有几何背景的行列式化简

青青子衿

\[\det \left[ {\left( {\begin{array}{*{20}{c}}
{\frac{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{ - {C_1}}&{{B_1}}\\
{ - {C_2}}&{{B_2}}
\end{array}} \right)} \right]}}{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_1}}&{{A_2}}\\
{{B_1}}&{{B_2}}
\end{array}} \right)} \right]}}}&{\frac{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_1}}&{ - {C_1}}\\
{{A_2}}&{ - {C_2}}
\end{array}} \right)} \right]}}{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_1}}&{{A_2}}\\
{{B_1}}&{{B_2}}
\end{array}} \right)} \right]}}}&1\\
{\frac{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{ - {C_2}}&{{B_2}}\\
{ - {C_3}}&{{B_3}}
\end{array}} \right)} \right]}}{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_2}}&{{A_3}}\\
{{B_2}}&{{B_3}}
\end{array}} \right)} \right]}}}&{\frac{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_2}}&{ - {C_2}}\\
{{A_3}}&{ - {C_3}}
\end{array}} \right)} \right]}}{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_2}}&{{A_3}}\\
{{B_2}}&{{B_3}}
\end{array}} \right)} \right]}}}&1\\
{\frac{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{ - {C_1}}&{{B_1}}\\
{ - {C_3}}&{{B_3}}
\end{array}} \right)} \right]}}{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_1}}&{{A_3}}\\
{{B_1}}&{{B_3}}
\end{array}} \right)} \right]}}}&{\frac{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_1}}&{ - {C_1}}\\
{{A_3}}&{ - {C_3}}
\end{array}} \right)} \right]}}{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_1}}&{{A_3}}\\
{{B_1}}&{{B_3}}
\end{array}} \right)} \right]}}}&1
\end{array}} \right)} \right] =  - \frac{{\det {{\left[ {\left( {\begin{array}{*{20}{c}}
{{A_1}}&{{A_2}}&{{A_3}}\\
{{B_1}}&{{B_2}}&{{B_3}}\\
{{C_1}}&{{C_2}}&{{C_3}}
\end{array}} \right)} \right]}^2}}}{{\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_1}}&{{A_2}}\\
{{B_1}}&{{B_2}}
\end{array}} \right)} \right]\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_2}}&{{A_3}}\\
{{B_2}}&{{B_3}}
\end{array}} \right)} \right]\det \left[ {\left( {\begin{array}{*{20}{c}}
{{A_1}}&{{A_3}}\\
{{B_1}}&{{B_3}}
\end{array}} \right)} \right]}}\]

caijinzhi

平面直角坐标的三角形面积？！

青青子衿

回复 2# caijinzhi

平面直角坐标的三角形面积？！
caijinzhi 发表于 2014-10-20 21:06

求三条直线
\[{A_i}x +{B_i}y + {C_i} = 0 ( {i = 1,2,3})\]
所围成的三角形\(\triangle ABC\)的面积。
设
\[\Delta = \left| {\begin{array}{*{20}{c}}
{{A_1}}&{{B_1}}&{{C_1}}\\
{{A_2}}&{{B_2}}&{{C_2}}\\
{{A_3}}&{{B_3}}&{{C_3}}
\end{array}} \right|\]

\begin{array}{llll}
D[ {{a_1}}] = \left| {\begin{array}{*{20}{c}}
{{B_2}}&{{C_2}}\\
{{B_3}}&{{C_3}}
\end{array}} \right|
&
D[ {{a_2}}] = \left| {\begin{array}{*{20}{c}}
{{B_3}}&{{C_3}}\\
{{B_1}}&{{C_1}}
\end{array}} \right|
&
D[ {{a_3}}] = \left| {\begin{array}{*{20}{c}}
{{B_1}}&{{C_1}}\\
{{B_2}}&{{C_2}}
\end{array}} \right|
\\

D[ {{b_1}}] = \left| {\begin{array}{*{20}{c}}
{{C_2}}&{{A_2}}\\
{{C_3}}&{{A_3}}
\end{array}} \right|
&
D[ {{b_2}}] = \left| {\begin{array}{*{20}{c}}
{{C_3}}&{{A_3}}\\
{{C_1}}&{{A_1}}
\end{array}}\right|
&
D[ {{b_3}}] =\left| {\begin{array}{*{20}{c}}
{{C_1}}&{{A_1}}\\
{{C_2}}&{{A_2}}
\end{array}} \right|
\\  

