2022考研数二第10题解析：向量组等价、初等变换、矩阵的秩

一、题目

»A« $\left\{0, 1\right\}$

»B« $\left\{\lambda \ \middle| \ \lambda \in \mathbb{R}, \lambda \neq -2\right\}$

»C« $\left\{\lambda \ \middle| \ \lambda \in \mathbb{R}, \lambda \neq -1, \lambda \neq -2\right\}$

»D« $\left\{\lambda \ \middle| \ \lambda \in \mathbb{R}, \lambda \neq -1\right\}$

二、解析

解法 1

首先：

$$
\begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4} \end{pmatrix} = \begin{pmatrix} \lambda & 1 & 1 & 1 \\ 1 & \lambda & 1 & \lambda \\ 1 & 1 & \lambda & \lambda^{2} \end{pmatrix}
$$

于是，通过初等行变换（为了不破坏 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4}$ 之间的相对位置，不能做初等列变换）：

$$
\begin{aligned}
& \ \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4} \end{pmatrix} \\ \\
= & \ \begin{pmatrix}
\lambda & 1 & 1 & 1 \\
1 & \lambda & 1 & \lambda \\
1 & 1 & \lambda & \lambda^{2}
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \textcolor{gray}{\text{第二行减去第一行}} \\ \\
= & \ \begin{pmatrix}
\lambda & 1 & 1 & 1 \\
0 & \lambda – 1 & 1 – \lambda & \lambda – \lambda^{2} \\
1 & 1 & \lambda & \lambda^{2}
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \textcolor{gray}{\text{第二行乘以} -1} \\ \\
= & \ \begin{pmatrix}
\lambda & 1 & 1 & 1 \\
0 & 1 – \lambda & \lambda – 1 & \lambda^{2} – \lambda \\
1 & 1 & \lambda & \lambda^{2}
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \textcolor{gray}{\text{第三行减去第二行}} \\ \\
= & \ \begin{pmatrix}
\lambda & 1 & 1 & 1 \\
0 & 1 – \lambda & \lambda – 1 & \lambda^{2} – \lambda \\
1 & \lambda & 1 & \lambda
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \textcolor{gray}{\text{第一行和第三行交换位置}} \\ \\
= & \ \begin{pmatrix}
1 & \lambda & 1 & \lambda \\
0 & 1 – \lambda & \lambda – 1 & \lambda^{2} – \lambda \\
\lambda & 1 & 1 & 1
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \textcolor{gray}{\text{第三行减去第一行的 } \lambda \text{ 倍}} \\ \\
= & \ \begin{pmatrix}
1 & \lambda & 1 & \lambda \\
0 & 1 – \lambda & \lambda – 1 & \lambda^{2} – \lambda \\
0 & 1 – \lambda^{2} & 1 – \lambda & 1 – \lambda^{2}
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \textcolor{gray}{\text{第三行减去第二行的 } 1+ \lambda \text{ 倍}} \\ \\
= & \ \textcolor{lightgreen}{ \begin{pmatrix}
1 & \lambda & 1 & \lambda \\
0 & 1 – \lambda & \lambda – 1 & \lambda^{2} – \lambda \\
0 & 0 & -\left(\lambda + 2\right)\left(\lambda – 1\right) & \left(1 + \lambda\right)\left(1 – \lambda^{2}\right)
\end{pmatrix} }
\end{aligned}
$$

接着，根据《等价向量组及其性质》这篇文章可知，如果向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 与 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 等价，则：

$$
\mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{pmatrix} = \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix}
$$

又因为，用向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 可以表示出向量 $\boldsymbol{\alpha}_{1}$, 还可以表示出向量 $\boldsymbol{\alpha}_{2}$, 还可以表示出向量 $\boldsymbol{\alpha}_{3}$, 因此，将向量 $\boldsymbol{\alpha}_{3}$ 添加到向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 中，并不会导致向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 的秩发生变化，因此，有：

$$
\textcolor{lightgreen}{
\mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{pmatrix} = \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix} = \mathbf{r} \left( \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4} \right)
} \tag{1}
$$

