kaoyan1basic 线性代数 第1题
📝 题目
### 【基础篇】第1题（选择题） 1．已知矩阵方程 $\boldsymbol{A}=\boldsymbol{B} \boldsymbol{C}$ ，其中 $\boldsymbol{A}=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1 & 0 \\ 0 & 1 & 1\end{array}\right]$ ，则 $\boldsymbol{B}, \boldsymbol{C}$ 可以是（ ） （A）$\displaystyle \left[\begin{array}{ccc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}}\end{array}\right],\left[\begin{array}{ccc}\sqrt{2} & 0 & 0 \\ 0 & \sqrt{3} & 0 \\ \frac{1}{\sqrt{2}} & \frac{2}{\sqrt{3}} & \frac{1}{\sqrt{6}}\end{array}\right]$ （B）$\displaystyle \left[\begin{array}{ccc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}}\end{array}\right],\left[\begin{array}{ccc}\sqrt{2} & 0 & \frac{1}{\sqrt{2}} \\ 0 & \sqrt{3} & \frac{2}{\sqrt{3}} \\ 0 & 0 & \frac{1}{\sqrt{6}}\end{array}\right]$ （C）$\displaystyle \left[\begin{array}{ccc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}}\end{array}\right],\left[\begin{array}{ccc}\sqrt{2} & 0 & 0 \\ 0 & \sqrt{3} & \frac{2}{\sqrt{3}} \\ 0 & 0 & \frac{1}{\sqrt{6}}\end{array}\right]$ （D）$\displaystyle \left[\begin{array}{ccc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}}\end{array}\right],\left[\begin{array}{ccc}\sqrt{2} & 0 & \frac{1}{\sqrt{2}} \\ 0 & \sqrt{3} & \frac{2}{\sqrt{3}} \\ 0 & 0 & \frac{1}{\sqrt{6}}\end{array}\right]$
💡 答案解析
**答案**：D **解析**： 步骤1：验证$A=BC$，计算各选项乘积是否等于$A$。 步骤2：选项D中，$B$为第一列$\displaystyle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0$，第二列$\displaystyle \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$，第三列$\displaystyle -\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}$；$C$为上三角矩阵。 步骤3：计算$BC$，得$A$，故D正确。
**难度**：★★★☆☆