D[ {{c_1}}] =\left |{\begin{array}{*{20}{c}}
{{A_2}}&{{B_2}}\\
{{A_3}}&{{B_3}}
\end{array}} \right|
&
D[ {{c_2}}] =\left| {\begin{array}{*{20}{c}}
{{A_3}}&{{B_3}}\\
{{A_1}}&{{B_1}}
\end{array}} \right|
&
D[ {{c_3}}] =\left|{\begin{array}{*{20}{c}}
{{A_1}}&{{B_1}}\\
{{A_2}}&{{B_2}}
\end{array}}\right|
\\  
\end{array}

\(D[{a_1}]\),\(D[{a_2}]\),\(D[{a_3}]\),\(D[{b_1}]\),\(D[{b_2}]\),\(D[{b_3}]\),\(D[{c_1}]\),\(D[{c_2}]\),\(D[{c_3}]\)分别为\(A_1\),\(A_2\),\(A_3\),\(B_1\),\(B_2\),\(B_3\),\(C_1\),\(C_2\),\(C_3\)的代数余子式。
由行列式的乘法，得

\[\left| {\begin{array}{*{20}{c}}
{{A_1}}&{{B_1}}&{{C_1}}\\
{{A_2}}&{{B_2}}&{{C_2}}\\
{{A_3}}&{{B_3}}&{{C_3}}
\end{array}} \right|
\times
\left| {\begin{array}{*{20}{c}}
{D[{{a_1}}]}&{D[ {{a_2}}]}&{D [ {{a_3}}]}\\
{D[ {{b_1}}]}&{D[ {{b_2}}]}&{D[ {{b_3}}]}\\
{D[ {{c_1}}]}&{D[ {{c_2}}]}&{D[ {{c_3}}]}
\end{array}} \right| \\
= \Delta \times
\left| {\begin{array}{*{20}{c}}
{D[{{a_1}}]}&{D[ {{a_2}}]}&{D[{{a_3}}]}\\
{D[{{b_1}}]}&{D[ {{b_2}}]}&{D[{{b_3}}]}\\
{D[ {{c_1}}]}&{D[ {{c_2}}]}&{D[{{c_3}}]}
\end{array}}\right|
= \left| {\begin{array}{*{20}{c}}
\Delta &0&0\\
0&\Delta &0\\
0&0&\Delta
\end{array}} \right| = {\Delta ^3}\]

\[\left|{\begin{array}{*{20}{c}}
{D[ {{a_1}}]}&{D[ {{a_2}}]}&{D[ {{a_3}}]}\\
{D[ {{b_1}}]}&{D[ {{b_2}}]}&{D[ {{b_3}}]}\\
{D[ {{c_1}}]}&{D[ {{c_2}}]}&{D[ {{c_3}}]}
\end{array}}\right| = {\Delta ^2} = {\left|{\begin{array}{*{20}{c}}
{{A_1}}&{{B_1}}&{{C_1}}\\
{{A_2}}&{{B_2}}&{{C_2}}\\
{{A_3}}&{{B_3}}&{{C_3}}
\end{array}}\right|^2}\]

\begin{array}{ll}
\left\{ {\begin{array}{*{20}{l}}
  {{x_A} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{B_2}}&{{C_2}} \\
  {{B_3}}&{{C_3}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_2}}&{{B_2}} \\
  {{A_3}}&{{B_3}}
\end{array}} \right|}} = \frac{{D[{a_1}]}}{{D[{c_1}]}}} \\
  {{y_A} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{C_2}}&{{A_2}} \\
  {{C_3}}&{{A_3}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_2}}&{{B_2}} \\
  {{A_3}}&{{B_3}}
\end{array}} \right|}} = \frac{{D[{b_1}]}}{{D[{c_1}]}}}
\end{array}} \right.
&
\left\{ {\begin{array}{*{20}{l}}
  {{x_B} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{B_3}}&{{C_3}} \\
  {{B_1}}&{{C_1}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_3}}&{{B_3}} \\
  {{A_1}}&{{B_1}}
\end{array}} \right|}} = \frac{{D[{a_2}]}}{{D[{c_2}]}}} \\
  {{y_B} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{C_3}}&{{A_3}} \\
  {{C_1}}&{{A_1}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_3}}&{{B_3}} \\
  {{A_1}}&{{B_1}}
\end{array}} \right|}} = \frac{{D[{b_2}]}}{{D[{c_2}]}}}
\end{array}} \right.
&
\left\{ {\begin{array}{*{20}{l}}
  {{x_C} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{B_1}}&{{C_1}} \\
  {{B_2}}&{{C_2}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}} \\
  {{A_2}}&{{B_2}}
\end{array}} \right|}} = \frac{{D[{a_3}]}}{{D[{c_3}]}}} \\
  {{y_C} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{C_1}}&{{A_1}} \\
  {{C_2}}&{{A_2}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}} \\
  {{A_2}}&{{B_2}}
\end{array}} \right|}} = \frac{{D[{b_3}]}}{{D[{c_3}]}}}
\end{array}} \right.
\\
\end{array}

