行列式“剥洋葱”：对于行或者列之间存在普遍规律的行列式可以尝试先提取其“公共部分”

一、题目

已知 $a_i$ $\ne$ $0$ ($i$ $=$ $1$, $2$, $3$, $4$), 则：

$$|V|=\left|\begin{array}{cccccc}& a_1^3& a_1^2{b}_1& a_1{b}_1^2& b_1^3& \\ \\ & a_2^3& a_2^2{b}_2& a_2{b}_2^2& b_2^3& \\ \\ & a_3^3& a_3^2{b}_3& a_3{b}_3^2& b_3^3& \\ \\ & a_4^3& a_4^2{b}_4& a_4{b}_4^2& b_4^3& \end{array}\right|=?$$

难度评级：

二、解析

观察可知，题目中所给的行列式 $V$ 是一个行与行之间非常有规律的一个行列式，如果我们提取出来一个 $a_1$, 那么，行列式 $|V|$ 的第一行将变为：

$$a_1^2{a}_1{b}_1{b}_1^2\frac{b_1^3}{a_1}$$

如果我们提取出来一个 $a_1^2$, 那么，行列式 $|V|$ 的第一行将变为：

$$a_1{b}_1\frac{b_1^2}{a_1}\frac{b_1^3}{a_1^2}$$

如果我们提取出来一个 $a_1^3$, 那么，行列式 $|V|$ 的第一行将变为：

$$1\frac{b_1}{a_1}\frac{b_1^2}{a_1^2}\frac{b_1^3}{a_1^3}$$

于是可知，我们可以对行列式 $|V|$ 做如下变形：

$$\begin{array}{rl}& \left|\begin{array}{cccccc}& a_1^3& a_1^2{b}_1& a_1{b}_1^2& b_1^3& \\ \\ & a_2^3& a_2^2{b}_2& a_2{b}_2^2& b_2^3& \\ \\ & a_3^3& a_3^2{b}_3& a_3{b}_3^2& b_3^3& \\ \\ & a_4^3& a_4^2{b}_4& a_4{b}_4^2& b_4^3& \end{array}\right|\\ \\ \Rightarrow a_1^3& \left|\begin{array}{cccccc}& 1& \frac{b_1}{a_1}& \frac{b_1^2}{a_1^2}& \frac{b_1^3}{a_1^3}& \\ \\ & a_2^3& a_2^2{b}_2& a_2{b}_2^2& b_2^3& \\ \\ & a_3^3& a_3^2{b}_3& a_3{b}_3^2& b_3^3& \\ \\ & a_4^3& a_4^2{b}_4& a_4{b}_4^2& b_4^3& \end{array}\right|\\ \\ \Rightarrow a_1^3{a}_2^3& \left|\begin{array}{cccccc}& 1& \frac{b_1}{a_1}& \frac{b_1^2}{a_1^2}& \frac{b_1^3}{a_1^3}& \\ \\ & 1& \frac{b_2}{a_2}& \frac{b_2^2}{a_2^2}& \frac{b_2^3}{a_2^3}& \\ \\ & a_3^3& a_3^2{b}_3& a_3{b}_3^2& b_3^3& \\ \\ & a_4^3& a_4^2{b}_4& a_4{b}_4^2& b_4^3& \end{array}\right|\\ \\ \Rightarrow a_1^3{a}_2^3{a}_3^3& \left|\begin{array}{cccccc}& 