2026年考研数学一第21题
📝 题目
（本题满分 12 分）
已知向量组 $\alpha_1=\begin{pmatrix}1 \\ 0 \\ -1 \\ -1\end{pmatrix}$, $\alpha_2=\begin{pmatrix}1 \\ -1 \\ 0 \\ -2\end{pmatrix}$, $\alpha_3=\begin{pmatrix}0 \\ -1 \\ 1 \\ -1\end{pmatrix}$, $\alpha_4=\begin{pmatrix}0 \\ 1 \\ -1 \\ 1\end{pmatrix}$，记
$$A=(\alpha_1, \alpha_2, \alpha_3, \alpha_4), \quad G=(\alpha_1, \alpha_2)$$
（1）证明：$\alpha_1, \alpha_2$ 是 $\alpha_1, \alpha_2, \alpha_3, \alpha_4$ 的极大线性无关组；
（2）求矩阵 $H$ 使得 $A=GH$，并求 $A^{10}$。
💡 答案解析
答案: 见解析
解析:
【解】 $\boldsymbol{A}=\left(\boldsymbol{\alpha}{1}, \boldsymbol{\alpha}{2}, \boldsymbol{\alpha}{3}, \boldsymbol{\alpha}{4}\right)=\left(\begin{array}{cccc}1 & 1 & 0 & 0 \ 0 & -1 & -1 & 1 \ -1 & 0 & 1 & -1 \ -1 & -2 & -1 & 1\end{array}\right) \rightarrow\left(\begin{array}{cccc}1 & 1 & 0 & 0 \ 0 & -1 & -1 & 1 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0\end{array}\right)$ ， $r(\boldsymbol{A})=2$ 且 $\boldsymbol{\alpha}{1}, \boldsymbol{\alpha}{2}$ 线性无关．
又 $\boldsymbol{\alpha}{3}=-\boldsymbol{\alpha}{1}+\boldsymbol{\alpha}{2}, \boldsymbol{\alpha}{4}=\boldsymbol{\alpha}{1}-\boldsymbol{\alpha}{2}$ ，所以 $\boldsymbol{\alpha}{1}, \boldsymbol{\alpha}{2}$ 是 $\boldsymbol{\alpha}{1}, \boldsymbol{\alpha}{2}, \boldsymbol{\alpha}{3}, \boldsymbol{\alpha}{4}$ 的极大线性无关组。故
$$
\begin{gathered}
\boldsymbol{A}=\left(\boldsymbol{\alpha}{1}, \boldsymbol{\alpha}{2},-\boldsymbol{\alpha}{1}+\boldsymbol{\alpha}{2}, \boldsymbol{\alpha}{1}-\boldsymbol{\alpha}{2}\right){1 \times 4}=\left(\boldsymbol{\alpha}{1}, \boldsymbol{\alpha}{2}\right){1 \times 2}\left(\begin{array}{cccc}
1 & 0 & -1 & 1 \
0 & 1 & 1 & -1
\end{array}\right)_{2 \times 4}, \
\boldsymbol{A}^{10}=\boldsymbol{G H G H} \cdots \boldsymbol{G H}=\boldsymbol{G D} \boldsymbol{D}^{9} \boldsymbol{H},
\end{gathered}
$$
其中 $\boldsymbol{D}=\left(\begin{array}{cccc}1 & 0 & -1 & 1 \ 0 & 1 & 1 & -1\end{array}\right){2 \times 4}\left(\begin{array}{cc}1 & 1 \ 0 & -1 \ -1 & 0 \ -1 & -2\end{array}\right){4 \times 2}=\left(\begin{array}{cc}1 & -1 \ 0 & 1\end{array}\right)_{2 \times 2}$ ．
又 $\boldsymbol{D}^{2}=\left(\begin{array}{cc}1 & -2 \ 0 & 1\end{array}\right), \boldsymbol{D}^{3}=\left(\begin{array}{cc}1 & -3 \ 0 & 1\end{array}\right)$ ，由此可推出 $\boldsymbol{D}^{9}=\left(\begin{array}{cc}1 & -9 \ 0 & 1\end{array}\right)$ ．
$$
\boldsymbol{A}^{10}=\boldsymbol{G} \boldsymbol{D}^{9} \boldsymbol{H}=\left(\begin{array}{cccc}
1 & 8 & -9 & 9 \
0 & -1 & -1 & 1 \
-1 & 9 & 10 & -10 \
-1 & 7 & 8 & -8
\end{array}\right)
$$