Why $\mathrm{adj}(A)\cdot A = A\cdot\mathrm{adj}(A)$? I know that $A\cdot\mathrm{adj}(A) = \det(A) \cdot I$, but why $\mathrm{adj}(A)\cdot A = A\cdot\mathrm{adj}(A)$?

Let $A\in M_n(K)$. If $K=\mathbb{C}$, then use the Travis argument. If $K$ is a commutative ring with unity, then use the reference  (Bill's answer) 
Sylvester's determinant identity
given above by Bigbear. 
EDIT. @  user3697301 


*

*For $K=\mathbb{C}$, Travis in his comment below, gave the key for a complete proof of (*): $adj(A).A=A.adj(A)=\det(A)I_n$. If you do not work, then you cannot do mathematics.

*That is interesting is that (*) holds also when $K$ is a commutative ring with unity and the key of the proof is in the MSE reference above.
