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$\newcommand{\adj}{\operatorname{adj}}\newcommand{\rank}{\operatorname{rank}}$If $\adj(A)$ denotes the classical adjoint and we are given that, for an $n \times n$ matrix $A$ over $\mathbb{R}$,

$$\rank(\adj(A)) = \begin{cases} n& \rank(A)=n\\1& \rank(A)=n-1\\0& \rank(A)<n-1\end{cases}$$

show that for $n \geq 2$

  1. $\det(\adj(A)) = \det(A)^{n-1};$
  2. $\adj(\adj(A)) = \det(A)^{n-2}A.$

The first is easy, actually. If $\rank(A) = n$, $\adj(A)A = \det(A)I$; taking determinants of both sides easily yields the result; if $\rank(A) < n$, $\det(A) = 0 = \det(\adj(A))$.

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marked as duplicate by user1551 linear-algebra Jan 4 '15 at 19:22

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For (1), if $rank(A) = n$, $adj(A)A = det(A)I$; taking determinants of both sides easily yields the result; if $rank(A) < n$, $det(A) = 0 = det(adj(A))$.

For (2), we use (1). If $rank A = n$, then

$$\begin{align} adj(adj(A))adj(A) &= det(adj(A))I\\ &= det(adj(A)) (1/det(A))adj(A)A\\ &= det(adj(A))^{n-2} adj(A)A adj(adj(A)) = det(adj(A))^{n-2}A, \end{align} $$

and, if $rank(A) < 0$, both sides are again zero.

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  • $\begingroup$ Some inaccuracies and minor bugs here. You probably don't want $rank(A) < 0$. Moreover, if $n = 2$, then $rank(A) < n$ does not imply $\operatorname{adj} A = 0$. $\endgroup$ – darij grinberg Sep 13 '16 at 16:26

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