设 $\alpha_{1}=\left(\begin{array}{l}\lambda \\ 1 \\ 1\end{array}\right), \alpha_{2}=\left(\begin{array}{l}1 \\ \lambda \\ 1\end{array}\right), \alpha_{3}=\left(\begin{array}{l}1 \\ 1 \\ \lambda\end{array}\right), \alpha_{4}=\left(\begin{array}{l}1 \\ \lambda \\ \lambda^{2}\end{array}\right)$, 若 $\alpha_{1}, \alpha_{2}, \alpha_{3}$ 与 $\alpha_{1}, \alpha_{2}, \alpha_{4}$ 等价, 则 $\lambda \in(\quad)$.
A. $\{\lambda \mid \lambda \in \mathbb{R}\}$
B. $\{\lambda \mid \lambda \in \mathbb{R}, \lambda \neq-1\}$
C. $\{\lambda \mid \lambda \in \mathbb{R}, \lambda \neq-1, \lambda \neq-2\}$
D. $\{\lambda \mid \lambda \in \mathbb{R}, \lambda \neq-2\}$
由于
$$
\begin{aligned}
& \left|\alpha_1, \alpha_2, \alpha_3\right|=\left|\begin{array}{lll}
\lambda & 1 & 1 \\
1 & \lambda & 1 \\
1 & 1 & \lambda
\end{array}\right|=\lambda^3-3 \lambda+2=(\lambda-1)^2(\lambda+2), \\
& \left|\alpha_1, \alpha_2, \alpha_4\right|=\left|\begin{array}{lll}
\lambda & 1 & 1 \\
1 & \lambda & \lambda \\
1 & 1 & \lambda^2
\end{array}\right|=\lambda^4-2 \lambda^2+1=(\lambda-1)^2(\lambda+1)^2 .
\end{aligned}
$$

 当 $\lambda=1$ 时, $\alpha_1=\alpha_2=\alpha_3=\alpha_4=\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)$, 此时 $\alpha_1, \alpha_2, \alpha_3$ 与 $\alpha_1, \alpha_2, \alpha_4$ 等价.
当 $\lambda=-2$ 时, $2=r\left(\alpha_1, \alpha_2, \alpha_3\right)<r\left(\alpha_1, \alpha_2, \alpha_4\right)=3 ， \alpha_1, \alpha_2, \alpha_3$ 与 $\alpha_1, \alpha_2, \alpha_4$ 不等价.当 $\lambda=-1$ 时, $3=r\left(\alpha_1, \alpha_2, \alpha_3\right)>r\left(\alpha_1, \alpha_2, \alpha_4\right)=1, \alpha_1, \alpha_2, \alpha_3$ 与 $\alpha_1, \alpha_2, \alpha_4$ 不等价. 因 此当 $\lambda=-2$ 或 $\lambda=-1$ 时, $\alpha_1, \alpha_2, \alpha_3$ 与 $\alpha_1, \alpha_2, \alpha_4$ 不等价等价, 所以 $\lambda$ 的取值范围为 $\{\lambda \mid \lambda \in \mathbb{R}, \lambda \neq-1, \lambda \neq-2\}$.

因此答案为：C