TITLE: Does $\mathrm{adj}(A)=\mathrm{adj}(B)$ imply $A=B$?
QUESTION [9 upvotes]: If $A$ and $B$ are any two square matrices of same order and if $\mathrm{adj}(A)=\mathrm{adj}(B)$, does it imply $A=B$?

I am pretty sure if $A$ and $B$ are invertible and if $A^{-1}=B^{-1}$, then $A=B$. 
So is it true for Adjoint?

REPLY [2 votes]: Indeed  
if $A^{-1}=B^{-1} $ then $AA^{-1}B=AB^{-1}B$  i.e.  $B=A $.  
If  $A,B$ are non-invertible then $\det(A)=\det(B)=0$ and
$\text{adj}(A)A=\det(A)I=0$ and $\text{adj}(B)B=\det(B)I=0$.
This can be satisfied without $A=B$.
Hence counterexamples for supposed theorem are, for example  
$\begin{bmatrix}  0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0  & 0 \ \end{bmatrix}$,
$\begin{bmatrix}  0 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0  & 0 \ \end{bmatrix}$.
Here of course $\text{adj}(A)=\text{adj}(B)$.