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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?
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.
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
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, $$A. adj(A)=A. C^T=\sum_{k=1}^{n} a_{ik}c_{kj}^T=\sum_{k=1}^n a_{ik}c_{jk}$$$$=\det(A) \delta_{i,j}= \det(A) I$$
