# 线性无关的向量组「乘以」线性相关的向量组会得到一个线性相关的向量组

## 一、题目

(A) $\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}+\boldsymbol{\alpha}_{4}, \boldsymbol{\alpha}_{4}+\boldsymbol{\alpha}_{1}$.

(B) $\boldsymbol{\alpha}_{1}-\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}-\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}-\boldsymbol{\alpha}_{4}, \boldsymbol{\alpha}_{4}-\boldsymbol{\alpha}_{1}$.

(C) $\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}-\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}-\boldsymbol{\alpha}_{4}, \boldsymbol{\alpha}_{4}+\boldsymbol{\alpha}_{1}$.

(D) $\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}-\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}-\boldsymbol{\alpha}_{4}, \boldsymbol{\alpha}_{4}-\boldsymbol{\alpha}_{1}$.

## 二、解析

$$( \boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}-\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}-\boldsymbol{\alpha}_{4}, \boldsymbol{\alpha}_{4}-\boldsymbol{\alpha}_{1} ) =$$

$$\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right)\left[\begin{array}{cccc}1 & 0 & 0 & -1 \\ 1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1\end{array}\right] \Rightarrow$$

$$\left|\begin{array}{cccc}1 & 0 & 0 & -1 \\ 1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1\end{array}\right|=\left|\begin{array}{ccc}1 & 1 & 0 \\ 0 & -1 & 1 \\ 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right|=$$

$$1+1=2 \neq 0$$