于是：

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $\lambda = -1$ 时，由 $\begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4} \end{pmatrix} = \textcolor{lightgreen}{ \begin{pmatrix} 1 & \lambda & 1 & \lambda \\ 0 & 1 – \lambda & \lambda – 1 & \lambda^{2} – \lambda \\ 0 & 0 & -\left(\lambda + 2\right)\left(\lambda – 1\right) & \left(1 + \lambda\right)\left(1 – \lambda^{2}\right) \end{pmatrix} }$ 可知：

$$
\begin{aligned}
& \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{pmatrix} = 3 \\
& \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix} = 2
\end{aligned}
$$

与前面的 $(1)$ 式矛盾，所以：

$$
\textcolor{lightgreen}{
\lambda \neq -1
}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $\lambda = -2$ 时，由 $\begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4} \end{pmatrix} = \textcolor{lightgreen}{ \begin{pmatrix} 1 & \lambda & 1 & \lambda \\ 0 & 1 – \lambda & \lambda – 1 & \lambda^{2} – \lambda \\ 0 & 0 & -\left(\lambda + 2\right)\left(\lambda – 1\right) & \left(1 + \lambda\right)\left(1 – \lambda^{2}\right) \end{pmatrix} }$ 可知：

$$
\begin{aligned}
& \mathbf{r} \left( \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{5} \right) = 2 \\
& \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix} = 3
\end{aligned}
$$

与前面的 $(1)$ 式矛盾，所以：

$$
\textcolor{lightgreen}{
\lambda \neq -2
}
$$

结合上面的结论，可以排除 »B« 选项和 »D« 选项.

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $\lambda = 2$ 时，由 $\begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4} \end{pmatrix} = \textcolor{lightgreen}{ \begin{pmatrix} 1 & \lambda & 1 & \lambda \\ 0 & 1 – \lambda & \lambda – 1 & \lambda^{2} – \lambda \\ 0 & 0 & -\left(\lambda + 2\right)\left(\lambda – 1\right) & \left(1 + \lambda\right)\left(1 – \lambda^{2}\right) \end{pmatrix} }$ 可知：

$$
\mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{pmatrix} = \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix} = 3
$$

因此，可以排除 »A« 选项.

即：

$$
\textcolor{lightgreen}{
\left\{\lambda \ \middle| \ \lambda \in \mathbb{R}, \ \lambda \neq -1, \ \lambda \neq -2\right\}
}
$$

综上可知，本 题 应 选 C

解法 2

$\textcolor{lightgreen}{\blacktriangleright}$ 首先，当 $\lambda = 1$ 的时候，有：

$$
\boldsymbol{\alpha}_{1} = \boldsymbol{\alpha}_{2} = \boldsymbol{\alpha}_{3} = \boldsymbol{\alpha}_{4} = \left( 1, 1, 1\right)^{\top}
$$

此时，一定有向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 与 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 等价.

同时，当 $\lambda = 1$ 的时候，»A«, »B«, »C«, »D« 四个选项都正确.

$\textcolor{lightgreen}{\blacktriangleright}$ 接着，当 $\lambda \neq 1$ 的时候，考虑矩阵：

$$
\begin{aligned}
\boldsymbol{A} & = \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4} \end{pmatrix} \\ \\
& = \begin{pmatrix}
\lambda & 1 & 1 & 1 \\
1 & \lambda & 1 & \lambda \\
1 & 1 & \lambda & \lambda^{2}
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{第一行和第二行交换位置}} \\ \\
& = \begin{pmatrix}
1 & \lambda & 1 & \lambda \\
\lambda & 1 & 1 & 1 \\
1 & 1 & \lambda & \lambda^{2}
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{第二行和第三行交换位置}} \\ \\
& = \begin{pmatrix}
1 & \lambda & 1 & \lambda \\
1 & 1 & \lambda & \lambda^{2} \\
\lambda & 1 & 1 & 1
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{第二行减去第一行}} \\ \\
& = \begin{pmatrix}
1 & \lambda & 1 & \lambda \\
0 & 1-\lambda & \lambda-1 & \lambda^{2}-\lambda \\
\lambda & 1 & 1 & 1
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{第三行减去第一行的 } \lambda \text{ 倍}} \\ \\
& = \begin{pmatrix}
1 & \lambda & 1 & \lambda \\
0 & 1 – \lambda & \lambda – 1 & \lambda^{2} – \lambda \\
0 & 1 – \lambda^{2} & 1 – \lambda & 1 – \lambda^{2}
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{第二行乘以 } \frac{1}{1-\lambda}} \\ \\
& = \begin{pmatrix}
1 & \lambda & 1 & \lambda \\
0 & 1 & -1 & -\lambda \\
0 & 1 – \lambda^{2} & 1 – \lambda & 1 – \lambda^{2}
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{第三行乘以 } \frac{1}{1-\lambda}} \\ \\
& = \begin{pmatrix}
1 & \lambda & 1 & \lambda \\
0 & 1 & -1 & -\lambda \\
0 & 1 + \lambda & 1 & 1 + \lambda
\end{pmatrix} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{第三行减去第二行的 } 1+\lambda \text{ 倍}} \\ \\
& = \textcolor{lightgreen}{ \begin{pmatrix}
1 & \lambda & 1 & \lambda \\
0 & 1 & -1 & -\lambda \\
0 & 0 & \lambda + 2 & \left( \lambda + 1 \right)^{2}
\end{pmatrix} }
\end{aligned}
$$

