再談克拉瑪公式的證明

\begin{aligned} a_{11}x_1+a_{12}x_2&=b_1\\ a_{21}x_1+a_{22}x_2&=b_2\end{aligned}

$x_1=\displaystyle\frac{\begin{vmatrix} b_1&a_{12}\\ b_2&a_{22} \end{vmatrix}}{\begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{vmatrix}},~ x_2=\frac{\begin{vmatrix} a_{11}&b_1\\ a_{21}&b_2 \end{vmatrix}}{\begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{vmatrix}}$

$\begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \end{bmatrix}=\begin{bmatrix} b_1\\ b_2 \end{bmatrix}$

$x_1\mathbf{a}_1+x_2\mathbf{a}_2=\mathbf{b}$

\begin{aligned} (x_1\mathbf{a}_1+x_2\mathbf{a}_2)\cdot\mathbf{p}&=x_1\mathbf{a}_1\cdot\mathbf{p}+x_2\mathbf{a}_2\cdot\mathbf{p}\\ &=x_1\mathbf{a}_1\cdot\mathbf{p}=\mathbf{b}\cdot\mathbf{p}\end{aligned}

\begin{aligned} x_1&=\displaystyle\frac{\mathbf{b}\cdot\mathbf{p}}{\mathbf{a}_1\cdot\mathbf{p}}=\frac{b_1a_{22}-b_2a_{12}}{a_{11}a_{22}-a_{21}a_{12}}=\frac{\begin{vmatrix} b_1&a_{12}\\ b_2&a_{22} \end{vmatrix}}{\begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{vmatrix}}\end{aligned}

$x_1\mathbf{a}_1+x_2\mathbf{a}_2+x_3\mathbf{a}_3=\mathbf{b}$

$\mathbf{a}_2\times\mathbf{a}_3\equiv\begin{vmatrix} a_{22}&a_{32}\\ a_{23}&a_{33} \end{vmatrix}\mathbf{e}_1+\begin{vmatrix} a_{32}&a_{12}\\ a_{33}&a_{13} \end{vmatrix}\mathbf{e}_2+\begin{vmatrix} a_{12}&a_{22}\\ a_{13}&a_{23} \end{vmatrix}\mathbf{e}_3$

$\mathbf{a}_2\times\mathbf{a}_3=\begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix}\mathbf{e}_1-\begin{vmatrix} a_{12}&a_{13}\\ a_{32}&a_{33} \end{vmatrix}\mathbf{e}_2+\begin{vmatrix} a_{12}&a_{13}\\ a_{22}&a_{23} \end{vmatrix}\mathbf{e}_3$

$\mathbf{a}_2\times\mathbf{a}_3=\begin{vmatrix} \mathbf{e}_1&a_{12}&a_{13}\\ \mathbf{e}_2&a_{22}&a_{23}\\ \mathbf{e}_3&a_{32}&a_{33} \end{vmatrix}$

$\mathbf{v}\cdot(\mathbf{a}_2\times\mathbf{a}_3)=\begin{vmatrix} v_1&a_{12}&a_{13}\\ v_2&a_{22}&a_{23}\\ v_3&a_{32}&a_{33} \end{vmatrix}$

$x_1\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)=\mathbf{b}\cdot(\mathbf{a}_2\times\mathbf{a}_3)$

$x_1=\displaystyle\frac{\mathbf{b}\cdot(\mathbf{a}_2\times\mathbf{a}_3)}{ \mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)}$

$\mathbf{b}\cdot(\mathbf{a}_2\times\mathbf{a}_3)=\begin{vmatrix} b_1&a_{12}&a_{13}\\ b_2&a_{22}&a_{23}\\ b_3&a_{32}&a_{33} \end{vmatrix}$

\begin{aligned} \mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)&=\begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{vmatrix}=\det A\end{aligned}

$x_1\mathbf{a}_1+x_2\mathbf{a}_2+\cdots+x_n\mathbf{a}_n=\mathbf{b}$

$\mathbf{p}=\begin{vmatrix} \mathbf{e}_1&a_{12}&\ldots&a_{1n}\\ \mathbf{e}_2&a_{22}&\ldots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ \mathbf{e}_n&a_{n2}&\ldots&a_{nn} \end{vmatrix}$

$\mathbf{v}\cdot\mathbf{p}=\begin{vmatrix} v_1&a_{12}&\ldots&a_{1n}\\ v_2&a_{22}&\ldots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ v_n&a_{n2}&\ldots&a_{nn} \end{vmatrix}$

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

$x_iL_i(\mathbf{a}_i)=L_i(\mathbf{b})$

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

Dan Kalman, Cramer’s Rule via Selective Annihilation, College Mathematics Journal, 18, 136-137, 1987.