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

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

#### 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.