由于，当 $\lambda \neq 1$ 的时候，矩阵 $\boldsymbol{A}$ 存在 $2$ 阶非零子式：

$$
\begin{vmatrix} 1 & \lambda \\ 0 & 1 \end{vmatrix} = 1 \neq 0
$$

所以：

$$
\mathbf{r} \left( \boldsymbol{A} \right) \geqslant 2
$$

又因为，不存在 $\lambda$ 的任何取值，使得下式成立：

$$
\lambda + 2 = \left( \lambda + 1 \right)^{2} = 0
$$

所以：

$$
\mathbf{r} \left( \boldsymbol{A} \right) = 3
$$

接着，从上面的计算结果 $\boldsymbol{A} = \textcolor{lightgreen}{ \begin{pmatrix} 1 & \lambda & 1 & \lambda \\ 0 & 1 & -1 & -\lambda \\ 0 & 0 & \lambda + 2 & \left( \lambda + 1 \right)^{2} \end{pmatrix} }$ 可知：

$\textcolor{orange}{\blacktriangleright}$ 当且仅当 $\lambda \neq -2$ 时，有：

$$
r \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{pmatrix} = 3
$$

$\textcolor{orange}{\blacktriangleright}$ 当且仅当 $\lambda \neq -1$ 时，有：

$$
\mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix} = 3
$$

$\textcolor{orange}{\blacktriangleright}$ 因此，当且仅当 $\lambda \neq -2$ 且 $\lambda \neq -1$ 时，有：

$$
\textcolor{lightgreen}{
\mathbf{r} \left( \boldsymbol{A} \right) = \mathbf{r} \begin{pmatrix}
\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4}
\end{pmatrix} = \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{pmatrix} = \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix}
} \tag{2}
$$

其中，当 $\lambda = 1$ 时，上面的 $(2)$ 式也成立.

接着，根据《等价向量组及其性质》这篇文章可知，如果上面的 $(2)$ 式成立，则向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 与 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 等价.

综上可知，本 题 应 选 C

解法 3

首先，根据《等价向量组及其性质》这篇文章可知，向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 与 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 等价的充分必要条件是：

$$
\mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{pmatrix} = \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix} = \mathbf{r} \left( \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \right)
$$

又因为：

$$
\begin{aligned}
& \ \begin{pmatrix}
\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}
\end{pmatrix} \\ \\
= & \ \left( \begin{array}{ccc:ccc}
\lambda & 1 & 1 & \lambda & 1 & 1 \\
1 & \lambda & 1 & 1 & \lambda & \lambda \\
1 & 1 & \lambda & 1 & 1 & \lambda^{2}
\end{array} \right) \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{交换第一行和第三行}} \\ \\
= & \ \left( \begin{array}{ccc:ccc}
1 & 1 & \lambda & 1 & 1 & \lambda^{2} \\
1 & \lambda & 1 & 1 & \lambda & \lambda \\
\lambda & 1 & 1 & \lambda & 1 & 1
\end{array} \right) \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{第二行减第一行}} \\ \\
= & \ \left( \begin{array}{ccc:ccc}
1 & 1 & \lambda & 1 & 1 & \lambda^{2} \\
0 & \lambda-1 & 1-\lambda & 0 & \lambda-1 & \lambda-\lambda^{2} \\
\lambda & 1 & 1 & \lambda & 1 & 1
\end{array} \right) \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{第三行减第一行的 } \lambda \text{ 倍}} \\ \\
= & \ \left( \begin{array}{ccc:ccc}
1 & 1 & \lambda & 1 & 1 & \lambda^{2} \\
0 & \lambda – 1 & 1 – \lambda & 0 & \lambda – 1 & \lambda \left( 1 – \lambda \right) \\
0 & 1 – \lambda & 1 – \lambda^{2} & 0 & 1 – \lambda & 1 – \lambda^{3}
\end{array} \right) \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{第三行加上第二行}} \\ \\
= & \ \left( \begin{array}{ccc:ccc}
1 & 1 & \lambda & 1 & 1 & \lambda^{2} \\
0 & \lambda – 1 & 1 – \lambda & 0 & \lambda – 1 & \lambda \left( 1 – \lambda \right) \\
0 & 0 & \left( 1 – \lambda \right) \left( 2 + \lambda \right) & 0 & 0 & \left( 1 – \lambda \right) \left( \lambda + 1 \right)^{2}
\end{array} \right)
\end{aligned}
$$

