---
tag: 4GY0RA4W28
kind: proof
name: Proof of Cramer's rule by the adjugate
msc: [15A06, 15A15]
proves: 0977KZ6PQB
uses: [0ENVM493VK, 0GMPYM0NSR, 06AMCJMMVM, 0W2QAVJQ1W, 0TWTS88TWN, 073E4XC016, 076JGP6R3T, 016PE4VZ8Q, 0QA8CHH8MG, 0JTRMBQ1SA]
sources:
  - url: https://en.wikipedia.org/wiki/Cramer's_rule
    locator: Proof
---

## 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|0QA8CHH8MG]] $F$ 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|016PE4VZ8Q]] 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|0ENVM493VK]], $A$ is invertible with $A^{-1} = (\det A)^{-1} \operatorname{adj}(A)$.
5. Existence: let $x = A^{-1} b$. By [[associativity|073E4XC016]], $A x = (A A^{-1}) b = I_n b = b$. So $x$ is a solution of the [[linear system|0JTRMBQ1SA]] $Ax = b$.
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|0W2QAVJQ1W]] of $A$. By the definition of the [[adjugate|0GMPYM0NSR]] and the [[matrix product|076JGP6R3T]], $x_i = (\det A)^{-1} \sum_{j=1}^{n} C_{ji} b_j$.
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 $i$|06AMCJMMVM]] 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 $1 \le i \le n$.

## Idea

The solution is $x = A^{-1} b = (\det A)^{-1} \operatorname{adj}(A)\, b$. Coordinate $i$ of $\operatorname{adj}(A)\, b$ pairs $b$ with the cofactors of column $i$. That pairing is exactly the Laplace expansion of $\det A_i$ along its new column $i$.