TITLE: Proof for $A^{-1}=\frac{adj(A_j)}{det(A)}$
QUESTION [5 upvotes]: I have a proof for the method to find the inverse by taking the adjugate of the matrix over the determinant. I believe that it is correct; however, I would like to have it checked over before I submit it for bonus marks. The proof goes as follows:
Let $A\in\mathbb{M}_{n\times{n}}(\mathbb{R})$ be an $n\times{n}$ invertible matrix. Let $A^{-1}=\begin{bmatrix}\vec{x_1}\cdots\vec{x_n}\end{bmatrix}$.
$AA^{-1}=A\begin{bmatrix}\vec{x_1}\cdots\vec{x_n}\end{bmatrix}=\begin{bmatrix}A\vec{x_1}\\\vdots\\A\vec{x_n}\end{bmatrix}=\begin{bmatrix}\vec{e_1}^T\\\vdots\\\vec{e_n}^T\end{bmatrix}$.
$\therefore A\vec{x_i}=\vec{e_i}^T\text{for } 1\leq{i}\leq{n}$
By Cramer's Rule, $x_{ji}=\frac{det(A_j)}{det(A)}\text{for }1\leq{i,j}\leq{n}$ where $A_j$ is the matrix formed by replacing the jth column of $A$ with the solution vector; in this case, $\vec{e_i}$.
By properties of determinants, we can add a multiple of a column to another column without changing the determinant.
$\therefore det(A_j)\\=det(\begin{bmatrix}\vec{a_1}\cdots\vec{a_{j-1}}\space\space\space\space\vec{e_i}\space\space\space\space\vec{a_{j+1}}\cdots\vec{a_n}\end{bmatrix})\\=det(\begin{bmatrix}\vec{a_1}-a_{1i}\vec{e_i}\cdots\vec{a_{j-1}}-a_{j-1,i}\vec{e_i}\space\space\space\space\vec{e_i}\space\space\space\space\vec{a_{j+1}}-a_{j+1,i}\vec{e_i}\cdots\vec{a_n}-a_{ni}\vec{e_i}\end{bmatrix})$
This matrix has all 0s in the ith row except for a 1 in the jth column. Thus, if we expand $det(A_j)$ along the ith row, we get:
$det(A_j)\\=0\space det(A_{i,1})\times(-1)^{i+1}+\cdots+1\space det(A_{i,j})\times(-1)^{i+j}+\cdots+0\space det(A_{i,n})\times(-1)^{i+n}\\=det(A_{i,j})\times(-1)^{i+j}$
where $A{i,j}$ is the matrix formed by removing the ith row and jth column of A.
$\therefore x_{ji}=\frac{det(A_j)}{det(A)}$
$\therefore A^{-1}=\frac{adj(A_j)}{det(A)}$
QED
My questions are:
- Can I use Cramer's rule in the proof?
- Is this proof correct?
- Are there any terminology issues or blatant inconsistencies/errors?
Thanks.

REPLY [2 votes]: It is perfectly fine to use Cramer's rule for such a purpose. Here it is a self-contained approach.

Lemma 1. If $A$ is an invertible $n\times n$ matrix, $v=(x_1,x_2,\ldots,x_n)^T$ and $w=(y_1,y_2,\ldots,y_n)^T$, the solution of $Av=w$ is given by 
  $x_k = \frac{\det(A_k)}{\det(A)} $, where $A_k$ is the original $A$ matrix with the $k$-th column being replaced by $w$.

Proof: Let $e_1,e_2,\ldots,e_n$ be the canonical base of $\mathbb{R}^n$ and $a_1,a_2,\ldots,a_n$ be the columns of $A$.
The linear map $f$ represented by $A$ sends $e_j$ into $a_j$ for any $j\in[1,n]$.
The linear map $f_k$ represented by $A_k$ does the same, with the exception of sending $e_k$ into $w$.
Let us consider what $f^{-1} f_k$ does: for any $j\neq k$, it sends $e_j$ into $e_j$. $f^{-1}f_k$ sends $e_k$ into $v$, hence the matrix representing $f^{-1} f_k$ has the $k$-th column equals to $v$ and the remaining columns equal to the columns of the identity matrix. In particular $\det(f^{-1}f_k)=x_k$ and the claim follows from Binet's formula $\det(AB)=\det(A)\det(B)$.

Lemma 2. If $A$ is an invertible matrix, in order to find $A^{-1}$ it is enough to find $A^{-1}e_1,\ldots,A^{-1}e_n$. That can be done through Lemma 1 and a Laplace expansion along the proper column, leading to the adjugate formula.