于是：

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $\lambda = 1$ 时：

$$
\mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{pmatrix} = \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix} = \mathbf{r} \left( \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \right) = 1
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $\lambda \neq 1$, $\lambda \neq -1$ 且 $\lambda \neq -2$ 时：

$$
\mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{pmatrix} = \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix} = \mathbf{r} \begin{pmatrix}
\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}
\end{pmatrix} = 3
$$

因此，若使向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 与 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 等价，必须有：

$$
\left\{ \lambda \mid \lambda \in \mathbb{R}, \lambda \neq -1 \text{ 且 } \lambda \neq -2 \right\}
$$

综上可知，本 题 应 选 C

解法 4

首先，当 $\lambda = 1$ 时，$\boldsymbol{\alpha}_{1} = \boldsymbol{\alpha}_{2} = \boldsymbol{\alpha}_{3} = \boldsymbol{\alpha}_{4} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$，此时 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 与 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 等价.

接着，由题可知：

$$
\begin{aligned}
\begin{vmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{vmatrix} & = \begin{vmatrix}
\lambda & 1 & 1 \\
1 & \lambda & 1 \\
1 & 1 & \lambda
\end{vmatrix} = \lambda^{3} – 3 \lambda + 2 \textcolor{gray}{ = \left( \lambda – 1 \right)^{2} \left( \lambda + 2 \right) } \\ \\
\begin{vmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{vmatrix} & = \begin{vmatrix}
\lambda & 1 & 1 \\
1 & \lambda & \lambda \\
1 & 1 & \lambda^{2}
\end{vmatrix} = \lambda^{4} – 2 \lambda^{2} + 1 \textcolor{gray}{ = \left( \lambda – 1 \right)^{2} \left( \lambda + 1 \right)^{2} }
\end{aligned}
$$

于是：

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $\lambda = -2$ 时：

$$
\begin{aligned}
& \ \textcolor{lightgreen}{ 0 = \begin{vmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{vmatrix} } \neq \textcolor{orange}{ \begin{vmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{vmatrix} \neq 0 } \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{pmatrix} \neq \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix}
\end{aligned}
$$

所以，此时向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 与 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 不等价.

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $\lambda = -1$ 时：

$$
\begin{aligned}
& \ \textcolor{lightgreen}{ 0 \neq \begin{vmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{vmatrix} } \neq \textcolor{orange}{ \begin{vmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{vmatrix} = 0 } \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \end{pmatrix} \neq \mathbf{r} \begin{pmatrix} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4} \end{pmatrix}
\end{aligned}
$$

所以，此时向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 与 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 不等价.

因此当 $\lambda = -2$ 或 $\lambda = -1$ 时，向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 与 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{4}$ 不等价，所以 $\lambda$ 的取值范围为：

$$
\left\{ \lambda \mid \lambda \in \mathbb{R}, \lambda \neq -1, \lambda \neq -2 \right\}
$$

综上可知，本 题 应 选 C

三、总结

前面的解法 1、解法 2、解法 3 实际上是同一个思路的解法，只是构造的矩阵（或者说向量组）以及具体实施的行变换化简步骤不同；解法 4 则是利用本题“特性”的一个带有技巧性的解法，这个“特性”就是线性相关与否的两个向量组对应的行列式在不同的取值下刚好表现出来了等于零和不等于零的结果.

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