How to prove that if $\det(A)=0$ then $\det(\operatorname{adj}(A))=0$? 
How to prove that if $\det(A)=0$ then $\det(\operatorname{adj}(A))=0$?

I have been trying to solve this but I can't use $$\det(A^{-1})=\det \Big(\frac{1}{\det(A)} \operatorname{adj}(A) \Big)$$ because $\det(A)=0$ and $\frac{1}{0}$ is not allowed.
 A: Let's write $A'$ for the adjoint of $A$. $AA'=(\det A)I$, so $\det A\det A'=(\det A)^n$ (where $A$ is an $n\times n$ matrix). If $\det A\ne0$, this yields $\det A'=(\det A)^{n-1}$. By continuity, this last equation is true even when $\det A=0$, and you're done. 
A: Let $B$ denote the adjugate matrix of $A$. Suppose for the sake of contradiction that $\det(B) \neq 0$. Then $B$ is invertible. Since the equation $AB =(\det{A})I = 0$ is true, we have then
$$AB\vec{v} = \vec{0}\ \ \forall \vec{v}$$
which implies $A$ is the zero matrix. But then the adjugate of the $0$ matrix is clearly $0$ itself which contradicts the fact that $B$ was invertible.
A: Hereafter, $A^*$ stands for adjoint, $|A|$ stands for the determinant of $A$.
Actually we have the conclusion that
$$
\operatorname{rank}(A^*) = \begin{cases}
    n   & \operatorname{rank}(A) = n   \\
    1 & \operatorname{rank}(A) = n-1 \\
    0   & \operatorname{rank}(A) = 0
  \end{cases}
$$
To prove the first one, notice that $A A^* = |A|E$ where $E$ is the identity matrix and $\operatorname{rank}(A)=n$ means $|A|\neq 0$. So $A^*$ is invertible and $\operatorname{rank}(A^*)=n$
To prove the second one, notice that $A A^* = |A|E = \mathbf{0}$. This means every column of $A^*$ is in the null space of $A$, which has dimension one. But the alternative definition of rank (see below) tells us $A^* \neq \mathbf{0}$, therefore $\operatorname{rank}(A^*)=1$.
To prove last one, simply use an alternative definition (or calculation) of rank of matrix: $\operatorname{rank}(A)$ is the size of its largest non-vanishing minor. See more at this wikipedia page under the subsection Determinantal rank – size of largest non-vanishing minor.
