Prove that if A is singular, then adj(A) is also singular Prove that if A is singular, then adj(A) is also singular.
How do you prove this without proving by contradiction?
 A: We know that $A\;adj(A)=det(A)\;I=diag(det(A))$. Since $A$ is singular, we have $det(A)=0$, which implies that $A\; adj(A)=0$. 


*

*If $A=0$ (null and singular), then $adj(A)=0$ and hence $adj(A)$ is
singular too.

*If $A\neq 0$ (nonnull and singular), then $A$ contains a nonnull row,
say the $i$th row $a'_i$. It follows that $$a'_i\; adj(A)=0$$ which
implies that the rows of $adj(A)$ are linearly dependent, and hence
$adj(A)$ is singular.

A: Assuming you are talking about the adugate matrix. If $A$ is an $n\times n$ matrix, then if $A$ is singular 
$$
\operatorname{adj}A A=\det A1_{n\times n} = 0.
$$
There are two cases, if $\operatorname{rank}A=0$ the assertion is trivial. If $\operatorname{rank}A\neq0$ then $\operatorname{null}\operatorname{adj}A>0$, so it can't be invertible.
A: From the basic equation: $A\,\text{adj}(A)=\det(A)I$ we have:
$$
\det(A\,\text{adj}A)=\det(\det(A)I)\\
\det(A)\det(\text{adj}(A))=\det(A)^n \det(I)
$$
$\det(\text{adj}(A))=\det(A)^{n-1}$
Since $A$ is singular,
$$
\det(\text{adj}(A))=0 
$$
and then $\text{adj}(A)$ is also singular 
