## 不說廢話──克拉瑪公式的證明

You know that I write slowly. This is chiefly because I am never satisfied until I have said as much as possible in a few words, and writing briefly takes far more time than writing at length.
― Carl Friedrich Gauss

$A=\begin{bmatrix} \mathbf{a}_1&\cdots&\mathbf{a}_n \end{bmatrix}$ 為一個 $n\times n$ 階矩陣且 $\mathbf{b}$ 為一個 $n$ 維行向量 (column vector)。若 $A$ 是可逆的，克拉瑪公式 (Cramer’s rule) 給出線性方程 $A\mathbf{x}=\mathbf{b}$ 的解 $\mathbf{x}=(x_1,\ldots,x_n)^T$，如下 (見“克拉瑪公式的證明”)：

$\displaystyle x_i=\frac{\det A_i(\mathbf{b})}{\det A},~~i=1,\ldots,n$

$A_i(\mathbf{b})=\begin{bmatrix} ~&~&~&~&~&~&~\\ \mathbf{a}_1&\cdots&\mathbf{a}_{i-1}&\mathbf{b}&\mathbf{a}_{i+1}&\cdots&\mathbf{a}_n\\ ~&~&~&~&~&~&~ \end{bmatrix}$

\displaystyle \begin{aligned} \frac{\det A_i(\mathbf{b})}{\det A}&=(\det A^{-1})(\det A_i(\mathbf{b}))=\det \left(A^{-1}A_i(\mathbf{b})\right)\\ &=\det\left(A^{-1}\begin{bmatrix} ~&~&~&~&~&~&~\\ \mathbf{a}_1&\cdots&\mathbf{a}_{i-1}&\mathbf{b}&\mathbf{a}_{i+1}&\cdots&\mathbf{a}_n\\ ~&~&~&~&~&~&~ \end{bmatrix}\right)\\ &=\det\begin{bmatrix} ~&~&~&~&~&~&~\\ \mathbf{e}_1&\cdots&\mathbf{e}_{i-1}&\mathbf{x}&\mathbf{e}_{i+1}&\cdots&\mathbf{e}_n\\ ~&~&~&~&~&~&~ \end{bmatrix}=x_i, \end{aligned}

\displaystyle \begin{aligned} \det A_i(\mathbf{b})&=\det\left(A+(\mathbf{b}-\mathbf{a}_i)\mathbf{e}_i^T\right)\\ &=(\det A)\det\left(I+A^{-1}(\mathbf{b}-\mathbf{a}_i)\mathbf{e}_i^T\right)\\ &=(\det A)\det\left(I+(\mathbf{x}-\mathbf{e}_i)\mathbf{e}_i^T\right)\\ &=(\det A)\left(1+\mathbf{e}_i^T(\mathbf{x}-\mathbf{e}_i)\right)\\ &=(\det A)\,x_i. \end{aligned}

### 2 Responses to 不說廢話──克拉瑪公式的證明

1. Meiyue Shao 說道：

这个证明是 Stephen Robinson 于 1970 年给出的，所以只有相对比较新的教材才会采用，而且还和教材本身的逻辑次序有关。

2. ccjou 說道：

補充另一個證明，但需要套用公式 $\det (I+uv^T)=1+v^Tu$