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 $α_{1} = (\begin{array}{l} 1 \\ 2 \\ 3 \end{array}), α_{2} = (\begin{array}{l} 2 \\ 1 \\ 1 \end{array}), β_{1} = (\begin{array}{l} 2 \\ 5 \\ 9 \end{array}), β_{2} = (\begin{array}{l} 1 \\ 0 \\ 1 \end{array})$ , 若 $γ$ 既可由 $α_{1}, α_{2}$ 线性表示,也可由 $β_{1}, β_{2}$ 线性表示, 则 $γ = ()$

(A) $k (\begin{array}{l} 3 \\ 3 \\ 4 \end{array}), k \in R$

(B) $k (\begin{matrix} 3 \\ 5 \\ 10 \end{matrix}), k \in R$

(D) $k (\begin{array}{l} 1 \\ 5 \\ 8 \end{array}), k \in R$

难度评级：

二、解析

方法一：反代法

根据 D 选项，取：

$γ = (1, 5, 8)^{⊤}$

于是：

$| α_{1}, α_{2}, γ | = | \begin{array}{lll} 1 & 2 & 1 \\ 2 & 1 & 5 \\ 3 & 1 & 8 \end{array} | =$

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

且：

$| β_{1}, β_{2}, γ | = | \begin{array}{lll} 2 & 1 & 1 \\ 5 & 0 & 5 \\ 9 & 1 & 8 \end{array} | =$

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

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：

$y = x_{1} α_{1} + x_{2} α_{2} = x_{3} β_{1} + x_{4} β_{2} \Rightarrow$

$x_{1} α_{1} + x_{2} α_{1} - x_{3} β_{1} - x_{4} β_{2} = 0 \Rightarrow$

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

${\begin{cases} x_{1} + 2 x_{2} - 2 x_{3} - x_{4} = 0 \\ 2 x_{1} + x_{2} - 5 x_{3} = 0 \\ 3 x_{1} + x_{2} - 9 x_{3} - x_{4} = 0 \end{cases} \Rightarrow$

${\begin{cases} 2 x_{1} + x_{2} - 5 x_{3} = 0 \\ 2 x_{1} - x_{2} - 7 x_{3} = 0 \end{cases} \Rightarrow$

${\begin{cases} \frac{14}{5} x_{1} + \frac{7}{5} x_{2} - 7 x_{3} = 0 \\ 2 x_{1} - x_{2} - 7 x_{3} = 0 \end{cases} \Rightarrow$

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

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

${\begin{cases} x_{1} = - 3 k \\ x_{2} = k \end{cases}$

于是：

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

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

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

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

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：

$y = x_{1} α_{1} + x_{2} α_{2} = x_{3} β_{1} + x_{4} β_{2} \Rightarrow$

$x_{1} α_{1} + x_{2} α_{2} - x_{3} β_{1} - x_{4} β_{2} = 0 \Rightarrow$

$x_{1} α_{1} + x_{2} α_{2} + x_{3} (- β_{1}) + x_{4} (- β_{2}) = 0 \Rightarrow$

$(α_{1}, α_{2}, - β_{1}, - β_{2}) = [\begin{array}{cccc} 1 & 2 & - 2 & - 1 \\ 2 & 1 & - 5 & 0 \\ 3 & 1 & - 9 & - 1 \end{array}] \Rightarrow$

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

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

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

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

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

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

$[\begin{array}{cccc} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 1 \end{array}]$

于是：

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

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

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

$γ = k (\begin{array}{l} 1 \\ 5 \\ 8 \end{array})$

综上可知，D 选项正确。

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