Given a matrix $A$ of rank $n$, show that $\det(\operatorname{adj}(A))=\det(A)^{n-1}$ and $\operatorname{adj}(\operatorname{adj}(A))=(\det A)^{n-2}A$

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  • Prove that if $A$ is regular then $\operatorname{adj}(\operatorname{adj}(A)) = (\det A)^{n-2} A$

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Solutions Collecting From Web of "Given a matrix $A$ of rank $n$, show that $\det(\operatorname{adj}(A))=\det(A)^{n-1}$ and $\operatorname{adj}(\operatorname{adj}(A))=(\det A)^{n-2}A$"

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.