Question: Given a matrix $A$ which is invertible and with determinant>0, then $ \operatorname{adj} A$ is the matrix such that:

$$ (\operatorname{adj} A) A = A (\operatorname{adj} A) = (\det A) I \tag{1}$$

The construction of $\operatorname{adj} A$ is that we take the cofactor matrix then it's transpose. How would we show that the construction would be the one that leads to property (1) intuitively?

Answer: 
The definition of the adjugate matrix is usually over the cofactor matrix, which is unique and defined for every square matrix $A$ (even if $\det A = 0)$. More precisely, denote by $M\_{ij}$ the submatrix of $A$, which is obtained by removing the $i$th row and $j$th column of $A$. Then, the $ij$th entry of the adjugate matrix is defined to be
$$ [\mathrm{adj}(A)]\_{ij} = (-1)^{i+j} \det M\_{ji}. $$
This way, we can directly proof that (1) is fulfilled. Looking at the diagonal entries of $\mathrm{adj}(A)A$, we obtain
$$ [A~\mathrm{adj}(A)]\_{ii} = \sum\_{j} [A]\_{ij}[\mathrm{adj}(A)]\_{ji} = \sum\_{j} (-1)^{i+j} [A]\_{ij} \det M\_{ij} = \det(A), $$
where in the last step, we identified the sum as the Laplace expansion of the determinant, see e.g., <https://en.wikipedia.org/wiki/Determinant#Laplace_expansion>. Similarly, the non-diagonal elements
$$ [A~\mathrm{adj}(A)]\_{ij} = \sum\_{k} [A]\_{ik}[\mathrm{adj}(A)]\_{kj} = \sum\_{k} (-1)^{k+j} [A]\_{ik} \det M\_{jk} = \det(A\_{j\to i}), $$
where $A\_{j\to i}$ is the matrix obtained from $A$ by replacing the $j$-th row of $A$ by the $i$-th row. This matrix contains twice the $i$-th row and thus is singular. Hence, $\det(A\_{j\to i}) = 0$ and the non-diagonal elements of $A~\mathrm{adj}(A)$ vanish. This proves that
$$ A~\mathrm{adj}(A) = I\det A. $$
The proof for $\mathrm{adj}(A)A$ works analogously.
One can understand this connestrong textction in simple terms as:

Suppose $A=[a\_{ij}]\_{n \times n}$ and let its cofactor matrix be $[c\_{ij}]\_{n \times n}$, then we know
that $$\sum\_{i=1}^{n} a\_{p,j} c\_{q,j}=\det(A) \delta\_{p.q},$$
where $\delta\_{p,q}=1, ~if~ p=q; 0$, otherwise.
Next, $$A. adj(A)=A. C^T=\sum\_{k=1}^{n} a\_{ik}c\_{kj}^T=\sum\_{k=1}^n a\_{ik}c\_{jk}$$$$=\det(A) \delta\_{i,j}= \det(A) I$$