2026年考研数学三第21题
📝 题目
（本题满分 12 分）设矩阵 $A=\left(\begin{array}{ccccc}1 & -1 & 3 & 0 & -1 \\ -1 & 0 & -2 & -a & -1 \\ 1 & 1 & a & 2 & 3\end{array}\right)$ 的秩为 2 ． （1）求 $a$ 的值． （2）求 $A$ 的列向量组的一个极大线性无关组 $\alpha, \beta$ ，并求矩阵 $H$ ，使得 $A=G H$ ，其中 $G=(\alpha, \beta)$ ．
💡 答案解析
答案: 见解析
解析:
已知向量组 $\alpha_{1}=\left(\begin{array}{c}1 \ 0 \ -1 \ -1\end{array}\right), \alpha_{2}=\left(\begin{array}{c}1 \ -1 \ 0 \ -2\end{array}\right), \alpha_{3}=\left(\begin{array}{c}0 \ -1 \ 1 \ -1\end{array}\right), \alpha_{4}=\left(\begin{array}{c}0 \ 1 \ -1 \ 1\end{array}\right)$ ，记 $A=\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right)$ ，
$G=\left(\alpha_{1}, \alpha_{2}\right)$.
（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 $A=G H$ ，并求 $A^{10}$ ．
【解析】（1）由 $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right)=\left(\begin{array}{cccc}1 & 1 & 0 & 0 \ 0 & -1 & -1 & 1 \ -1 & 0 & 1 & -1 \ -1 & -2 & -1 & 1\end{array}\right) \rightarrow\left(\begin{array}{cccc}1 & 0 & -1 & 1 \ 0 & 1 & 1 & -1 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0\end{array}\right)$
故 $r\left(\alpha_{1}, \alpha_{2}\right)=r\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right)=2$ ，故极大线性无关组中有 2 个向量，又由 $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ 均可由 $\alpha_{1}, \alpha_{2}$ 线性表示，故 $\alpha_{1}, \alpha_{2}$ 为向量组 $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ 的一个极大线性无关组．
（2）由（1）知 $\alpha_{3}=-\alpha_{1}+\alpha_{2}, \alpha_{4}=\alpha_{1}-\alpha_{2}$ ，故 $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right)=\left(\alpha_{1}, \alpha_{2}\right)\left(\begin{array}{cccc}1 & 0 & -1 & 1 \ 0 & 1 & 1 & -1\end{array}\right)$ ，
故 $H=\left(\begin{array}{cccc}1 & 0 & -1 & 1 \ 0 & 1 & 1 & -1\end{array}\right)$ ，由于 $A=G H$ ，
故 $A^{10}=G H \cdot G H \cdot G H \cdot \mathrm{~L} G H=G(H G)^{9} H$ ，
由 $H G=\left(\begin{array}{cccc}1 & 0 & -1 & 1 \ 0 & 1 & 1 & -1\end{array}\right)\left(\begin{array}{cc}1 & 1 \ 0 & -1 \ -1 & 0 \ -1 & -2\end{array}\right)=\left(\begin{array}{cc}1 & -1 \ 0 & 1\end{array}\right)$ ，
故 $(H G)^{9}=\left(\begin{array}{cc}1 & -9 \ 0 & 1\end{array}\right)$ ，
则 $A^{10}=G(H G)^{9} H=\left(\begin{array}{cc}1 & 1 \ 0 & -1 \ -1 & 0 \ -1 & -2\end{array}\right)\left(\begin{array}{cc}1 & -9 \ 0 & 1\end{array}\right)\left(\begin{array}{cccc}1 & 0 & -1 & 1 \ 0 & 1 & 1 & -1\end{array}\right)$
$=\left(\begin{array}{cccc}1 & -8 & -9 & 9 \ 0 & -1 & -1 & 1 \ -1 & 9 & 10 & -10 \ -1 & 7 & 8 & -8\end{array}\right)$ ．