Cramer's Rule

Theorem
Let $n \in \N$.

Let $b_1, b_2, \dots, b_n$ be real numbers.

Let $\mathbf b = \tuple {b_1, b_2, \dots, b_n}^T$.

Let $x_1, x_2, \dots, x_n$ be real numbers.

Let $\mathbf x = \tuple {x_1, x_2, \dots, x_n}^T$.

Let $A$ be an invertible $n \times n$ matrix whose elements are in $\R$.

For each $i \in \set {1, \dots, n}$, let $A_i$ be the matrix obtained by replacing the $i$th column with $\mathbf b$.

Let:
 * $A \mathbf x = \mathbf b$

Then:
 * $\mathbf x_i = \dfrac {\map \det {A_i} } {\map \det A}$

for each $i \in \set {1, \dots, n}$.

Proof
Let $C$ be the cofactor matrix of $A$.

By definition, $C^T$ is the adjugate matrix of $A$.

Therefore by Matrix Product with Adjugate Matrix:
 * $A \cdot C^T = \det A \cdot I_n$

Because $A$ is invertible, $A^{-1}$ exists.

From Matrix is Invertible iff Determinant has Multiplicative Inverse, $1 / \det A$ also exists.

Therefore:
 * $A^{-1} = \dfrac 1 {\det A} \cdot C^T$

Since $A \mathbf x = \mathbf b$:
 * $\mathbf x = A^{-1} \mathbf b$

Therefore:
 * $\mathbf x = \paren {\dfrac 1 {\det A} C^T} \mathbf b$

For each $r, s \in \set {1, 2, \ldots, n}$, let $A_{r s}$ denote the element of $C$ whose index is $\tuple {r, s}$.

Then by the definition of transpose:


 * $C^T = \begin {bmatrix} A_{11} & A_{21} & \dots & A_{n1} \\

A_{12} & A_{22} & \dots & A_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ A_{1n} & A_{2n} & \dots & A_{nn} \end{bmatrix}$

By the definition of cofactor matrix, $A_{r s}$ is the cofactor of the element of $A$ whose index is $\tuple {r, s}$.

Let $i \in \set {1, 2, \ldots, n}$.

Then by definition of matrix product:
 * $\ds x_i = \dfrac 1 {\det A} \paren {\sum_{j \mathop = 1}^n A_{ji} b_j}$

Recall that $A_i$ is the matrix obtained from $A$ by replacing the $i$th column with $\mathbf b$.

Then if:
 * $A = \begin {bmatrix} a_{1,1} & a_{1,2} & \dots & a_{1,n} \\

a_{2,1} & a_{2,2} & \dots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \dots & a_{n,n} \end{bmatrix}$

then:
 * $A_i = \begin{bmatrix}

a_{1,1} & a_{1,2} & \dots & a_{1, i - 1} & b_1 & a_{1, i + 1} & \dots & a_{1,n} \\ a_{2,1} & a_{2,2} & \dots & a_{2, i - 1} & b_2 & a_{2, i + 1} & \dots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \dots & a_{n, i - 1} & b_n & a_{n, i + 1} & \dots & a_{n,n} \end{bmatrix}$

For all $j \in \set {1, 2, \dots, n}$ the matrix obtained by deleting the $j$th row and the $i$th column of $A$ is equal to the matrix obtained by deleting the $j$th row and $i$th column of $A_i$.

Therefore, by definition, the cofactor of $\sqbrk {A_i}_{j,i}$ is equal to $A_{ji}$.

By the Expansion Theorem for Determinants, we now have:
 * $\ds \det{A_i} = \sum_{j \mathop = 1}^n A_{ji} b_j$

Therefore:
 * $x_i = \dfrac {\det {A_i} } {\det A}$

as desired.