不說廢話──克拉瑪公式的證明

本文的閱讀等級:初級

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

其中 A_i(\mathbf{b}) 表示以 \mathbf{b} 取代 A 的第 i 行 (即 \mathbf{a}_i) 而得的 n\times n 階矩陣,

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}

 
證明 1.

\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}

其中 \mathbf{e}_1,\ldots,\mathbf{e}_nn 維標準單位向量。

 
證明 2.

\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}

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This entry was posted in 線性代數專欄, 行列式 and tagged , . Bookmark the permalink.

2 Responses to 不說廢話──克拉瑪公式的證明

  1. Meiyue Shao says:

    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 says:

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

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