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 $\mathrm{adj}(A)\cdot A = A\cdot\mathrm{adj}(A)$?

Since $A^{-1}=\frac{1}{\mathrm{det}(A)}\mathrm{adj}(A)$ and $A$ represents a finite dimensional linear operator, so a left inverse is also a right inverse i.e. $AA^{-1}=A^{-1}A$. From this is follows easily that $\mathrm{adj}(A)\cdot A = A\cdot\mathrm{adj}(A)$ by canceling the scalar $\frac{1}{\mathrm{det}(A)}$ from both side.
