What is the $n$-time iterated adjugate of an $n\times n$ matrix $A$? 
What is $\underbrace{\text{adj}\Big(\text{adj}\big(\ldots(\text{adj}}_{n\text{ adj}}\ A)\ldots\big)\Big)$, where $\text{adj}$ is written $n$ times, and the order of the matrix $A$ is $n\times n$?

Can you show the proof for each $n$ (I mean by induction)!!
 A: Hint. We know that
$$
\mathrm{adj}\,A\cdot A=\det A\cdot I, \tag{1}
$$
and hence, if $\det A\ne 0$,
$$
\mathrm{adj}\,A=\det A \cdot A^{-1},
$$
and
$$
\det(\mathrm{adj}\,A)\cdot \det A=(\det A)^n.
$$
Hence $\det(\mathrm{adj}\,A)=(\det A)^{n-1}$.
Now $(1)$ implies that
$$
\mathrm{adj}(\mathrm{adj}\,A)\cdot \mathrm{adj}\,A=\det (\mathrm{adj}\,A)\cdot I
=(\det A)^{n-1}I,
$$
and hence
$$
\mathrm{adj}(\mathrm{adj}\,A)=(\det A)^{n-1}\cdot(\mathrm{adj}\,A)^{-1}=(\det A)^n\cdot A
$$
and
$$
\det\big(\mathrm{adj}(\mathrm{adj}\,A)\big)=\big(\det(\mathrm{adj}\,A)\big)^{n-1}=(\det A)^{(n-1)^2}, \quad \det(\mathrm{adj}^k\,A)=(\det A)^{(n-1)^k}
\tag{2}$$
and
$$
\mathrm{adj}^{k+1}\,A\cdot \mathrm{adj}^{k}\,A=\det (\mathrm{adj}^k\,A)\cdot I
=(\det A)^{(n-1)^k}I, \tag{3}
$$
Finally, we obtain that:
$$
\mathrm{adj}^{2k}A=c_{k}A, \quad \mathrm{adj}^{2k+1}A=d_{k}A^{-1}
$$
where $c_k,d_k$ can be obtained from (2) and (3).
A: Hint.  For any integer $n\geq 2$, for any $n$-by-$n$ matrix $X$, and for each scalar $k$, show that
$$\text{adj}(k\,X)=k^{n-1}\,\text{adj}(X)\,,$$ $$\det\big(\text{adj}(X)\big)=\big(\det(X)\big)^{n-1}\,,$$ and $$\text{adj}\big(\text{adj}(X)\big)=\big(\det(X)\big)^{n-2} X\,.$$  Here, $0^0:=1$.  (For $n=1$, $\text{adj}(X)=1$ always.)

  For each $m\in\mathbb{Z}_{\geq 0}$, let $\text{adj}^m$ be the $m$-time iteration of $\text{adj}$.  Then, for any $n$-by-$n$ matrix $A$, $$\text{adj}^m(A)=\begin{cases}\big(\det(A)\big)^{\frac{(n-1)^{m}-1}{n}}\,A&\text{if }m\text{ is even}\,, \\ \big(\det(A)\big)^{\frac{(n-1)^{m}-(n-1)}{n}}\,\text{adj}(A)&\text{if }m\text{ is odd}\,. \end{cases}\tag{*}$$  For $n=1$, $$\text{adj}^0(A)=A\,,\text{ and }\text{adj}^m(A)=1\text{ for any }m\in\mathbb{Z}_{>0}\,.$$  For $n=2$,  $$\text{adj}^m(A)=\begin{cases}A&\text{if }m\text{ is even}\,, \\ \text{adj}(A)&\text{if }m\text{ is odd}\,. \end{cases}$$    Clearly, if $n\geq 3$ and $\det(A)=0$, then $$\text{adj}^0(A)=A\,,\,\, \text{adj}^1(A)=\text{adj}(A)\,,\text{ and }\text{adj}^m(A)=0\text{ for all }m\geq 2\,.$$   If $\det(A)\neq 0$, then you may simplify (*) to $$\text{adj}^m(A)=\big(\det(A)\big)^{\frac{(n-1)^{m}-(-1)^m}{n}}\,A^{(-1)^m}\text{ for every }m\in\mathbb{Z}_{\geq 0}\,.$$  In particular, $$\text{adj}^n(A)=\begin{cases}\big(\det(A)\big)^{\frac{(n-1)^{n}-1}{n}}\,A&\text{if }n\text{ is even}\,, \\ \big(\det(A)\big)^{\frac{(n-1)^{n}-(n-1)}{n}}\,\text{adj}(A)&\text{if }n\text{ is odd}\,. \end{cases}$$

