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 adjugate matrix can be constructed the following way:

$$ adj( \begin{bmatrix} p &q & r \end{bmatrix}) = \begin{bmatrix} q \times r & p \times r & p \times q \end{bmatrix}$$

Where $\times$ is the cross product operator. It is clear that if we multiply ,

$$\begin{bmatrix} q \times r \\ p \times r \\ p \times q \end{bmatrix} \begin{bmatrix} p & q &r \end{bmatrix} = \det A$$

If one doesn't see the above immediately, expand everything out in components and do it. The above construction makes everything clear, and in my personal opinion, also easier to compute.

extracted from [here](https://math.stackexchange.com/questions/3294679/what-exactly-is-minors-adjoints-and-adjugates/4453894#4453894)
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, $$A. adj(A)=A. C^T=\sum\_{k=1}^{n} a\_{ik}c\_{kj}^T=\sum\_{k=1}^n a\_{ik}c\_{jk}$$$$=\det(A) \delta\_{i,j}= \det(A) I$$