Adjoint of a Square Matrix

Let $A=[a_{ij}]$ be a square matrix of order $n$, with determinant $|A|$ (also written $\det(A)$). Deleting row $i$ and column $j$ of $A$ leaves a sub-matrix of order $(n-1)$; its determinant is the minor $M_{ij}$ of the entry $a_{ij}$. The cofactor of $a_{ij}$ is the signed minor

Key row/column fact. The sum of the products of a row's entries with their own cofactors reproduces the determinant: $a_{i1}A_{i1}+a_{i2}A_{i2}+\cdots+a_{in}A_{in}=|A|$ (Laplace expansion along row $i$). But the sum of a row's entries against a different row's cofactors is always $0$: $a_{i1}A_{k1}+a_{i2}A_{k2}+\cdots+a_{in}A_{kn}=0$ for $i\ne k$ (this determinant would have two identical rows, hence vanish).

Definition (adjoint). Replace every entry $a_{ij}$ of $A$ by its cofactor $A_{ij}$ to get the matrix of cofactors; the adjoint of $A$, written $\operatorname{adj}A$, is the transpose of the matrix of cofactors:

So the $(i,j)$ entry of $\operatorname{adj}A$ is the cofactor $A_{ji}$ of the transposed position.

Theorem 1.1 (the central identity). For every square matrix $A$ of order $n$,

Proof idea. The $(i,i)$ entry of the product $A(\operatorname{adj}A)$ is $\sum_k a_{ik}A_{ik}=|A|$ (own-row-cofactor sum), while every off-diagonal $(i,k)$ entry, $i\ne k$, is $\sum_j a_{ij}A_{kj}=0$ (mismatched-row-cofactor sum) -- so the product is exactly $|A|$ down the diagonal and $0$ elsewhere, i.e. $|A|I_n$. The same argument applied column-wise gives $(\operatorname{adj}A)A=|A|I_n$ too. …