做这道题不需要事先知道待求解的矩阵是几行几列

一、题目

若 $\boldsymbol{\alpha}_{1}=(1,0,0)^{\mathrm{\top}}, \boldsymbol{\alpha}_{2}=(1,2,-1)^{\mathrm{\top}}, \boldsymbol{\alpha}_{3}=(-1,1,0)^{\mathrm{\top}}$ 且 $\boldsymbol{A} \boldsymbol{\alpha}_{1}=(2,1)^{\mathrm{\top}}, A \boldsymbol{\alpha}_{2}=$ $(-1,1)^{\mathrm{\top}}, \boldsymbol{A} \boldsymbol{\alpha}_{3}=(3,-4)^{\mathrm{\top}}$, 则 $\boldsymbol{A}=?$

难度评级：

二、解析

由题可知：

$$
A\left[\alpha_{1}, \alpha_{2}, \alpha_{3}\right]=\left[\begin{array}{ccc}2 & -1 & 3 \\ 1 & 1 & -4\end{array}\right]
$$

$$
\left|\alpha_{1}, \alpha_{2}, \alpha_{3}\right|=\left|\begin{array}{ccc}1 & 1 & -1 \\ 0 & 2 & 1 \\ 0 & -1 & 0\end{array}\right|=1 \neq 0
$$

求 $\left[\alpha_{1}, \alpha_{2}, \alpha_{3}\right]$ 的逆矩阵：

$$
\left[\begin{array}{cccccc}1 & 1 & -1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 0 & 1\end{array}\right] \Rightarrow
$$

$$
\left[\begin{array}{cccccc}1 & 0 & -1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 0 & 0 & -1\end{array}\right] \Rightarrow
$$

$$
\left[\begin{array}{lllllc}1 & 0 & 0 & 1 & 1 & 3 \\ 0 & 1 & 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 & 1 & 2\end{array}\right] \Rightarrow
$$

$$
\textcolor{springgreen}{
\left[\alpha_{1}, \alpha_{2}, \alpha_{3}\right]^{-1} = \left[\begin{array}{ccc}1 & 1 & 3 \\ 0 & 0 & -1 \\ 0 & 1 & 2\end{array}\right]
}
$$

于是：

$$
A=\left[\begin{array}{ccc}2 & -1 & 3 \\ 1 & 1 & -4\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 3 \\ 0 & 0 & -1 \\ 0 & 1 & 2\end{array}\right] \Rightarrow
$$

$$
A=\left[\begin{array}{ccc}2 & 5 & 13 \\ 1 & -3 & -6\end{array}\right]
$$

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