题目
设n维列向量组alpha_(1),alpha_(2),...,alpha_(m) (mlt n)线性无关,则n维列向量组beta_(1),beta_(2),...,beta_(m)线性无关的充分必要条件为（ ）。A. 向量组alpha_(1),alpha_(2),...,alpha_(m)可由向量组beta_(1),beta_(2),...,beta_(m)线性表示B. 向量组beta_(1),beta_(2),...,beta_(m)可由向量组alpha_(1),alpha_(2),...,alpha_(m)线性表示C. 向量组alpha_(1),alpha_(2),...,alpha_(m)与向量组beta_(1),beta_(2),...,beta_(m)等价D. 矩阵A=(alpha_(1),alpha_(2),...,alpha_(m))与矩阵B=(beta_(1),beta_(2),...,beta_(m))等价
设$n$维列向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$ $(m\lt n)$线性无关$,$则$n$维列向量组$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性无关的充分必要条件为$（ \quad ）$。
A. 向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$可由向量组$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性表示
B. 向量组$\beta_{1},\beta_{2},\cdots,\beta_{m}$可由向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$线性表示
C. 向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$与向量组$\beta_{1},\beta_{2},\cdots,\beta_{m}$等价
D. 矩阵$A=(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})$与矩阵$B=(\beta_{1},\beta_{2},\cdots,\beta_{m})$等价
题目解答
答案
D. 矩阵$A=(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})$与矩阵$B=(\beta_{1},\beta_{2},\cdots,\beta_{m})$等价
解析
步骤 1：理解线性无关的定义
线性无关是指一组向量中，没有一个向量可以由其他向量线性表示。即，对于向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$，如果存在一组不全为零的数$k_{1},k_{2},\cdots,k_{m}$，使得$k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{m}\alpha_{m}=0$，则向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$线性相关；否则，线性无关。
步骤 2：分析选项
A. 向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$可由向量组$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性表示，这并不意味着$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性无关，因为$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$线性无关，但$\beta_{1},\beta_{2},\cdots,\beta_{m}$可能线性相关。
B. 向量组$\beta_{1},\beta_{2},\cdots,\beta_{m}$可由向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$线性表示，这也不意味着$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性无关，因为$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$线性无关，但$\beta_{1},\beta_{2},\cdots,\beta_{m}$可能线性相关。
C. 向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$与向量组$\beta_{1},\beta_{2},\cdots,\beta_{m}$等价，意味着它们可以互相线性表示，但并不直接说明$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性无关。
D. 矩阵$A=(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})$与矩阵$B=(\beta_{1},\beta_{2},\cdots,\beta_{m})$等价，意味着它们可以通过初等变换互相转换，这说明$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$与$\beta_{1},\beta_{2},\cdots,\beta_{m}$等价，从而$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性无关。
步骤 3：选择正确答案
根据上述分析，选项D是正确的，因为矩阵$A=(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})$与矩阵$B=(\beta_{1},\beta_{2},\cdots,\beta_{m})$等价，意味着它们可以通过初等变换互相转换，从而$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性无关。
线性无关是指一组向量中，没有一个向量可以由其他向量线性表示。即，对于向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$，如果存在一组不全为零的数$k_{1},k_{2},\cdots,k_{m}$，使得$k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{m}\alpha_{m}=0$，则向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$线性相关；否则，线性无关。
步骤 2：分析选项
A. 向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$可由向量组$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性表示，这并不意味着$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性无关，因为$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$线性无关，但$\beta_{1},\beta_{2},\cdots,\beta_{m}$可能线性相关。
B. 向量组$\beta_{1},\beta_{2},\cdots,\beta_{m}$可由向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$线性表示，这也不意味着$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性无关，因为$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$线性无关，但$\beta_{1},\beta_{2},\cdots,\beta_{m}$可能线性相关。
C. 向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$与向量组$\beta_{1},\beta_{2},\cdots,\beta_{m}$等价，意味着它们可以互相线性表示，但并不直接说明$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性无关。
D. 矩阵$A=(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})$与矩阵$B=(\beta_{1},\beta_{2},\cdots,\beta_{m})$等价，意味着它们可以通过初等变换互相转换，这说明$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$与$\beta_{1},\beta_{2},\cdots,\beta_{m}$等价，从而$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性无关。
步骤 3：选择正确答案
根据上述分析，选项D是正确的，因为矩阵$A=(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})$与矩阵$B=(\beta_{1},\beta_{2},\cdots,\beta_{m})$等价，意味着它们可以通过初等变换互相转换，从而$\beta_{1},\beta_{2},\cdots,\beta_{m}$线性无关。