Is adjoint of singular matrix singular? What would be its rank? Let $A$ be a square and singular matrix of order $n$.
Is $\operatorname{adj}(A)$ necessarily singular? What would be the rank of $\operatorname{adj}(A)$?
 A: Hint : $$A \cdot\mbox{adj}(A)=\det(A) I$$
If $\det(A)=0$, we get  $A \cdot\mbox{adj}(A)=0$. Can $\mbox{adj}(A)$ be invertible?
For the rank, if you are familiar with linear transformations, prove that the above relation implies that the image of the transformation defined by $A$ must be in the kernel of the transformation defined by $\mbox{adj}(A)$. This yields an inequality of ranks.
A: No, the adjugate of a singular matrix can be non-singular. But it happens only for the $1\times 1$ zero matrix.
Here is a complete classification, referring to this answer and this one. One always has $$\def\adj{\operatorname{adj}}A \cdot \adj(A) = \det(A) I_n.$$


*

*If $A$ has rank$~n$, then it is invertible, and so is $\det(A)$, and $\adj(A)=\det(A)A^{-1}$ is invertible too, and has rank$~n$.

*If $A$ has rank$~n-1$ then at least one $(n-1)\times(n-1)$ minor is nonzero, and so $\adj(A)\neq0$. On the other hand by the given relation the image of $\adj(A)$ is contained in the kernel of $A$ which has dimension$~1$ by rank-nullity; it follows that $\adj(A)$ has rank$~1$ in this case.

*If $A$ has rank${}<n-1$ then all $(n-1)\times(n-1)$ minors are equal to zero, and so $\adj(A)$ has rank$~0$.
The cases where $\adj(A)$ has rank$~n$ are the first case for any$~n$, and the second case for $n=1$. (And, I would be inclined to say, the last case for $n=0$; but that of course cannot happen at all.) So the only case where $A$ is singular but $\adj(A)$ is not, is the case $A=(0)$ (with $n=1$).
