化简行列式的两种方式：消零和拆项

一、题目

$$
\begin{vmatrix}
\boldsymbol{D}
\end{vmatrix} = \begin{vmatrix}
1 & -1 & 1 & x-1 \\ 1 & -1 & x+1 & -1 \\ 1 & x-1 & 1 & -1 \\ x+1 & -1 & 1 & -1
\end{vmatrix} = ?
$$

二、解析

解法一：提取公因数+消零

观察可知，如果将行列式 $\begin{vmatrix}
\boldsymbol{D}
\end{vmatrix}$ 的第二列、第三列和第四列都加到第一列，则第一列就会变成：

$$
\begin{pmatrix}
x \\
x \\
x \\
x
\end{pmatrix}
$$

提取公因数之后，第一列就会变成：

$$
\begin{pmatrix}
1 \\
1 \\
1 \\
1
\end{pmatrix}
$$

由于此时行列式 $\begin{vmatrix}
\boldsymbol{D}
\end{vmatrix}$ 中的相同元素变多了，所以可以尝试使用初等变化消出来 $0$ 元素，从而简化计算难度（事实上，可以得到一个上三角行列式，或者一个下三角行列式，计算难度会变得很低）.

具体计算步骤如下：

$$
\begin{aligned}
\begin{vmatrix}
\boldsymbol{D}
\end{vmatrix} = & \begin{vmatrix}
1 & -1 & 1 & x-1 \\
1 & -1 & x+1 & -1 \\
1 & x-1 & 1 & -1 \\
x+1 & -1 & 1 & -1
\end{vmatrix} \\ \\
= & \begin{vmatrix}
x & -1 & 1 & x-1 \\
x & -1 & x+1 & -1 \\
x & x-1 & 1 & -1 \\
x & -1 & 1 & -1
\end{vmatrix} \\ \\
= x & \begin{vmatrix}
1 & -1 & 1 & x-1 \\
1 & -1 & x+1 & -1 \\
1 & x-1 & 1 & -1 \\
1 & -1 & 1 & -1
\end{vmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \text{第二列} + \text{第一列}, \text{第三列} – \text{第一列}, \text{第四列} + \text{第一列} \\ \\
= x & \begin{vmatrix}
1 & \textcolor{gray}{0} & \textcolor{gray}{0} & x \\
1 & \textcolor{gray}{0} & x & \textcolor{gray}{0} \\
1 & x & \textcolor{gray}{0} & \textcolor{gray}{0} \\
1 & \textcolor{gray}{0} & \textcolor{gray}{0} & \textcolor{gray}{0}
\end{vmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \text{第一行和第四行交换位置}, \text{第二行和第三行交换位置} \\ \\
= x & \begin{vmatrix}
\textcolor{lightgreen}{1} & \textcolor{gray}{0} & \textcolor{gray}{0} & \textcolor{gray}{0} \\
1 & \textcolor{lightgreen}{x} & \textcolor{gray}{0} & \textcolor{gray}{0} \\
1 & \textcolor{gray}{0} & \textcolor{lightgreen}{x} & \textcolor{gray}{0} \\
1 & \textcolor{gray}{0} & \textcolor{gray}{0} & \textcolor{lightgreen}{x}
\end{vmatrix} \\ \\
= & \textcolor{lightgreen}{ x^{4} }
\end{aligned}
$$

解法二：拆项化简

要想计算比较复杂的行列式，就需要对行列式进行化简，让行列式中出现尽可能多的零元素. 除了解法一所示的用初等变换在行列式中进行消零操作之外，另一个方法就是在本解法中所示的，用先补充零元素，后拆项的方法进行行列式的化简，从而简化行列式的计算.

