## 克拉瑪公式的證明

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

$A$ 是可逆矩陣，即 $\det A\neq 0$，針對方程式 $A\mathbf{x}=\mathbf{b}$，克拉瑪公式給出下面的求解算式：

$x_i=\displaystyle\frac{\det A_i(\mathbf{b})}{\det A},~~i=1,2,\ldots,n$

$I=\begin{bmatrix} \mathbf{e}_1&\cdots&\mathbf{e}_n \end{bmatrix}$

$\mathbf{x}$ 取代 $I$ 的第 $i$ 行，再左乘 $A$

$A I_i(\mathbf{x})=A\begin{bmatrix} \mathbf{e}_1&\cdots&\mathbf{x}&\cdots&\mathbf{e}_n \end{bmatrix}$

\begin{aligned} A I_i(\mathbf{x})&=\begin{bmatrix} A\mathbf{e}_1&\cdots&A\mathbf{x}&\cdots&A\mathbf{e}_n \end{bmatrix}\\ &=\begin{bmatrix} \mathbf{a}_1&\cdots&\mathbf{b}&\cdots&\mathbf{a}_n \end{bmatrix}=A_i(\mathbf{b}).\end{aligned}

$\det(AX)=(\det A)(\det X)=(\det A)(\det I_i(\mathbf{x}))=\det A_i(\mathbf{b})$

$\begin{vmatrix} 1&x_1&0&0\\ 0&x_2&0&0\\ 0&x_3&1&0\\ 0&x_4&0&1 \end{vmatrix}=x_2\begin{vmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{vmatrix}=x_2\cdot 1=x_2$

$x_i=\displaystyle\frac{\det A_i(\mathbf{b})}{\det A}$

$\det(AB)=(\det A)(\det B)$

[1] David C. Lay, Linear Algebra and its Applications，3rd edition, 2006.
[2] Gilbert Strang, Introduction to Linear Algebra, 3rd edition, 2003.
[3] Spence, Insel, Friedberg, Elementary Linear Algebra, 2003.

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