求矩阵 $\left(\begin{array}{ccccc}1 & 2 & 3 & -2 & -1 \\ 4 & 3 & 6 & -3 & 2 \\ 2 & -1 & 0 & 1 & 4 \\ 5 & 0 & 3 & 2 & 7\end{array}\right)$ 的秩
解: $D=\left(\begin{array}{ccccc}1 & 2 & 3 & -2 & -1 \\ 4 & 3 & 6 & -3 & 2 \\ 2 & -1 & 0 & 1 & 4 \\ 5 & 0 & 3 & 2 & 7\end{array}\right) \rightarrow\left(\begin{array}{ccccc}1 & 2 & 3 & -2 & -1 \\ 0 & -5 & -6 & 5 & 6 \\ 0 & -5 & -6 & 5 & 6 \\ 0 & -10 & -12 & 12 & 12\end{array}\right)$ $$ \rightarrow\left(\begin{array}{ccccc} 1 & 2 & 3 & -2 & -1 \\ 0 & -5 & -6 & 5 & 6 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \end{array}\right)=D^{\prime}, \mathbf{r}(\mathbf{D})=\mathbf{r}\left(\mathbf{D}^{\prime}\right)=3 $$
求解齐次线性方程组 $\left\{\begin{array}{l}3 x_{1}+4 x_{2}-5 x_{3}+7 x_{4}=0 \\ 2 x_{1}-3 x_{2}+3 x_{3}-2 x_{4}=0 \\ 4 x_{1}+11 x_{2}-13 x_{3}+16 x_{4}=0 \\ 7 x_{1}-2 x_{2}+x_{3}+3 x_{4}=0\end{array}\right.$
解: $\rightarrow\left(\begin{array}{cccc}1 & 0 & -\frac{3}{17} & \frac{13}{17} \\ 0 & 1 & -\frac{19}{17} & \frac{20}{17} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{array}\right)$ $\left(\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{array}\right)=k_{1}\left(\begin{array}{l}\frac{3}{17} \\ \frac{19}{17} \\ 1 \\ 0\end{array}\right)+k_{2}\left(\begin{array}{l}-\frac{13}{17} \\ -\frac{20}{17} \\ 1 \\ 0\end{array}\right)$
非次线性方程组 $\left\{\begin{array}{l}-2 x_{1}+x_{2}+x_{3}=-2 \\ x_{1}-2 x_{2}+x_{3}=\lambda \\ x_{1}-x_{2}-2 x_{3}=\lambda^{2}\end{array}\right.$ 当 $\lambda$ 取何值时有解? 并求出它的解.
解: $(A b)=\left(\begin{array}{cccc}-2 & 1 & 1 & -2 \\ 1 & -2 & 1 & \lambda \\ 1 & 1 & -2 & \lambda^{2}\end{array}\right) \rightarrow\left(\begin{array}{cccc}1 & -2 & 1 & \lambda \\ -2 & 1 & 1 & -2 \\ 1 & 1 & -2 & \lambda^{2}\end{array}\right)$ $\rightarrow\left(\begin{array}{cccc}1 & -2 & 1 & \lambda \\ 0 & -3 & 3 & 2(\lambda-1) \\ 0 & 3 & -3 & \lambda^{2}-\lambda\end{array}\right) \rightarrow\left(\begin{array}{cccc}1 & -2 & 1 & \lambda \\ 0 & -3 & 3 & 2(\lambda-1) \\ 0 & 0 & 0 & (\lambda-1)(\lambda+2)\end{array}\right)$
当 $(\lambda-1)(\lambda+2)=0$, 即 $\lambda=1$ 或 $\lambda=-2$ 时有解.
(1) $\lambda=1 .\left\{\begin{array}{l}-2 x_{1}+x_{2}+x_{3}=-2 \\ x_{1}-2 x_{2}+x_{3}=1 \\ x_{1}+x_{2}-2 x_{3}=1\end{array},\left(\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right)=k\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)+\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)\right.$.
(2) $\lambda=-2,\left\{\begin{array}{l}-2 x_{1}+x_{2}+x_{3}=-2 \\ x_{1}-2 x_{2}+x_{3}=-2 \\ x_{1}+x_{2}-2 x_{3}=4\end{array},\left(\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right)=k\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)+\left(\begin{array}{l}2 \\ 2 \\ 0\end{array}\right)\right.$