it makes sense to define $\operatorname{Adj}(A)=1$ and the relation is still true in the form $\det(A)\operatorname{Adj}(A)=A$.
A: The relation that is always true is
$$\text{Adj}\,A\cdot A=|A|\cdot I\implies |\text{Adj}\,A||A|=|A|^n\,,\,\,\text{and from here that if $A$ isn't invertible} :$$
$$ \text{Adj}\,A\cdot A=0\cdot I$$