1& \frac{b_1}{a_1}& \frac{b_1^2}{a_1^2}& \frac{b_1^3}{a_1^3}& \\ \\ & 1& \frac{b_2}{a_2}& \frac{b_2^2}{a_2^2}& \frac{b_2^3}{a_2^3}& \\ \\ & 1& \frac{b_3}{a_3}& \frac{b_3^2}{a_3^2}& \frac{b_3^3}{a_3^3}& \\ \\ & a_4^3& a_4^2{b}_4& a_4{b}_4^2& b_4^3& \end{array}\right|\\ \\ \Rightarrow a_1^3{a}_2^3{a}_3^3{a}_4^3& \left|\begin{array}{cccccc}& 1& \frac{b_1}{a_1}& \frac{b_1^2}{a_1^2}& \frac{b_1^3}{a_1^3}& \\ \\ & 1& \frac{b_2}{a_2}& \frac{b_2^2}{a_2^2}& \frac{b_2^3}{a_2^3}& \\ \\ & 1& \frac{b_3}{a_3}& \frac{b_3^2}{a_3^2}& \frac{b_3^3}{a_3^3}& \\ \\ & 1& \frac{b_4}{a_4}& \frac{b_4^2}{a_4^2}& \frac{b_4^3}{a_4^3}& \end{array}\right|\\ \\ \Rightarrow a_1^3{a}_2^3{a}_3^3{a}_4^3& \left|\begin{array}{cccccc}& 1& \frac{b_1}{a_1}& \frac{b_1^2}{a_1^2}& \frac{b_1^3}{a_1^3}& \\ \\ & 0& \frac{b_2}{a_2}–\frac{b_1}{a_1}& \frac{b_2^2}{a_2^2}–\frac{b_1^2}{a_1^2}& \frac{b_2^3}{a_2^3}–\frac{b_1^3}{a_1^3}& \\ \\ & 0& \frac{b_3}{a_3}–\frac{b_1}{a_1}& \frac{b_3^2}{a_3^2}–\frac{b_1^2}{a_1^2}& \frac{b_3^3}{a_3^3}–\frac{b_1^3}{a_1^3}& \\ \\ & 0& \frac{b_4}{a_4}–\frac{b_1}{a_1}& \frac{b_4^2}{a_4^2}–\frac{b_1^2}{a_1^2}& \frac{b_4^3}{a_4^3}–\frac{b_1^3}{a_1^3}& \end{array}\right|\\ \\ \Rightarrow a_1^3{a}_2^3{a}_3^3{a}_4^3\cdot 1& \left|\begin{array}{ccccc}& \frac{b_2}{a_2}–\frac{b_1}{a_1}& \frac{b_2^2}{a_2^2}–\frac{b_1^2}{a_1^2}& \frac{b_2^3}{a_2^3}–\frac{b_1^3}{a_1^3}& \\ \\ & \frac{b_3}{a_3}–\frac{b_1}{a_1}& \frac{b_3^2}{a_3^2}–\frac{b_1^2}{a_1^2}& \frac{b_3^3}{a_3^3}–\frac{b_1^3}{a_1^3}& \\ \\ & \frac{b_4}{a_4}–\frac{b_1}{a_1}& \frac{b_4^2}{a_4^2}–\frac{b_1^2}{a_1^2}& \frac{b_4^3}{a_4^3}–\frac{b_1^3}{a_1^3}& \end{array}\right|\end{array}$$

