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 $a _{ i }$ $\neq$ $0$ ($i$ $=$ $1$, $2$, $3$, $4$), 则：

$$
|V| =
\begin{vmatrix}
& a_{1}^{3} & a_{1}^{2}b_{1} & a_{1}b_{1}^{2} & b_{1}^{3} & \\ \\
& a_{2}^{3} & a_{2}^{2}b_{2} & a_{2}b_{2}^{2} & b_{2}^{3} & \\ \\
& a_{3}^{3} & a_{3}^{2}b_{3} & a_{3}b_{3}^{2} & b_{3}^{3} & \\ \\
& a_{4}^{3} & a_{4}^{2}b_{4} & a_{4}b_{4}^{2} & b_{4}^{3} &
\end{vmatrix} ⁢= ?
$$

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：

$$
a_{1} ^{2} \quad a_{1} b_{1} \quad b_{1} ^{2} \quad \frac{b_{1} ^{3}}{a_{1}}
$$

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：

$$
a_{1} \quad b_{1} \quad \frac{b_{1} ^{2}}{a_{1}} \quad \frac{b_{1} ^{3}}{a_{1} ^{2}}
$$

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：

$$
1 \quad \frac{b_{1}}{a_{1}} \quad \frac{b_{1} ^{2}}{a_{1} ^{2}} \quad \frac{b_{1} ^{3}}{a_{1} ^{3}}
$$

于是可知，我们可以对行列式 $|V|$ 做如下变形：

$$
\begin{aligned}
& \begin{vmatrix}
& 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{vmatrix} \\ \\
\textcolor{pink}{ \Rightarrow } a_{1} ^{3} & \begin{vmatrix}
& 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{vmatrix} \\ \\
\textcolor{pink}{ \Rightarrow } a_{1} ^{3} a_{2} ^{3} & \begin{vmatrix}
& 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{vmatrix} \\ \\
\textcolor{pink}{ \Rightarrow } a_{1} ^{3} a_{2} ^{3} a_{3} ^{3} & \begin{vmatrix}
& 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{vmatrix} \\ \\
\textcolor{pink}{ \Rightarrow } a_{1} ^{3} a_{2} ^{3} a_{3} ^{3} a_{4} ^{3} & \begin{vmatrix}
& 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{vmatrix} \\ \\
\textcolor{pink}{ \Rightarrow } a_{1} ^{3} a_{2} ^{3} a_{3} ^{3} a_{4} ^{3} & \begin{vmatrix}
& \textcolor{orangered}{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{vmatrix} \\ \\
\textcolor{pink}{ \Rightarrow } a_{1} ^{3} a_{2} ^{3} a_{3} ^{3} a_{4} ^{3} \cdot \textcolor{orangered}{1} & \textcolor{springgreen}{ \begin{vmatrix}
& \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{vmatrix} }
\end{aligned}
$$

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{aligned}
& a_{1} ^{3} a_{2} ^{3} a_{3} ^{3} a_{4} ^{3} \textcolor{springgreen}{ \begin{vmatrix} & \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{vmatrix} } \\ \\ \\
\textcolor{pink}{ \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 \\ & \begin{vmatrix}
& 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{vmatrix} \\ \\ \\
\textcolor{pink}{ \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 \\ & \begin{vmatrix}
& 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{vmatrix} \\ \\ \\
\textcolor{pink}{ \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 \\ & \begin{vmatrix}
& 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{vmatrix} \\ \\ \\
\textcolor{pink}{ \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 \\ & \begin{vmatrix}
& \textcolor{orangered}{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{vmatrix} \\ \\ \\
\textcolor{pink}{ \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 \textcolor{orangered}{1} \times \\ & \begin{vmatrix}
& \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{vmatrix} \\ \\ \\
\textcolor{pink}{ \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 \\ & \begin{vmatrix}
& \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{vmatrix} \\ \\ \\
\textcolor{pink}{ \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 \\ & \begin{vmatrix}
& 1 & \frac{b_{3}}{a_{3}} – \frac{b_{2}}{a_{2}} & \\ \\
& 1 & \frac{b_{4}}{a_{4}} – \frac{b_{2}}{a_{2}} &
\end{vmatrix} \\ \\ \\
\textcolor{pink}{ \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 \\ & \begin{vmatrix}
& 1 & \frac{b_{3}}{a_{3}} – \frac{b_{2}}{a_{2}} & \\ \\
& 0 & \frac{b_{4}}{a_{4}} – \frac{b_{3}}{a_{3}} &
\end{vmatrix} \\ \\ \\
\textcolor{pink}{ \Rightarrow } & \textcolor{green}{\boldsymbol{ \left( a_{1} a_{2} a_{3} a_{4} \right) ^{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) \left( \frac{b_{4}}{a_{4}} – \frac{b_{3}}{a_{3}} \right) }} \\ \\ \\
\textcolor{pink}{ \Rightarrow } & \left( a _ { 1 } a _ { 2 } a _ { 3 } a _ { 4 } \right) ^ { 3 } \prod _ { 1 \leqslant j < i \leqslant 4 } \left( \frac { b _ { i } } { a _ { i } } – \frac { b _ { j } } { a _ { j } } \right)
\end{aligned}
$$

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