In the proof of adjoints, it is sufficient to show $A(\frac{1}{det(A)} adj(A)) = I$.
It went through a series of co-factor expansion along the $i^{th} row$ and $j^{th} row$, and this resulted in
$A[adj(a)]=(b_{ij})_{n×n}$
$$ = \begin{bmatrix} det(A) & 0 & \cdots & 0 \\ 0 & det(A) & \ddots & 0 \\ 1 & \vdots & \ddots & 0 \\ 1 & \cdots & 0 & det(A) \\ \end{bmatrix} $$
$$ = det(A)I $$
and that's how it gets to $A(\frac{1}{det(A)} adj(A))$.
Now, the whole problem is I don't get the whole idea of the co-factor portion and also how does $(b_{ij})_{n×n}$ leads to the chunk of $det(A)$ identity matrix as shown above.
Thanks for the help. ^^