设向量组alpha_1, alpha_2, alpha_3, alpha_4线性无关，则()。A. alpha_1 + alpha_2, alpha_2 + alpha_3, alpha_3 + alpha_4, alpha_4 - alpha_1，线性无关B. alpha_1 + alpha_2, alpha_2 + alpha_3, alpha_3 - alpha_4, alpha_4 - alpha_1，线性无关C. alpha_1 + alpha_2, alpha_2 + alpha_3, alpha_3 + alpha_4, alpha_4 - alpha_1，线性无关D. alpha_1 - alpha_2, alpha_2 - alpha_3, alpha_3 - alpha_4, alpha_4 - alpha_1，线性无关
设向量组$\alpha_1, \alpha_2, \alpha_3, \alpha_4$线性无关，则()。
A. $\alpha_1 + \alpha_2, \alpha_2 + \alpha_3, \alpha_3 + \alpha_4, \alpha_4 - \alpha_1$，线性无关
B. $\alpha_1 + \alpha_2, \alpha_2 + \alpha_3, \alpha_3 - \alpha_4, \alpha_4 - \alpha_1$，线性无关
C. $\alpha_1 + \alpha_2, \alpha_2 + \alpha_3, \alpha_3 + \alpha_4, \alpha_4 - \alpha_1$，线性无关
D. $\alpha_1 - \alpha_2, \alpha_2 - \alpha_3, \alpha_3 - \alpha_4, \alpha_4 - \alpha_1$，线性无关
题目解答
答案
答案：A
解析：
对于选项A，向量组为 $a_1 + a_2, a_2 + a_3, a_3 + a_4, a_4 - a_1$。设 $k_1(a_1 + a_2) + k_2(a_2 + a_3) + k_3(a_3 + a_4) + k_4(a_4 - a_1) = 0$，合并同类项得：
$(k_1 - k_4)a_1 + (k_1 + k_2)a_2 + (k_2 + k_3)a_3 + (k_3 + k_4)a_4 = 0$
由线性无关性，系数全为零，解得 $k_1 = k_2 = k_3 = k_4 = 0$，故线性无关。
选项B中，向量组为 $a_1 + a_2, a_2 + a_3, a_3 - a_4, a_4 - a_1$，类似分析可得存在非零解，线性相关。
选项C与A相同，线性无关。
选项D中，向量组为 $a_1 - a_2, a_2 - a_3, a_3 - a_4, a_4 - a_1$，类似分析可得存在非零解，线性相关。
结论：
正确选项为 $\boxed{A}$（或 $\boxed{C}$，但题目中选项A和C相同，故选A）。
解析
本题考查向量组线性无关的判定知识。解题思路是通过设线性组合等于零向量，然后合并同类项，根据已知向量组线性无关得到系数的方程组，求解系数，若系数全为零则向量组线性线性无关，若存在非零解则向量组线性相关。
对于选项A，向量组为 $\alpha_1 + \alpha_2, \text{, } \alpha_2 + \alpha_3 \text{, } \alpha_3 + \alpha_4 \text \ \alpha_4 - \alpha_1$。设 $k_1(\alpha_1 + \alpha_2) + k_2(\alpha_2 + \alpha_3) + k__3(\alpha_3 + \alpha_4) + k_4(\alpha_4 - \alpha_1) = 0$，合并同类项得：
$(k_1 - k_4)\alpha_1 + (k_1 + k_2)\alpha_2 + (k_2 + k_3)\alpha_3 + (k_3 + k_4)\alpha_4 = 0$
由线性无关性，系数全为零，即 $\begin{cases}k_1 - k_4 = 0\\k_1 + k_2 =0\\k_2 + k_3 =0\\k_3 + k_4 =0\end{cases}$，解得 $k_1 = k_2 = k_3 = k_4 = 0$，故线性无关。
对于选项B中，向量组为 $\alpha_1 + \alpha_2, \alpha_2 + \alpha_3, \alpha_3 - \alpha_4, \alpha_4 - \alpha_1$，设 $k_1(\alpha_1 + \alpha_2) + k_2(\alpha_2 + \alpha_3) + k_3(\alpha_3 - \alpha_4) + k_4(\alpha_4 - \alpha_1) = 0$，合并同类项得 $(k_1 - k_4)\alpha_1 + (k_1 + k_2)\alpha_2 + (k_2 + k_3)\alpha_3 + (k_3 + k_4)\alpha_4 = 0$，得到系数方程组 $\begin{cases}k_1 - k_4 =0\\k_1 + k_2 =0\\k_2 + k_3 =0\\k_3 + k_4 =0{cases}$，该方程组存在非零解，所以线性相关。
选项C与A相同，向量组为 $\alpha_1 + \alpha_2, \alpha_2 + \alpha_3, \alpha_3 + \alpha_4, \alpha_4 - \alpha_1$，设 $k_1(\alpha_1 + \alpha_2) + k_2(\alpha_2 + \alpha_3) + k_3(\alpha_3 + \alpha_4) + k_4(\alpha_4 - \alpha_1) = 0$，合并同类项得 $(k_1 - k_4)\alpha_1 + (k_1 + k_2)\alpha_2 + (k_2 + k_3)\alpha_3 + (k_3 + k_4)\alpha_4 = 0$，得到系数方程组 $\begin{cases}k_1 - k_4 =0\\k_1 + k_2 =0\\k_2 + k_3 =0\\k_3 + k_4 =0{cases}$，解得 $k_1 = k_2 = k_3 = k_4 = 0$，故线性无关。
选项D中，向量组为 $\alpha_1 - \alpha_2, \alpha_2 - \alpha_3, \alpha_3 - \alpha_4, \alpha_4 - \alpha_1$，设 $k_1(\alpha_1 - \alpha_2) + k_2(\alpha_2 - \alpha_3) + k_3(\alpha_3 - \alpha_4) + k_4(\alpha_4 - \alpha_1) = 0$，合并同类项得 $(k_1 - k_4)\alpha_1 + (-k_1 + k_2)\alpha_2 + (-k_2 + k_3)\alpha_3 + (-k_3 + k_4)\alpha_4 = 0$，得到系数方程组 $\begin{cases}k_1 - k_4 =0\\-k_1 + k_2 =0\\k_2 + k_3 =0\\k_3 + k_4 =0{cases}$，该方程组有非零解，所以线性相关。