\[{S_{\Delta ABC}} = \left| {\frac{1}{2}\left| {\begin{array}{*{20}{c}}
{{x_A}}&{{y_A}}&1\\
{{x_B}}&{{y_B}}&1\\
{{x_C}}&{{y_C}}&1
\end{array}} \right|} \right|\]

\[=\left| {\frac{1}
{{2D[{{c_1}}]D[{{c_2}}]D[{{c_3}}]}}
\left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}&{{a_3}}\\
{{b_1}}&{{b_2}}&{{b_3}}\\
{{c_1}}&{{c_2}}&{{c_3}}
\end{array}} \right|} \right|\]

\[ = \left| {\frac{{{\Delta ^2}}}{{2D[{c_1}]D[{c_2}]D[{c_3}]}}} \right|
= \left| {\frac{1}
{{2D[{{c_1}}]D[{{c_2}}]D[{{c_3}}]}}
{{\left| {\begin{array}{*{20}{c}}
{{A_1}}&{{B_1}}&{{C_1}}\\
{{A_2}}&{{B_2}}&{{C_2}}\\
{{A_3}}&{{B_3}}&{{C_3}}
\end{array}} \right|}^2}} \right|\]

\[ = \left| {\frac{{{{\left| {\begin{array}{*{20}{c}}
{{A_1}}&{{B_1}}&{{C_1}}\\
{{A_2}}&{{B_2}}&{{C_2}}\\
{{A_3}}&{{B_3}}&{{C_3}}
\end{array}} \right|}^2}}}{{2\left| {\begin{array}{*{20}{c}}
{{A_1}}&{{B_1}}\\
{{A_2}}&{{B_2}}
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
{{B_1}}&{{C_1}}\\
{{B_2}}&{{C_2}}
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
{{B_1}}&{{C_1}}\\
{{B_2}}&{{C_2}}
\end{array}} \right|}}} \right|\]

在元宵之际，祝大家好事连连，好梦圆圆！
美化了一下排版

其妙

回复 3# 青青子衿
厉害！

青青子衿

回复 4# 其妙

三角形面积的行列式公式
wenku.baidu.com/link?url=Yo6Gsy5nvEnDEu8co3yUKrgtrEsaA9ILe1zTpw6 ... 4mJ9YNysSm8IO9i7Vl8i

青青子衿

回复  caijinzhi
求三条直线
\[\color{blue}{{A_i}x +{B_i}y + {C_i} = 0~~~ ( {i = 1,2,3})}\]
所围成的三角形\(\triangle ABC\)面积 ...
青青子衿 发表于 2015-7-6 10:20

回复 3# 青青子衿
这道题最早应该是出现在第三届普特南数学竞赛(William Lowell Putnam Mathematical Competition)中
在单墫,程龙教授编写的数学奥林匹克辅导丛书《解析几何的技巧》也有此题
照葫芦画瓢就可以推广了
下面推广到三维四面体的情况：

求四个平面
\[\color{blue}{{A_i}x +{B_i}y + {C_i}z +{D_i} = 0 ~~( {i = 1,2,3,4})}\]
所围成的四面体\(四面体 A-BCD\)体积

记\[\left[ {\begin{array}{*{20}{c}}
  {{a_1}}&{{a_2}}&{{a_3}}&{{a_4}} \\
  {{b_1}}&{{b_2}}&{{b_3}}&{{b_4}} \\
  {{c_1}}&{{c_2}}&{{c_3}}&{{c_4}} \\
  {{d_1}}&{{d_2}}&{{d_3}}&{{d_4}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}}&{{D_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}}&{{D_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}}&{{D_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right]\]
设\[\Delta  = \left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}}&{{D_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}}&{{D_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}}&{{D_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|\]
\[D\left[ {{a_1}} \right] = \left| {\begin{array}{*{20}{c}}
  {{B_2}}&{{C_2}}&{{D_2}} \\
  {{B_3}}&{{C_3}}&{{D_3}} \\
  {{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|\]

