2026年考研数二第22题解析：极大线性无关组、矩阵乘法运算

一、题目

二、解析

第 (1) 问

首先：

$$
\begin{aligned}
\boldsymbol{A} & = (\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4}) \\ \\
& = \begin{pmatrix}
1 & 1 & 0 & 0 \\
0 & -1 & -1 & 1 \\
-1 & 0 & 1 & -1 \\
-1 & -2 & -1 & 1
\end{pmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{pmatrix}
1 & 0 & -1 & 1 \\
0 & 1 & 1 & -1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\end{aligned}
$$

于是可知，$\mathrm{r}(\boldsymbol{A}) = 2$ 且 $\boldsymbol{\alpha}_{1}$, $\boldsymbol{\alpha}_{2}$ 线性无关，同时可知：

$$
\begin{aligned}
\boldsymbol{\alpha}_{3} & = -\boldsymbol{\alpha}_{1} + \boldsymbol{\alpha}_{2} \\
\boldsymbol{\alpha}_{4} & = \boldsymbol{\alpha}_{1} – \boldsymbol{\alpha}_{2}
\end{aligned}
$$

综上可知，$\boldsymbol{\alpha}_{1}$, $\boldsymbol{\alpha}_{2}$ 是 $\boldsymbol{\alpha}_{1}$, $\boldsymbol{\alpha}_{2}$, $\boldsymbol{\alpha}_{3}$, $\boldsymbol{\alpha}_{4}$ 的极大线性无关组.

第 (2) 问

由题及上面的第 (1) 问，可知：

$$
\begin{aligned}
\boldsymbol{A} & = (\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4}) \\ \\
& = (\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},-\boldsymbol{\alpha}_{1} + \boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{1} – \boldsymbol{\alpha}_{2})_{4 \times 4} \\ \\
& =(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2})_{4 \times 2}\begin{pmatrix} 1 & 0 & -1 & 1 \\ 0 & 1 & 1 & -1 \end{pmatrix}_{2\times4}
\end{aligned}
$$

又因为，$\boldsymbol{A} = \boldsymbol{GH}$, $\boldsymbol{G}$ $=$ $(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2})$, 所以：

$$
\textcolor{lightgreen}{
\boldsymbol{H} = \begin{pmatrix} 1 & 0 & -1 & 1 \\ 0 & 1 & 1 & -1 \end{pmatrix}
}
$$

接着，有：

$$
\begin{aligned}
\boldsymbol{A}^{10} & = (\boldsymbol{GH})^{10} \\ \\
& = \boldsymbol{GHGH} \cdots \boldsymbol{GH} \\ \\
& = \boldsymbol{G} (\boldsymbol{HG})^{9} \boldsymbol{H}
\end{aligned}
$$

其中：

$$
\boldsymbol{ HG } = \begin{pmatrix}1 & 0 & -1 & 1 \\ 0 & 1 & 1 & -1\end{pmatrix} \begin{pmatrix}1 & 1 \\ 0 & -1 \\ -1 & 0 \\ -1 & -2\end{pmatrix} = \begin{pmatrix}1 & -1 \\ 0 & 1\end{pmatrix}
$$

又因为：

$$
\begin{aligned}
& \boldsymbol{D}^{2}=\begin{pmatrix}1 & -2 \\ 0 & 1\end{pmatrix} \\ \\
& \boldsymbol{D}^{3}=\begin{pmatrix}1 & -3 \\ 0 & 1\end{pmatrix}
\end{aligned}
$$

由此，递推可知：

$$
\boldsymbol{D}^{9} = \begin{pmatrix}1 & -9 \\ 0 & 1\end{pmatrix}
$$

综上可得：

$$
\begin{aligned}
\textcolor{lightgreen}{ \boldsymbol{A}^{10} } & = \boldsymbol{GD}^{9} \boldsymbol{H} \\ \\
& = \begin{pmatrix} 1 & 1 \\ 0 & -1 \\ -1 & 0 \\ -1 & -2 \end{pmatrix}\begin{pmatrix} 1 & -9 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & -1 & 1 \\ 0 & 1 & 1 & -1 \end{pmatrix} \\ \\
& = \textcolor{lightgreen}{ \begin{pmatrix}1 & 8 & -9 & 9 \\ 0 & -1 & -1 & 1 \\ -1 & 9 & 10 & -10 \\ -1 & 7 & 8 & -8\end{pmatrix} }
\end{aligned}
$$

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