Proof of Cramer's rule by the adjugate

Proof

We solve the system with the inverse matrix and read off each coordinate by a Laplace expansion.

1. Let $A = (a_{kl})$ be an $n \times n$ matrix over the field $F$ with $\det A \ne 0$, and let $b = (b_1, \dots, b_n) \in F^n$.
2. First let $n = 1$. Then $A = (a_{11})$ with $a_{11} \ne 0$, by the Leibniz formula with one term. The equation $a_{11} x_1 = b_1$ has the single solution $x_1 = b_1 / a_{11}$.
3. For $n = 1$, $A_1 = (b_1)$ and $\det A_1 = b_1$. So $x_1 = \det A_1 / \det A$, as claimed. From now on let $n \ge 2$.
4. In a field, the nonzero element $\det A$ is a unit. By the inverse in terms of the adjugate, $A$ is invertible with $A^{-1} = (\det A)^{-1} \operatorname{adj}(A)$.
5. Existence: let $x = A^{-1} b$. By associativity, $A x = (A A^{-1}) b = I_n b = b$. So $x$ is a solution of the linear system $Ax = b$.
6. Uniqueness: if $A y = b$, then $y = (A^{-1} A) y = A^{-1} (A y) = A^{-1} b = x$, by associativity.
7. Let $C_{kl}$ be the cofactors of $A$. By the definition of the adjugate and the matrix product, $x_i = (\det A)^{-1} \sum_{j=1}^{n} C_{ji} b_j$.
8. Fix $i$. The matrix $A_i$ differs from $A$ only in column $i$, and it has entry $b_j$ in position $(j, i)$.
9. A cofactor in position $(j, i)$ deletes column $i$. So the cofactors of $A_i$ in the positions $(j, i)$ equal the cofactors $C_{ji}$ of $A$.
10. By the Laplace expansion along column $i$ of $A_i$ and steps 8 and 9, $\det A_i = \sum_{j=1}^{n} b_j C_{ji}$.
11. By steps 7 and 10, $x_i = \det A_i / \det A$ for $1 \le i \le n$.