交换一次行或者列的位置，行列式的值就要变一次正负号

一、题目

已知 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\beta}_{1}, \boldsymbol{\beta}_{2}$ 均为四维列向量, $\left|\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\beta}_{1}\right|=a$, $\left|\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\beta}_{2}, \boldsymbol{\alpha}_{3}\right|=b$, 则 $\left|\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{1}, \boldsymbol{\beta}_{1}+2 \boldsymbol{\beta}_{2}\right|=?$

难度评级：

二、解析

$$
\left(\alpha_{3}, \alpha_{2}, \alpha_{1}, \beta_{1}+2 \beta_{2}\right)=
$$

$$
\left(\alpha_{3}, \alpha_{2}, \alpha_{1}, \beta_{1}\right)+\left(\alpha_{3}, \alpha_{2}, \alpha_{1}, 2 \beta_{2}\right)=
$$

$\left(\alpha_{3}, \alpha_{2}, \alpha_{1}, \beta_{1}\right)$ 需要经过 $3$ 次列变换变成 $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \beta_{1}\right)$, 需要加负号，而 $\left(\alpha_{3}, \alpha_{2}, \alpha_{1}, 2 \beta_{2}\right)$ 需要经过 $4$ 次列变换变成 $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, 2\beta_{2}\right)$, 不需要加负号。

$$
-\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \beta_{1}\right)+2\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \beta_{2}\right)=
$$

$$
-a+2 b=2 b-a
$$

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