$\alpha_1+\alpha_2, \alpha_2+\alpha_3, \alpha_3+\alpha_1$ 也是 $\boldsymbol{A x}=\mathbf{0}$ 的一个基础解系.

b.向量组 $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \cdots, \boldsymbol{\alpha}_n$ 线性无关 $\Leftrightarrow \mathrm{r}\left(\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \cdots, \boldsymbol{\alpha}_n\right)=n$.

$\alpha_1+\alpha_2, \alpha_2+\alpha_3, \alpha_3+\alpha_1$ 也是 $A x=0$ 的解;
$\alpha_1, \alpha_2, \alpha_3$ 作为 $\boldsymbol{A x}=\mathbf{0}$ 的一个基础解系, 因此线性无关。

$$\left(\alpha_1+\alpha_2, \boldsymbol{\alpha}_2+\boldsymbol{\alpha}_3, \boldsymbol{\alpha}_3+\boldsymbol{\alpha}_1\right)=\left(\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3\right)\left(\begin{array}{lll} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right),$$
$\boldsymbol{B}=\left(\alpha_1+\alpha_2, \alpha_2+\alpha_3, \alpha_3+\alpha_1\right), \boldsymbol{A}=\left(\alpha_1, \alpha_2, \alpha_3\right), \boldsymbol{K}=\left(\begin{array}{lll}1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right)$,

$\text { 计算 得 }|\boldsymbol{K}|=2 \neq 0$ $\text { 知 } \boldsymbol{K}=\left(\begin{array}{lll} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right) \text { 可逆 }$

$r(B)=r(A)$

$\mathrm{r}(\boldsymbol{B})=\mathrm{r}(\boldsymbol{A})=3$, 向量组 $\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2, \boldsymbol{\alpha}_2+\alpha_3, \boldsymbol{\alpha}_3+\boldsymbol{\alpha}_1$ 的秩 $\mathrm{r}\left(\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2, \boldsymbol{\alpha}_2+\alpha_3, \boldsymbol{\alpha}_3+\boldsymbol{\alpha}_1\right)=\mathrm{r}(\boldsymbol{B})=3$, 因此(析 $\left.\mathbf{b}\right)$