使用相同的提取方法，继续对上面绿色部分的行列式进行转换，可得：

$$\begin{array}{rl}& a_1^3{a}_2^3{a}_3^3{a}_4^3\left|\begin{array}{ccccc}& \frac{b_2}{a_2}–\frac{b_1}{a_1}& \frac{b_2^2}{a_2^2}–\frac{b_1^2}{a_1^2}& \frac{b_2^3}{a_2^3}–\frac{b_1^3}{a_1^3}& \\ \\ & \frac{b_3}{a_3}–\frac{b_1}{a_1}& \frac{b_3^2}{a_3^2}–\frac{b_1^2}{a_1^2}& \frac{b_3^3}{a_3^3}–\frac{b_1^3}{a_1^3}& \\ \\ & \frac{b_4}{a_4}–\frac{b_1}{a_1}& \frac{b_4^2}{a_4^2}–\frac{b_1^2}{a_1^2}& \frac{b_4^3}{a_4^3}–\frac{b_1^3}{a_1^3}& \end{array}\right|\\ \\ \\ \Rightarrow & a_1^3{a}_2^3{a}_3^3{a}_4^3\left(\frac{b_2}{a_2}–\frac{b_1}{a_1}\right)\times \\ & \left|\begin{array}{ccccc}& 1& \frac{b_2}{a_2}–\frac{b_1}{a_1}& \frac{b_2^2}{a_2^2}–\frac{b_1^2}{a_1^2}& \\ \\ & \frac{b_3}{a_3}–\frac{b_1}{a_1}& \frac{b_3^2}{a_3^2}–\frac{b_1^2}{a_1^2}& \frac{b_3^3}{a_3^3}–\frac{b_1^3}{a_1^3}& \\ \\ & \frac{b_4}{a_4}–\frac{b_1}{a_1}& \frac{b_4^2}{a_4^2}–\frac{b_1^2}{a_1^2}& \frac{b_4^3}{a_4^3}–\frac{b_1^3}{a_1^3}& \end{array}\right|\\ \\ \\ \Rightarrow & a_1^3{a}_2^3{a}_3^3{a}_4^3\left(\frac{b_2}{a_2}–\frac{b_1}{a_1}\right)\left(\frac{b_3}{a_3}–\frac{b_1}{a_1}\right)\times \\ & \left|\begin{array}{ccccc}& 1& \frac{b_2}{a_2}–\frac{b_1}{a_1}& \frac{b_2^2}{a_2^2}–\frac{b_1^2}{a_1^2}& \\ \\ & 1& \frac{b_3}{a_3}–\frac{b_1}{a_1}& \frac{b_3^2}{a_3^2}–\frac{b_1^2}{a_1^2}& \\ \\ & \frac{b_4}{a_4}–\frac{b_1}{a_1}& \frac{b_4^2}{a_4^2}–\frac{b_1^2}{a_1^2}& \frac{b_4^3}{a_4^3}–\frac{b_1^3}{a_1^3}& \end{array}\right|\\ \\ \\ \Rightarrow & a_1^3{a}_2^3{a}_3^3{a}_4^3\left(\frac{b_2}{a_2}–\frac{b_1}{a_1}\right)\left(\frac{b_3}{a_3}–\frac{b_1}{a_1}\right)\left(\frac{b_4}{a_4}–\frac{b_1}{a_1}\right)\times \\ & \left|\begin{array}{ccccc}& 1& \frac{b_2}{a_2}–\frac{b_1}{a_1}& \frac{b_2^2}{a_2^2}–\frac{b_1^2}{a_1^2}& \\ \\ & 1& \frac{b_3}{a_3}–\frac{b_1}{a_1}& \frac{b_3^2}{a_3^2}–\frac{b_1^2}{a_1^2}& \\ \\ & 1& \frac{b_4}{a_4}–\frac{b_1}{a_1}& \frac{b_4^2}{a_4^2}–\frac{b_1^2}{a_1^2}& \end{array}\right|\\ \\ \\ \Rightarrow & a_1^3{a}_2^3{a}_3^3{a}_4^3\left(\frac{b_2}{a_2}–\frac{b_1}{a_1}\right)\left(\frac{b_3}{a_3}–\frac{b_1}{a_1}\right)\left(\frac{b_4}{a_4}–\frac{b_1}{a_1}\right)\times \\ & \left|\begin{array}{ccccc}& 1& \frac{b_2}{a_2}–\frac{b_1}{a_1}& \frac{b_2^2}{a_2^2}–\frac{b_1^2}{a_1^2}& \\ \\ & 0& \frac{b_3}{a_3}–\frac{b_2}{a_2}& \frac{b_3^2}{a_3^2}–\frac{b_2^2}{a_2^2}& \\ \\ & 0& \frac{b_4}{a_4}–\frac{b_2}{a_2}& \frac{b_4^2}{a_4^2}–\frac{b_2^2}{a_2^2}& \end{array}\right|\\ \\ \\ \Rightarrow & a_1^3{a}_2^3{a}_3^3{a}_4^3\left(\frac{b_2}{a_2}–\frac{b_1}{a_1}\right)\left(\frac{b_3}{a_3}–\frac{b_1}{a_1}\right)\left(\frac{b_4}{a_4}–\frac{b_1}{a_1}\right)\cdot 