已知 beta_1 = alpha_1 - alpha_2，beta_2 = alpha_2 - alpha_3，beta_3 = alpha_3 - alpha_1，gamma_1 = alpha_1 + alpha_2，gamma_2 = alpha_2 + alpha_3，gamma_3 = alpha_3 + alpha_1，下列结论中错误的是（）。A 若 alpha_1, alpha_2, alpha_3 线性相关，则 beta_1, beta_2, beta_3 线性相关。B 若 alpha_1, alpha_2, alpha_3 线性无关，则 beta_1, beta_2, beta_3 线性无关。C 若 alpha_1, alpha_2, alpha_3 线性相关，则 gamma_1, gamma_2, gamma_3 线性相关。D 若 alpha_1, alpha_2, alpha_3 线性无关，则 gamma_1, gamma_2, gamma_3 线性无关。

本题主要考查向量组线性相关与线性无关的概念及性质，解题的关键在于根据已知的向量组之间的线性关系，通过定义来判断向量组的线性相关性。

选项A

若$\alpha_1, \alpha_2, \alpha_3$线性相关，根据线性相关的定义，存在不全为零的实数$k_1, k_2, k_3$，使得$k_1\alpha_1 + k_2\alpha_2 + k_3\alpha_3 = 0$。
已知$\beta_1 = \alpha_1 - \alpha_2$，$\beta_2 = \alpha_2 - \alpha_3$，$\beta_3 = \alpha_3 - \alpha_1$，设$a_1\beta_1 + a_2\beta_2 + a_3\beta_3 = 0$，将$\beta_1, \beta_2, \beta_3$代入可得：
$\begin{align*}a_1(\alpha_1 - \alpha_2) + a_2(\alpha_2 - \alpha_3) + a_3(\alpha_3 - \alpha_1) &= 0\\a_1\alpha_1 - a_1\alpha_2 + a_2\alpha_2 - a_2\alpha_3 + a_3\alpha_3 - a_3\alpha_1 &= 0\\(a_1 - a_3)\alpha_1 + (-a_1 + a_2)\alpha_2 + (-a_2 + a_3)\alpha_3 &= 0\end{align*}$
因为$\alpha_1, \alpha_2, \alpha_3$线性相关，所以存在不全为零的$k_1, k_2, k_3$使得$k_1\alpha_1 + k_2\alpha_2 + k_3\alpha_3 = 0$，那么对于方程$(a_1 - a_3)\alpha_1 + (-a_1 + a_2)\alpha_2 + (-a_2 + a_3)\alpha_3 = 0$，一定存在不全为零的$a_1, a_2, a_3$使得该方程成立，即$\beta_1, \beta_2, \beta_3$线性相关，选项A正确。

选项B

已知$\beta_1 = \alpha_1 - \alpha_2$，$\beta_2 = \alpha_2 - \alpha_3$，$\beta_3 = \alpha_3 - \alpha_1$，则$\beta_1 + \beta_2 + \beta_3 = (\alpha_1 - \alpha_2) + (\alpha_2 - \alpha_3) + (\alpha_3 - \alpha_1) = 0$。
根据线性相关的定义，若存在不全为零的数（这里是$1,1,1$）使得向量组的线性组合为零向量，则该向量组线性相关，所以$\beta_1, \beta_2, \beta_3$线性相关，而不是线性无关，选项B错误。

选项C

若$\alpha_1, \alpha_2, \alpha_3$线性相关，存在不全为零的实数$k_1, k_2, k_3$，使得$k_1\alpha_1 + k_2\alpha_2 + k_3\alpha_3 = 0$。
已知$\gamma_1 = \alpha_1 + \alpha_2$，$\gamma_2 = \alpha_2 + \alpha_3$，$\gamma_3 = \alpha_3 + \alpha_1$，设$a_1\gamma_1 + a_2\gamma_2 + a_3\gamma_3 = 0$，将$\gamma_1, \gamma_2, \gamma_3$代入可得：
$\begin{align*}a_1(\alpha_1 + \alpha_2) + a_2(\alpha_2 + \alpha_3) + a_3(\alpha_3 + \alpha_1) &= 0\\a_1\alpha_1 + a_1\alpha_2 + a_2\alpha_2 + a_2\alpha_3 + a_3\alpha_3 + a_3\alpha_1 &= 0\\(a_1 + a_3)\alpha_1 + (a_1 + a_2)\alpha_2 + (a_2 + a_3)\alpha_3 &= 0\end{align*}$
因为$\alpha_1, \alpha_2, \alpha_3$线性相关，所以存在不全为零的$k_1, k_2, k_3$使得$k_1\alpha_1 + k_2\alpha_2 + k_3\alpha_3 = 0$，那么对于方程$(a_1 + a_3)\alpha_1 + (a_1 + a_2)\alpha_2 + (a_2 + a_3)\alpha_3 = 0$，一定存在不全为零的$a_1, a_2, a_3$使得该方程成立，即$\gamma_1, \gamma_2, \gamma_3$线性相关，选项C正确。

选项D

若$\alpha_1, \alpha_2, \alpha_3$线性无关，设$a_1\gamma_1 + a_2\gamma_2 + a_3\gamma_3 = 0$，将$\gamma_1 = \alpha_1 + \alpha_2$，$\gamma_2 = \alpha_2 + \alpha_3$，$\gamma_3 = \alpha_3 + \alpha_1$代入可得：
$\begin{align*}a_1(\alpha_1 + \alpha_2) + a_2(\alpha_2 + \alpha_3) + a_3(\alpha_3 + \alpha_1) &= 0\\a_1\alpha_1 + a_1\alpha_2 + a_2\alpha_2 + a_2\alpha_3 + a_3\alpha_3 + a_3\alpha_1 &= 0\\(a_1 + a_3)\alpha_1 + (a_1 + a_2)\alpha_2 + (a_2 + a_3)\alpha_3 &= 0\end{align*}$
因为$\alpha_1, \alpha_2, \alpha_3$线性无关，所以$\begin{cases}a_1 + a_3 = 0\\a_1 + a_2 = 0\\a_2 + a_3 = 0\end{cases}$，解这个方程组：
由$a_1 + a_3 = 0$可得$a_3 = -a_1$，将其代入$a_2 + a_3 = 0$可得$a_2 - a_1 = 0$，即$a_2 = a_1$，再将$a_2 = a_1$代入$a_1 + a_2 = 0$可得$2a_1 = 0$，所以$a_1 = 0$，进而可得$a_2 = 0$，$a_3 = 0$。
即只有当$a_1 = a_2 = a_3 = 0$时，$a_1\gamma_1 + a_2\gamma_2 + a_3\gamma_3 = 0$成立，所以$\gamma_1, \gamma_2, \gamma_3$线性无关，选项D正确。