Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant/Proof 2

Proof
Let $\mathbf A = \begin {bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end {bmatrix}$.

Let $\mathbf A^{-1} = \begin {bmatrix} b_{11} & \cdots & b_{1n} \\ \vdots & \ddots & \vdots \\ b_{n1} & \cdots & b_{nn} \end {bmatrix}$.

Let $\tuple {\mathbf e_1, \mathbf e_2, \cdots, \mathbf e_n}$ be the standard ordered basis of $\R^n$.

Let $T: \R^n \to \R^n, \mathbf x \mapsto \map T {\mathbf x}$ be a linear transformation.

From Linear Transformation as Matrix Product, let:

Let $p, q \in \set {1, \dots, n}$.

Let $\mathbf I_n$ be the unit matrix of order $n$.

Let $A_p$ be the matrix obtained by replacing the $p$th column with $\mathbf e_q$.

Let $C_{q p}$ be the cofactor of $a_{q p}$ in $\map \det {\mathbf A_p}$.

Hence,