题号:1722    题型:单选题    来源:2009年全国硕士研究生招生考试试题
设 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 是 3 维向量空间 $\mathbf{R}^{3}$ 的一组基,则由基 $\boldsymbol{\alpha}_{1}, \frac{1}{2} \boldsymbol{\alpha}_{2}, \frac{1}{3} \boldsymbol{\alpha}_{3}$ 到基 $\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}+\boldsymbol{\alpha}_{1}$ 的过渡矩阵为 ()
$A.$ $\left(\begin{array}{lll}1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 3 & 3\end{array}\right)$. $B.$ $\left(\begin{array}{lll}1 & 2 & 0 \\ 0 & 2 & 3 \\ 1 & 0 & 3\end{array}\right)$. $C.$ $\left(\begin{array}{rrr}\frac{1}{2} & \frac{1}{4} & -\frac{1}{6} \\ -\frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ -\frac{1}{2} & -\frac{1}{4} & \frac{1}{6}\end{array}\right)$. $D.$ $\left(\begin{array}{rrr}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & \frac{1}{4} & -\frac{1}{4} \\ -\frac{1}{6} & \frac{1}{6} & \frac{1}{6}-\end{array}\right)$.
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答案:
A

解析:

解 由 $\left(\boldsymbol{\alpha}_{1}, \frac{1}{2} \boldsymbol{\alpha}_{2}, \frac{1}{3} \boldsymbol{\alpha}_{3}\right)=\left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}\right)\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{3}\end{array}\right]$,
得 $\left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}\right)=\left(\boldsymbol{\alpha}_{1}, \frac{1}{2} \boldsymbol{\alpha}_{2}, \frac{1}{3} \boldsymbol{\alpha}_{3}\right)\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{array}\right]$,

$$
\begin{aligned}
&\left(\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}+\boldsymbol{\alpha}_{1}\right)=\left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}\right)\left[\begin{array}{lll}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{array}\right] \\
&=\left(\boldsymbol{\alpha}_{1}, \frac{1}{2} \boldsymbol{\alpha}_{2}, \frac{1}{3} \boldsymbol{\alpha}_{3}\right)\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{array}\right]=\left(\boldsymbol{\alpha}_{1}, \frac{1}{2} \boldsymbol{\alpha}_{2}, \frac{1}{3} \boldsymbol{\alpha}_{3}\right)\left[\begin{array}{lll}
1 & 0 & 1 \\
2 & 2 & 0 \\
0 & 3 & 3
\end{array}\right] .
\end{aligned}
$$
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