For given $n\times n$ matrix $A$ singular matrix, prove that $\operatorname{rank}(\operatorname{adj}A) \leq 1$
So from the properties of the adjugate matrix we know that
$$ A \cdot \operatorname{adj}(A) = \operatorname{det}(A)\cdot I$$
Since $A$ is singular we know that $\operatorname{det}(A) = 0$, thus
$$ A \cdot \operatorname{adj}(A) = 0$$
This is where I'm getting lost, I think I should say that for the above to happen one of the two, $A$ or $\operatorname{adj}(A)$ would have to be the $0$ matrix, but if $A = 0$ then $\operatorname{adj}(A) = 0$ for sure, which means I said nothing.
A leading hint is needed.
Best Answer
Since $A$ is singular, $\mbox{rank}A\leq n-1$.
Case 1: $\mbox{rank}A\leq n-2$. Then $A$ contains no invertible submatrix of order $n-1$. So every minor of order $n-1$ is zero. What can you conclude about $\mbox{adj}(A)$?
Case 2: $\mbox{rank}A= n-1$. By rank-nullity, we get $\dim\ker A=1$. Now $A\cdot \mbox{adj}(A)=0$ means that the range of $\mbox{adj}(A)$ is contained in $\ker A$. So...