1\times \\ & \left|\begin{array}{cccc}& \frac{b_3}{a_3}–\frac{b_2}{a_2}& \frac{b_3^2}{a_3^2}–\frac{b_2^2}{a_2^2}& \\ \\ & \frac{b_4}{a_4}–\frac{b_2}{a_2}& \frac{b_4^2}{a_4^2}–\frac{b_2^2}{a_2^2}& \end{array}\right|\\ \\ \\ \Rightarrow & a_1^3{a}_2^3{a}_3^3{a}_4^3\left(\frac{b_2}{a_2}–\frac{b_1}{a_1}\right)\left(\frac{b_3}{a_3}–\frac{b_1}{a_1}\right)\left(\frac{b_4}{a_4}–\frac{b_1}{a_1}\right)\times \\ & \left|\begin{array}{cccc}& \frac{b_3}{a_3}–\frac{b_2}{a_2}& \frac{b_3^2}{a_3^2}–\frac{b_2^2}{a_2^2}& \\ \\ & \frac{b_4}{a_4}–\frac{b_2}{a_2}& \frac{b_4^2}{a_4^2}–\frac{b_2^2}{a_2^2}& \end{array}\right|\\ \\ \\ \Rightarrow & a_1^3{a}_2^3{a}_3^3{a}_4^3\left(\frac{b_2}{a_2}–\frac{b_1}{a_1}\right)\left(\frac{b_3}{a_3}–\frac{b_1}{a_1}\right)\left(\frac{b_4}{a_4}–\frac{b_1}{a_1}\right)\left(\frac{b_3}{a_3}–\frac{b_2}{a_2}\right)\left(\frac{b_4}{a_4}–\frac{b_2}{a_2}\right)\times \\ & \left|\begin{array}{cccc}& 1& \frac{b_3}{a_3}–\frac{b_2}{a_2}& \\ \\ & 1& \frac{b_4}{a_4}–\frac{b_2}{a_2}& \end{array}\right|\\ \\ \\ \Rightarrow & a_1^3{a}_2^3{a}_3^3{a}_4^3\left(\frac{b_2}{a_2}–\frac{b_1}{a_1}\right)\left(\frac{b_3}{a_3}–\frac{b_1}{a_1}\right)\left(\frac{b_4}{a_4}–\frac{b_1}{a_1}\right)\left(\frac{b_3}{a_3}–\frac{b_2}{a_2}\right)\left(\frac{b_4}{a_4}–\frac{b_2}{a_2}\right)\times \\ & \left|\begin{array}{cccc}& 1& \frac{b_3}{a_3}–\frac{b_2}{a_2}& \\ \\ & 0& \frac{b_4}{a_4}–\frac{b_3}{a_3}& \end{array}\right|\\ \\ \\ \Rightarrow & {\left({\mathit{a}}_{\mathbf{1}}{\mathit{a}}_{\mathbf{2}}{\mathit{a}}_{\mathbf{3}}{\mathit{a}}_{\mathbf{4}}\right)}^{\mathbf{3}}\left(\frac{{\mathit{b}}_{\mathbf{2}}}{{\mathit{a}}_{\mathbf{2}}}–\frac{{\mathit{b}}_{\mathbf{1}}}{{\mathit{a}}_{\mathbf{1}}}\right)\left(\frac{{\mathit{b}}_{\mathbf{3}}}{{\mathit{a}}_{\mathbf{3}}}–\frac{{\mathit{b}}_{\mathbf{1}}}{{\mathit{a}}_{\mathbf{1}}}\right)\left(\frac{{\mathit{b}}_{\mathbf{4}}}{{\mathit{a}}_{\mathbf{4}}}–\frac{{\mathit{b}}_{\mathbf{1}}}{{\mathit{a}}_{\mathbf{1}}}\right)\left(\frac{{\mathit{b}}_{\mathbf{3}}}{{\mathit{a}}_{\mathbf{3}}}–\frac{{\mathit{b}}_{\mathbf{2}}}{{\mathit{a}}_{\mathbf{2}}}\right)\left(\frac{{\mathit{b}}_{\mathbf{4}}}{{\mathit{a}}_{\mathbf{4}}}–\frac{{\mathit{b}}_{\mathbf{2}}}{{\mathit{a}}_{\mathbf{2}}}\right)\left(\frac{{\mathit{b}}_{\mathbf{4}}}{{\mathit{a}}_{\mathbf{4}}}–\frac{{\mathit{b}}_{\mathbf{3}}}{{\mathit{a}}_{\mathbf{3}}}\right)\\ \\ \\ \Rightarrow & {\left(a_1{a}_2{a}_3{a}_4\right)}^3\prod _{1⩽j<i⩽4}\left(\frac{b_i}{a_i}–\frac{b_j}{a_j}\right)\end{array}$$

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