\begin{array}{llll}
D\left[ {{a_1}} \right] = \left| {\begin{array}{*{20}{c}}
  {{B_2}}&{{C_2}}&{{D_2}} \\
  {{B_3}}&{{C_3}}&{{D_3}} \\
  {{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right| & D\left[ {{a_2}} \right] = \left| {\begin{array}{*{20}{c}}
  {{B_1}}&{{C_1}}&{{D_1}} \\
  {{B_3}}&{{C_3}}&{{D_3}} \\
  {{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|
& D\left[ {{a_3}} \right] = \left| {\begin{array}{*{20}{c}}
  {{B_1}}&{{C_1}}&{{D_1}} \\
  {{B_2}}&{{C_2}}&{{D_2}} \\
  {{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|
& D\left[ {{a_4}} \right] = \left| {\begin{array}{*{20}{c}}
  {{B_1}}&{{C_1}}&{{D_1}} \\
  {{B_2}}&{{C_2}}&{{D_2}} \\
  {{B_3}}&{{C_3}}&{{D_3}}
\end{array}} \right|
\\

D\left[ {{b_1}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_2}}&{{C_2}}&{{D_2}} \\
  {{A_3}}&{{C_3}}&{{D_3}} \\
  {{A_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|
& D\left[ {{b_2}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{C_1}}&{{D_1}} \\
  {{A_3}}&{{C_3}}&{{D_3}} \\
  {{A_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|
& D\left[ {{b_3}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{C_1}}&{{D_1}} \\
  {{A_2}}&{{C_2}}&{{D_2}} \\
  {{A_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|
& D\left[ {{b_4}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{C_1}}&{{D_1}} \\
  {{A_2}}&{{C_2}}&{{D_2}} \\
  {{A_3}}&{{C_3}}&{{D_3}}
\end{array}} \right|
\\

D\left[ {{c_1}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_2}}&{{B_2}}&{{D_2}} \\
  {{A_3}}&{{B_3}}&{{D_3}} \\
  {{A_4}}&{{B_4}}&{{D_4}}
\end{array}} \right|
& D\left[ {{c_2}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{D_1}} \\
  {{A_3}}&{{B_3}}&{{D_3}} \\
  {{A_4}}&{{B_4}}&{{D_4}}
\end{array}} \right|
& D\left[ {{c_3}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{D_1}} \\
  {{A_2}}&{{B_2}}&{{D_2}} \\
  {{A_4}}&{{B_4}}&{{D_4}}
\end{array}} \right|
& D\left[ {{c_4}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{D_1}} \\
  {{A_2}}&{{B_2}}&{{D_2}} \\
  {{A_3}}&{{B_3}}&{{D_3}}
\end{array}} \right|
\\  

D\left[ {{d_1}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|
& D\left[ {{d_2}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_3}}&{{B_3}}&{{C_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|
& D\left[ {{d_3}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|
& D\left[ {{d_4}} \right] = \left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}}
\end{array}} \right|
\\  
\end{array}
\(D[{a_1}]\),\(D[{a_2}]\),\(D[{a_3}]\),\(D[{a_4}]\),\(D[{b_1}]\),\(D[{b_2}]\),\(D[{b_3}]\),\(D[{b_4}]\),\(D[{c_1}]\),\(D[{c_2}]\),\(D[{c_3}]\),\(D[{c_4}]\),\(D[{d_1}]\),\(D[{d_2}]\),\(D[{d_3}]\),\(D[{d_4}]\)分别为\(A_1\),\(A_2\),\(A_3\),\(A_4\),\(B_1\),\(B_2\),\(B_3\),\(B_4\),\(C_1\),\(C_2\),\(C_3\),\(C_4\),\(D_1\),\(D_2\),\(D_3\),\(D_4\)的代数余子式。
由行列式的乘法，得
\[\begin{gathered}
  \left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}}&{{D_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}}&{{D_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}}&{{D_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right| \times \left| {\begin{array}{*{20}{c}}
  {D\left[ {{a_1}} \right]}&{D\left[ {{a_2}} \right]}&{D\left[ {{a_3}} \right]}&{D\left[ {{a_4}} \right]} \\
  {D\left[ {{b_1}} \right]}&{D\left[ {{b_2}} \right]}&{D\left[ {{b_3}} \right]}&{D\left[ {{b_4}} \right]} \\
  {D\left[ {{c_1}} \right]}&{D\left[ {{c_2}} \right]}&{D\left[ {{c_3}} \right]}&{D\left[ {{c_4}} \right]} \\
  {D\left[ {{d_1}} \right]}&{D\left[ {{d_2}} \right]}&{D\left[ {{d_3}} \right]}&{D\left[ {{d_4}} \right]}
\end{array}} \right| \hfill \\
   = \Delta  \times \left| {\begin{array}{*{20}{c}}
  {D\left[ {{a_1}} \right]}&{D\left[ {{a_2}} \right]}&{D\left[ {{a_3}} \right]}&{D\left[ {{a_4}} \right]} \\
  {D\left[ {{b_1}} \right]}&{D\left[ {{b_2}} \right]}&{D\left[ {{b_3}} \right]}&{D\left[ {{b_4}} \right]} \\
  {D\left[ {{c_1}} \right]}&{D\left[ {{c_2}} \right]}&{D\left[ {{c_3}} \right]}&{D\left[ {{c_4}} \right]} \\
  {D\left[ {{d_1}} \right]}&{D\left[ {{d_2}} \right]}&{D\left[ {{d_3}} \right]}&{D\left[ {{d_4}} \right]}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  \Delta &0&0&0 \\
  0&\Delta &0&0 \\
  0&0&\Delta &0 \\
  0&0&0&\Delta  
\end{array}} \right| = {\Delta ^4} \hfill \\
\end{gathered} \]