$$
\begin{aligned}
\begin{vmatrix}
\boldsymbol{D}
\end{vmatrix} = \ & \begin{vmatrix}
1 & -1 & 1 & x-1 \\
1 & -1 & x+1 & -1 \\
1 & x-1 & 1 & -1 \\
x+1 & -1 & 1 & -1
\end{vmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \text{第一次构造拆项} \\ \\
= \ & \begin{vmatrix}
1+\textcolor{gray}{0} & -1 & 1 & x-1 \\
1+\textcolor{gray}{0} & -1 & x+1 & -1 \\
1+\textcolor{gray}{0} & x-1 & 1 & -1 \\
1+x & -1 & 1 & -1
\end{vmatrix} \\ \\
= \ & \begin{vmatrix}
1 & -1 & 1 & x-1 \\
1 & -1 & x+1 & -1 \\
1 & x-1 & 1 & -1 \\
1 & -1 & 1 & -1
\end{vmatrix} + \begin{vmatrix}
\textcolor{gray}{0} & -1 & 1 & x-1 \\
\textcolor{gray}{0} & -1 & x+1 & -1 \\
\textcolor{gray}{0} & x-1 & 1 & -1 \\
x & -1 & 1 & -1
\end{vmatrix} \\ \\
= \ & \begin{vmatrix}
1 & \textcolor{gray}{0} & \textcolor{gray}{0} & x \\
1 & \textcolor{gray}{0} & x & \textcolor{gray}{0} \\
1 & x & \textcolor{gray}{0} & \textcolor{gray}{0} \\
1 & \textcolor{gray}{0} & \textcolor{gray}{0} & \textcolor{gray}{0}
\end{vmatrix} + (-1)^{4+1} x \begin{vmatrix}
-1 & 1 & x-1 \\
-1 & x+1 & -1 \\
x-1 & 1 & -1
\end{vmatrix} \\ \\
= \ & \begin{vmatrix}
1 & \textcolor{gray}{0} & \textcolor{gray}{0} & \textcolor{gray}{0} \\
1 & x & \textcolor{gray}{0} & \textcolor{gray}{0} \\
1 & \textcolor{gray}{0} & x & \textcolor{gray}{0} \\
1 & \textcolor{gray}{0} & \textcolor{gray}{0} & x
\end{vmatrix} – x \begin{vmatrix}
-1 & 1 & x-1 \\
-1 & x+1 & -1 \\
x-1 & 1 & -1
\end{vmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \text{第二次构造拆项} \\ \\
= \ & x^{3} – x \begin{vmatrix}
\textcolor{gray}{0}-1 & 1 & x-1 \\
\textcolor{gray}{0}-1 & x+1 & -1 \\
x-1 & 1 & -1
\end{vmatrix} \\ \\
= \ & x^{3} – x \begin{vmatrix}
\textcolor{gray}{0} & 1 & x-1 \\
\textcolor{gray}{0} & x+1 & -1 \\
x & 1 & -1
\end{vmatrix} – x \begin{vmatrix}
-1 & 1 & x-1 \\
-1 & x+1 & -1 \\
-1 & 1 & -1
\end{vmatrix} \\ \\
= \ & x^{3}-x^{2} \begin{vmatrix}
1 & x-1 \\
x+1 & -1
\end{vmatrix} + x \begin{vmatrix}
1 & 1 & x-1 \\ 1 & x+1 & -1 \\ 1 & 1 & -1
\end{vmatrix} \\ \\
= \ & x^{3} + x^{4} + x \begin{vmatrix}
1 & \textcolor{gray}{0} & x \\
1 & x & \textcolor{gray}{0} \\
1 & \textcolor{gray}{0} & \textcolor{gray}{0}
\end{vmatrix} \\ \\
= \ & x^{3} + x^{4} – x \begin{vmatrix}
1 & \textcolor{gray}{0} & \textcolor{gray}{0} \\
1 & x & \textcolor{gray}{0} \\
1 & \textcolor{gray}{0} & x
\end{vmatrix} \\ \\
= \ & \textcolor{lightgreen}{ x^{4} }
\end{aligned}
$$

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