一、题目
已知 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 线性无关，则线性无关的向量组是：
(A) $\boldsymbol{\alpha}_{1}-\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}-\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}-\boldsymbol{\alpha}_{1}$
(B) $\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}-\boldsymbol{\alpha}_{1}$
(C) $\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}+\boldsymbol{\alpha}_{1}$
(D) $\boldsymbol{\alpha}_{1}-\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}+\boldsymbol{\alpha}_{1}$
难度评级：
二、解析 
只有 C 选项存在如下情况：
$$
(\alpha_{1}, \alpha_{2}, \alpha_{3}) \begin{bmatrix}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{bmatrix} \Rightarrow
$$
$$
\begin{vmatrix}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{vmatrix} = 1 + 1 = 2 \neq 0
$$
因此，本题应选 C.