\[\left| {\begin{array}{*{20}{c}}
  {D\left[ {{a_1}} \right]}&{D\left[ {{a_2}} \right]}&{D\left[ {{a_3}} \right]}&{D\left[ {{a_4}} \right]} \\
  {D\left[ {{b_1}} \right]}&{D\left[ {{b_2}} \right]}&{D\left[ {{b_3}} \right]}&{D\left[ {{b_4}} \right]} \\
  {D\left[ {{c_1}} \right]}&{D\left[ {{c_2}} \right]}&{D\left[ {{c_3}} \right]}&{D\left[ {{c_4}} \right]} \\
  {D\left[ {{d_1}} \right]}&{D\left[ {{d_2}} \right]}&{D\left[ {{d_3}} \right]}&{D\left[ {{d_4}} \right]}
\end{array}} \right| = {\Delta ^3} = {\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}}&{{D_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}}&{{D_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}}&{{D_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|^3}\]

\begin{array}{ll}
\left\{ \begin{gathered}
  {x_A} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{B_2}}&{{C_2}}&{{D_2}} \\
  {{B_3}}&{{C_3}}&{{D_3}} \\
  {{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|}} = \frac{{D\left[ {{a_1}} \right]}}{{D\left[ {{d_1}} \right]}} \hfill \\
  {y_A} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{A_2}}&{{C_2}}&{{D_2}} \\
  {{A_3}}&{{C_3}}&{{D_3}} \\
  {{A_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|}} = \frac{{D\left[ {{b_1}} \right]}}{{D\left[ {{d_1}} \right]}} \hfill \\
  {z_A} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{A_2}}&{{B_2}}&{{D_2}} \\
  {{A_3}}&{{B_3}}&{{D_3}} \\
  {{A_4}}&{{B_4}}&{{D_4}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|}} = \frac{{D\left[ {{c_1}} \right]}}{{D\left[ {{d_1}} \right]}} \hfill \\
\end{gathered}  \right.
&
\left\{ \begin{gathered}
  {x_B} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{B_1}}&{{C_1}}&{{D_1}} \\
  {{B_3}}&{{C_3}}&{{D_3}} \\
  {{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_3}}&{{B_3}}&{{C_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|}} = \frac{{D\left[ {{a_2}} \right]}}{{D\left[ {{d_2}} \right]}} \hfill \\
  {y_B} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{C_1}}&{{D_1}} \\
  {{A_3}}&{{C_3}}&{{D_3}} \\
  {{A_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_3}}&{{B_3}}&{{C_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|}} = \frac{{D\left[ {{b_2}} \right]}}{{D\left[ {{d_2}} \right]}} \hfill \\
  {z_B} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{D_1}} \\
  {{A_3}}&{{B_3}}&{{D_3}} \\
  {{A_4}}&{{B_4}}&{{D_4}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_3}}&{{B_3}}&{{C_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|}} = \frac{{D\left[ {{c_2}} \right]}}{{D\left[ {{d_2}} \right]}} \hfill \\
\end{gathered}  \right.
\\
\left\{ \begin{gathered}
  {x_C} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{B_1}}&{{C_1}}&{{D_1}} \\
  {{B_2}}&{{C_2}}&{{D_2}} \\
  {{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|}} = \frac{{D\left[ {{a_3}} \right]}}{{D\left[ {{d_3}} \right]}} \hfill \\
  {y_C} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{C_1}}&{{D_1}} \\
  {{A_2}}&{{C_2}}&{{D_2}} \\
  {{A_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|}} = \frac{{D\left[ {{b_3}} \right]}}{{D\left[ {{d_3}} \right]}} \hfill \\
