题目内容

设`\A`是方阵,如有矩阵关系式`\AB=AC`,则必有 ( )

A. `\A=O`
B. `\B \ne C`时`\A=O`
C. `\A \ne O`时`\B = C`
D. `\|A| \ne 0`时`\B = C`

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设有任意两个`n`维向量组`\alpha_1,\alpha_2,\cdots,\alpha_m`和`\beta_1,\beta_2,\cdots,\beta_m`. 若存在两组不全为零的数 `\lambda_1,\lambda_2,\cdots,\lambda_m`和`k_1,k_2,\cdots,k_m` 使 `(\lambda_1+k_1)\alpha_1+(\lambda_2+k_2)\alpha_2+\cdots+(\lambda_m+k_m)\alpha_m +(\lambda_1-k_1)\beta_1+(\lambda_2-k_2)\beta_2+\cdots+(\lambda_m-k_m)\beta_m=0`,则( )

A. ` \alpha_1,\alpha_2,\cdots,\alpha_m`和`\beta_1,\beta_2,\cdots,\beta_m`都线性相关;
B. ` \alpha_1,\alpha_2,\cdots,\alpha_m`和`\beta_1,\beta_2,\cdots,\beta_m`都线性无关;
C. ` \alpha_1+\beta_1,\alpha_2+\beta_2,\cdots,\alpha_m+\beta_m,\alpha_1-\beta_1,\alpha_2-\beta_2,\cdots,\alpha_m-\beta_m`线性无关;
D. ` \alpha_1+\beta_1,\alpha_2+\beta_2,\cdots,\alpha_m+\beta_m,\alpha_1-\beta_1,\alpha_2-\beta_2,\cdots,\alpha_m-\beta_m`线性相关。

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`线性无关的充要条件为( )

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。

已知向量组`\alpha_1=(1,2,3,4),\alpha_2=(2,3,4,5),\alpha_3=(3,4,5,6),\alpha_4=(4,5,6,7)`,则该向量组的秩是( )

A. `1;`
B. `2;`
C. `3;`
D. `4.`

已知`\alpha_1+\alpha_2,\alpha_2+\alpha_3,\alpha_3+\alpha_1`线性无关,下列证明`\alpha_1,\alpha_2,\alpha_3`线性无关的步骤错误是( )

A. 因为`\alpha_1+\alpha_2,\alpha_2+\alpha_3,\alpha_3+\alpha_1`线性无关,所以只有当`k_1=k_2=k_3=0`时,才有`k_1(\alpha_1+\alpha_2)+k_2(\alpha_2+\alpha_3)+k_3(\alpha_3+\alpha_1)=0`.
B. 上式整理得`(k_1+k_3)\alpha_1+(k_1+k_2)\alpha_2+(k_2+k_3)\alpha_3=0`。
C. 由`k_1=k_2=k_3=0`推知`k_1+k_3=0, k_1+k_2=0, k_2+k_3=0`。
D. 由第二第三两步可知`\alpha_1,\alpha_2,\alpha_3`线性无关。

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