2023年考研数二第10题解析：线性相关、齐次线性方程组

一、题目

已知向量 $\alpha_{1}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \alpha_{2}=\left(\begin{array}{l}2 \\ 1 \\ 1\end{array}\right), \beta_{1}=\left(\begin{array}{l}2 \\ 5 \\ 9\end{array}\right), \beta_{2}=\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)$, 若 $\gamma$ 既可由 $\alpha_{1}, \alpha_{2}$ 线性表示,也可由 $\beta_{1}, \beta_{2}$ 线性表示, 则 $\gamma = (\quad)$

难度评级：

二、解析

方法一：反代法

根据 D 选项，取：

$$
\gamma = (1,5,8)^{\top}
$$

于是：

$$
\left|\alpha_{1}, \alpha_{2}, \gamma \right|=\left|\begin{array}{lll}1 & 2 & 1 \\ 2 & 1 & 5 \\ 3 & 1 & 8\end{array}\right|=
$$

$$
8+30+2-3-32-5=0
$$

且：

$$
\left|\beta_{1}, \beta_{2}, \gamma \right|=\left|\begin{array}{lll}2 & 1 & 1 \\ 5 & 0 & 5 \\ 9 & 1 & 8\end{array}\right|=
$$

$$
0+45+5-0-40-10=0
$$

于是可知，D 选项正确。

方法二：求解齐次线性方程组的部分解

由题可知：

$$
y=x_{1} \alpha_{1}+x_{2} \alpha_{2}=x_{3} \beta_{1}+x_{4} \beta_{2} \Rightarrow
$$

$$
x_{1} \alpha_{1}+x_{2} \alpha_{1}-x_{3} \beta_{1}-x_{4} \beta_{2}=0 \Rightarrow
$$

$$
x_{1}\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)+x_{2}\left(\begin{array}{l}2 \\ 1 \\ 1\end{array}\right)-x_{3}\left(\begin{array}{l}2 \\ 5 \\ 9\end{array}\right)-x_{4}\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)=0 \Rightarrow
$$

$$
\left\{\begin{array}{l}x_{1}+2 x_{2}-2 x_{3}-x_{4}=0 \\ 2 x_{1}+x_{2}-5 x_{3}=0 \\ 3 x_{1}+x_{2}-9 x_{3}-x_{4}=0\end{array} \right.
\Rightarrow
$$

$$
\left\{\begin{array}{l}2 x_{1}+x_{2}-5 x_{3}=0 \\ 2 x_{1}-x_{2}-7 x_{3}=0\end{array} \right.\Rightarrow
$$

$$
\left\{\begin{array}{l}\frac{14}{5} x_{1}+\frac{7}{5} x_{2}-7 x_{3}=0 \\ 2 x_{1}-x_{2}-7 x_{3}=0\end{array} \right.
\Rightarrow
$$

$$
\frac{4}{5} x_{1}+\frac{12}{5} x_{2}=0 \Rightarrow
$$

$$
x_{1}+3 x_{2}=0 \Rightarrow
$$

$$
\left\{\begin{array}{l}x_{1}=-3 k \\ x_{2}=k\end{array} \right.
$$

于是：

$$
y=x_{1}\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)+x_{2}\left(\begin{array}{l}2 \\ 1 \\ 1\end{array}\right) \Rightarrow
$$

$$
y=-3 k\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)+k\left(\begin{array}{l}2 \\ 1 \\ 1\end{array}\right) \Rightarrow
$$

$$
y=k\left(\begin{array}{l}-1 \\ -5 \\ -8\end{array}\right) \Rightarrow k=-k \Rightarrow
$$

$$
y=k\left(\begin{array}{l}1 \\ 5 \\ 8\end{array}\right)
$$

综上可知，D 选项正确。

方法三：通过行阶梯矩阵求解齐次线性方程组的解

由题可知：

$$
y=x_{1} \alpha_{1}+x_{2} \alpha_{2}=x_{3} \beta_{1}+x_{4} \beta_{2} \Rightarrow
$$

$$
x_{1} \alpha_{1}+x_{2} \alpha_{2}-x_{3} \beta_{1}-x_{4} \beta_{2}=0 \Rightarrow
$$

$$
x_{1} \alpha_{1}+x_{2} \alpha_{2}+x_{3}\left(-\beta_{1}\right)+x_{4}\left(-\beta_{2}\right)=0 \Rightarrow
$$

$$
\left(\alpha_{1}, \alpha_{2},-\beta_{1},-\beta_{2}\right)=\left[\begin{array}{cccc}1 & 2 & -2 & -1 \\ 2 & 1 & -5 & 0 \\ 3 & 1 & -9 & -1\end{array}\right] \Rightarrow
$$

$$
\left[\begin{array}{cccc}1 & 2 & -2 & -1 \\ 0 & -3 & -1 & 2 \\ 0 & -5 & -3 & 2\end{array}\right] \Rightarrow
$$

$$
\left[\begin{array}{cccc}1 & 2 & -2 & -1 \\ 0 & 3 & 1 & -2 \\ 0 & 0 & \frac{-4}{3} & \frac{-4}{3} \end{array}\right] \Rightarrow
$$

$$
\left[\begin{array}{cccc}1 & 2 & -2 & -1 \\ 0 & 3 & 1 & -2 \\ 0 & 0 & 1 & 1 \end{array}\right] \Rightarrow
$$

$$
\left[\begin{array}{llll}1 & 2 & -2 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1\end{array}\right] \Rightarrow
$$

$$
\left[\begin{array}{cccc}1 & 0 & -2 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1\end{array}\right] \Rightarrow
$$

$$
\left[\begin{array}{llll}1 & 0 & 0 & 3 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{array}\right] \Rightarrow
$$

$$
\textcolor{springgreen}{
\left[\begin{array}{cccc}1 & 0 & 0 & 3 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1\end{array}\right]
}
$$

于是：

$$
\left(x_{1}, x_{2}, x_{3}, x_{4}\right)^{\top}=k(-3,1,-1,1)^{\top} \Rightarrow
$$

$$
\gamma = -3 k\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)+k\left(\begin{array}{l}2 \\ 1 \\ 1\end{array}\right) \Rightarrow
$$

$$
\gamma = k\left(\begin{array}{l}-1 \\ -5 \\ -8\end{array}\right) \Rightarrow k=-k \Rightarrow
$$

$$
\gamma = k\left(\begin{array}{l}1 \\ 5 \\ 8\end{array}\right)
$$

综上可知，D 选项正确。

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