题目 06
已知 $\boldsymbol{A}$ $=$ $\left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}\right)$ 为三阶矩阵, 且 $|\boldsymbol{A}|=3$, 则 $\left|\boldsymbol{\alpha}_{1}+2 \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}-3 \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}+2 \boldsymbol{\alpha}_{1}\right|=?$
解析 06
$$
\left(\alpha_{1}+2 \alpha_{2}, \alpha_{2}-3 \alpha_{3}, \alpha_{3}+2 \alpha_{1}\right)=
$$
$$
\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)\left(\begin{array}{ccc}1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & -3 & 1\end{array}\right) \Rightarrow
$$
$$
\left|\alpha_{1}+2 \alpha_{2}, \alpha_{2}-3 \alpha_{3}, \alpha_{3}+2 \alpha_{1}\right|=
$$
$$
|A| \cdot \left|\begin{array}{ccc}1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & -3 & 1\end{array}\right|=
$$
$$
3(1+0-12-0-0-0) =
$$
$$
3 \times(-11)=-33.
$$
或者：
$$
\left|\alpha_{1}+2 \alpha_{2}, \alpha_{2}-3 \alpha_{3}, \alpha_{3}+2 \alpha_{1}\right| =
$$
$$
\left|\alpha_{1}, \alpha_{2}-3 \alpha_{3}, \alpha_{3}+2 \alpha_{1}\right|+
$$
$$
\left|2 \alpha_{2}, \alpha_{2}-3 \alpha_{3}, \alpha_{3}+2 \alpha_{1}\right|=
$$
Tips:
通过初等行变换，可以将 $\left|\alpha_{1}, \alpha_{2}-3 \alpha_{3}, \alpha_{3}+2 \alpha_{1}\right|$ 化简为 $\left|\alpha_{1}, \alpha_{2}-3 \alpha_{3}, \alpha_{3}\right|$——
或者也可以将 $\left|\alpha_{1}, \alpha_{2}-3 \alpha_{3}, \alpha_{3}+2 \alpha_{1}\right|$ 拆分成 $\left|\alpha_{1}, \alpha_{2}-3 \alpha_{3}, \alpha_{3}\right|$ 和 $\left|\alpha_{1}, \alpha_{2}-3 \alpha_{3}, 2 \alpha_{1}\right|$——
由于 $\left|\alpha_{1}, \alpha_{2}-3 \alpha_{3}, 2 \alpha_{1}\right|$ 中的 $\alpha_{1}$ 和 $2 \alpha_{1}$ 成比例，因此 $\left|\alpha_{1}, \alpha_{2}-3 \alpha_{3}, 2 \alpha_{1}\right|$ $=$ $0$.
$$
\left|\alpha_{1}, \alpha_{2}-3 \alpha_{3}, \alpha_{3}\right|+
$$
$$
\left|\alpha_{2},-3 \alpha_{3}, 2 \alpha_{1}\right| =
$$
$$
\left|\alpha_{1}, \alpha_{2}, \alpha_{3}\right|-2 \times 3\left|\alpha_{2}, \alpha_{3}, 2 \alpha_{1}\right| =
$$
$$
\left|\alpha_{1}, \alpha_{2}, \alpha_{3}\right|-12\left|\alpha_{2}, \alpha_{3}, \alpha_{1}\right| =
$$
Tips:
将 $\left|\alpha_{2}, \alpha_{3}, \alpha_{1}\right|$ 中的 $\alpha_{2}$ 和 $\alpha_{1}$ 交换位置需要移动两次。
$$
\left|\alpha_{1}, \alpha_{2}, \alpha_{3}\right|-12 \times(-1)^{2}\left|\alpha_{1}, \alpha_{2}, \alpha_{3}\right| =
$$
$$
3-12 \times 3=(1-12) \times 3=-33.
$$