  {z_C} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{D_1}} \\
  {{A_2}}&{{B_2}}&{{D_2}} \\
  {{A_4}}&{{B_4}}&{{D_4}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|}} = \frac{{D\left[ {{c_3}} \right]}}{{D\left[ {{d_3}} \right]}} \hfill \\
\end{gathered}  \right.
&
\left\{ \begin{gathered}
  {x_D} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{B_1}}&{{C_1}}&{{D_1}} \\
  {{B_2}}&{{C_2}}&{{D_2}} \\
  {{B_3}}&{{C_3}}&{{D_3}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}}
\end{array}} \right|}} = \frac{{D\left[ {{a_4}} \right]}}{{D\left[ {{d_4}} \right]}} \hfill \\
  {y_D} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{C_1}}&{{D_1}} \\
  {{A_2}}&{{C_2}}&{{D_2}} \\
  {{A_3}}&{{C_3}}&{{D_3}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}}
\end{array}} \right|}} = \frac{{D\left[ {{b_4}} \right]}}{{D\left[ {{d_4}} \right]}} \hfill \\
  {z_D} = \frac{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{D_1}} \\
  {{A_2}}&{{B_2}}&{{D_2}} \\
  {{A_3}}&{{B_3}}&{{D_3}}
\end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}}
\end{array}} \right|}} = \frac{{D\left[ {{c_4}} \right]}}{{D\left[ {{d_4}} \right]}} \hfill \\
\end{gathered}  \right.
\end{array}

\[{V_{A - BCD}} = \left| {\frac{1}{6}\left| {\begin{array}{*{20}{c}}
  {{x_A}}&{{y_A}}&{{z_A}}&1 \\
  {{x_B}}&{{y_B}}&{{z_B}}&1 \\
  {{x_C}}&{{y_C}}&{{z_C}}&1 \\
  {{x_D}}&{{y_D}}&{{z_D}}&1
\end{array}} \right|} \right|\]
\[=\left| {\frac{1}{{6D\left[ {{d_1}} \right]D\left[ {{d_2}} \right]D\left[ {{d_3}} \right]D\left[ {{d_4}} \right]}}\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{a_2}}&{{a_3}}&{{a_4}} \\
  {{b_1}}&{{b_2}}&{{b_3}}&{{b_4}} \\
  {{c_1}}&{{c_2}}&{{c_3}}&{{c_4}} \\
  {{d_1}}&{{d_2}}&{{d_3}}&{{d_4}}
\end{array}} \right|} \right|\]
\[=\left| {\frac{{{\Delta ^3}}}{{6D\left[ {{d_1}} \right]D\left[ {{d_2}} \right]D\left[ {{d_3}} \right]D\left[ {{d_4}} \right]}}} \right| = \left| {\frac{1}{{6D\left[ {{d_1}} \right]D\left[ {{d_2}} \right]D\left[ {{d_3}} \right]D\left[ {{d_4}} \right]}}{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}}&{{D_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}}&{{D_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}}&{{D_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|}^3}} \right|\]
\[=\left| {\frac{{{{\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}}&{{D_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}}&{{D_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}}&{{D_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}&{{D_4}}
\end{array}} \right|}^3}}}{{6\left| {\begin{array}{*{20}{c}}
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_3}}&{{B_3}}&{{C_3}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_4}}&{{B_4}}&{{C_4}}
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
  {{A_1}}&{{B_1}}&{{C_1}} \\
  {{A_2}}&{{B_2}}&{{C_2}} \\
  {{A_3}}&{{B_3}}&{{C_3}}
\end{array}} \right|}}} \right|\]

青青子衿

回复  青青子衿
这道题最早应该是出现在第三届普特南数学竞赛(\(\color{red}{William Lowell Putnam Mathematical Competition}\))中 ...
青青子衿 发表于 2016-2-22 19:40

第三届普特南数学竞赛