# 2012 年研究生入学考试数学一选择题第 5 题解析

## 题目

( A ) $a_{1},a_{2},a_{3}.$

( B ) $a_{1},a_{2},a_{4}.$

( C ) $a_{1},a_{3},a_{4}.$

( D ) $a_{2},a_{3},a_{4}.$

## 解析

$n$ 个 $n$ 维向量 $a_{1},a_{2},\dots a_{n}$ 线性相关 $\Leftrightarrow$ 行列式 $|a_{1},a_{2},\dots,a_{n}|=0.$

$a=\begin{bmatrix}a_{1}\\ a_{2}\\ \vdots\\ a_{n}\end{bmatrix}.$

$b=\begin{bmatrix}b_{1},b_{2},\dots,b_{n}\end{bmatrix}.$

$\begin{bmatrix}\lambda_{1}& 0& 0\\ 0& \lambda_{2}&0\\ 0& 0& \lambda_{3}\end{bmatrix}=\begin{bmatrix}\lambda_{1}& \star& \star\\ 0& \lambda_{2}& \star\\ 0& 0& \lambda_{3}\end{bmatrix}=\begin{bmatrix}\lambda_{1}& 0& 0\\ \star& \lambda_{2}& 0 \\ \star& \star& \lambda_{3} \end{bmatrix}=\lambda_{1} \times \lambda_{2} \times \lambda_{3}.$

$\begin{bmatrix}0& 0& \lambda_{1}\\ 0& \lambda_{2}&0\\ \lambda_{3}& 0& 0\end{bmatrix}=\begin{bmatrix}\star& \star& \lambda_{1}\\ \star& \lambda_{2}& 0\\ \lambda_{3}& 0& 0\end{bmatrix}=\begin{bmatrix}0& 0& \lambda_{1}\\ 0& \lambda_{2}& \star \\ \lambda_{3}& \star& \star \end{bmatrix}=(-1)^{\frac{n(n-1)}{2}} \times \lambda_{1} \times \lambda_{2} \times \lambda_{3}.$

A 项：

$\begin{vmatrix}0& 0& 1\\ 0& 1& -1\\ c_{1}& c_{2}& c_{3}\end{vmatrix}=(-1)^{\frac{3 \times 2}{2}}\times1\times1\times c_{1}=-c_{1}.$

B 项：

$\begin{vmatrix}0& 0& -1\\ 0& 1& 1\\ c_{1}& c_{2}& c_{4}\end{vmatrix}=(-1)^{\frac{3\times2}{2}}\times (-1) \times 1 \times c_{1}=c_{1}.$

C 项：

$\begin{vmatrix}0& 1& -1\\ 0& -1& 1\\ c_{1}& c_{3}& c_{4}\end{vmatrix}=c_{1}-c_{1}=0, 恒成立.$

$a_{1},a_{3},a_{4}$ 的线性相关性恒成立。

D 项：

$\begin{vmatrix}0& 1& -1\\ 1& -1& 1\\ c_{2}& c_{3}& c_{4}\end{vmatrix}=c_{2}-c_{3}-c_{2}-c_{4}=-c_{3}-c_{4}.$

EOF