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 $A$ be an $n \times n$ matrix, $A_{i,j}$ the $(i,j)$-minor of $A$ and $C_{i,j}$ the $(i,j)$-cofactor of $A$, defined as:
$$
C_{i,j} = (-1)^{i+j}A_{i,j}.
$$
By definition we know that the adjungate of $A$ is:
$$
\operatorname{adj} A = [C_{j,i}].
$$
The cofactor expansion along rows gives for all $i,j=1,\dots,n$:
$$
\sum_{k=1}^{n} a_{i,k} C_{j,k} = \delta_{i,j}\det A,
$$
and along columns gives for all $i,j=1,\dots,n$:
$$
\sum_{k=1}^n a_{k,i}C_{k,j} = \delta_{i,j}\det A,
$$
where $\delta_{i,j}$ is the Kronecker delta.
You can express these equations using the definition of the adjungate matrix as the following:
$$
A \cdot \operatorname{adj} A = \det A \cdot I_n,
$$
and
$$
\operatorname{adj} A \cdot A = \det A \cdot I_n,
$$
where $I_n = [\delta_{i,j}]$ is the identitiy matrix of size $n \times n$. From here we have that
$$
A \cdot \operatorname{adj} A = \operatorname{adj} A \cdot A = \det A \cdot I_n.
$$
