2026年考研数学二第22题
📝 题目
(本题满分 12 分) 已知向量组 $\boldsymbol{\alpha}_1=\left(\begin{array}{c}1 \\ 0 \\ -1 \\ -1\end{array}\right), \boldsymbol{\alpha}_2=\left(\begin{array}{c}1 \\ -1 \\ 0 \\ -2\end{array}\right), \boldsymbol{\alpha}_3=\left(\begin{array}{c}0 \\ -1 \\ 1 \\ -1\end{array}\right), \boldsymbol{\alpha}_4=\left(\begin{array}{c}0 \\ 1 \\ -1 \\ 1\end{array}\right)$ ， 记 $A=\left(\alpha_1, \alpha_2, \alpha_3, \alpha_4\right), \quad G=\left(\alpha_1, \alpha_2\right)$. （1）证明： $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2$ 是 $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3, \boldsymbol{\alpha}_4$ 的极大线性无关组； （2）求矩阵 $\boldsymbol{H}$ 使得 $\boldsymbol{A}=\boldsymbol{G} \boldsymbol{H}$ ，并求 $\boldsymbol{A}^{10}$ ．
💡 答案解析
答案: 见解析
解析:
【解】 $\boldsymbol{A}=\left(\boldsymbol{\alpha}{1}, \boldsymbol{\alpha}{2},-\boldsymbol{\alpha}{1}+\boldsymbol{\alpha}{2}, \boldsymbol{\alpha}{1}-\boldsymbol{\alpha}{2}\right){1 \times 4}=\left(\boldsymbol{\alpha}{1}, \boldsymbol{\alpha}{2}\right){1 \times 2}\left(\begin{array}{cccc}1 & 0 & -1 & 1 \ 0 & 1 & 1 & -1\end{array}\right)_{2 \times 4}$ ，
$$
A^{10}=G H G H \cdots G H=G D^{9} H \text {, }
$$
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 $\boldsymbol{D}^{9}=\left(\begin{array}{cc}1 & -9 \ 0 & 1\end{array}\right)$ ．
$$
\boldsymbol{A}^{10}=\boldsymbol{G} \boldsymbol{D}^{9} \boldsymbol{H}=\left(\begin{array}{cccc}
1 & 8 & -9 & 9 \
0 & -1 & -1 & 1 \
-1 & 9 & 10 & -10 \
-1 & 7 & 8 & -8
\end{array}\right)
$$