Appendix_常见题型

型二 线性方程组解的结构与性质

1. 设 $\boldsymbol{A}$ 为四阶矩阵，$r(\boldsymbol{A}) = 3$，且 $\boldsymbol{A}$ 的每行元素之和为 $0$，求方程组 $\boldsymbol{AX} = \boldsymbol{0}$ 的解 $$\begin{array}{ll} \because & \boldsymbol{A}\text{的每行元素之和为}0 \\ \therefore & \boldsymbol{A} \left( \begin{matrix} 1 \\ 1 \\ 1 \\ 1 \end{matrix} \right) = \boldsymbol{0}\\ \therefore & \boldsymbol{AX} = \boldsymbol{0}\text{的通解} k \left( \begin{matrix} 1 \\ 1 \\ 1 \\ 1 \end{matrix} \right)\\ \end{array}$$
2. 设 $\boldsymbol{A}$ 为四阶矩阵，$r(\boldsymbol{A}) < 4$，且 $A_{21}\ne 0$，求方程组 $\boldsymbol{AX} = \boldsymbol{0}$ 的解 $$\begin{array}{ll} \because & \boldsymbol{A} \text{为四阶矩阵} \\ & r(\boldsymbol{A})< 4, A_{21} \ne 0 \\ \therefore & r(\boldsymbol{A})=3 \\ \therefore & \boldsymbol{A}^{*} = \left( \begin{matrix} A_{11} & A_{21} & A_{31} & A_{41} \\ A_{12} & A_{22} & A_{32} & A_{42} \\ A_{13} & A_{23} & A_{33} & A_{43} \\ A_{14} & A_{24} & A_{34} & A_{44} \\ \end{matrix} \right)\\ \because & \boldsymbol{AA}^{*} = \vert \boldsymbol{A} \vert \boldsymbol{E} \\ \therefore & \boldsymbol{AA}^{*} = \boldsymbol{0} \\ \because & A_{21} \ne 0 \\ \therefore & \boldsymbol{x} = k \left( \begin{matrix} A_{21} \\ A_{22} \\ A_{23} \\ A_{24} \\ \end{matrix} \right)\\ \end{array}$$
3. 设 $\boldsymbol{A}$ 为 $n$ 阶矩阵，且 $A_{11}\ne 0$，若 $\eta_{1}, \eta_{2}, \eta_{3}$ 为非其次线性方程组 $\boldsymbol{AX} = \boldsymbol{\beta}$ 的不同解，则非齐次线性方程组 $\boldsymbol{AX} = \boldsymbol{0}$ 的基础解系所含的线性无关解的向量个数为( ) $$\begin{array}{cccc} \text{A. }1 & \text{B. }2 & \text{C. } 3 & \text{D. }4 \end{array}$$ $$\begin{array}{ll} \because & A_{11}\ne 0\\ \therefore & r(\boldsymbol{A}) \ge n-1 \\ \because & \eta_{1}, \eta_{2},\eta_{3}\text{为非其次线性方程组}\boldsymbol{AX} = \boldsymbol{\beta}\text{的不同解} \\ \therefore & r(\boldsymbol{A}) = r(\bar{\boldsymbol{A}}) < n \\ \therefore & r(\boldsymbol{A}) = n-1 \\ \therefore & \boldsymbol{AX} = \boldsymbol{0}\text{基础解系所含线性无关向量个数为}1 \\ \end{array}$$
4. 设 $\eta_{1} = \left( \begin{smallmatrix} 1 \\ -2 \\ 1 \\ 2 \end{smallmatrix} \right) ,\eta_{2} = \left( \begin{smallmatrix} 0 \\ 1 \\ 0 \\ -1 \end{smallmatrix} \right) ,\eta_{3} = \left( \begin{smallmatrix} 2 \\ 1 \\ 3 \\ -2 \end{smallmatrix} \right)$ 为非齐次线性方程组 $\begin{cases} x_{1} + a_{2}x_{2} + 4x_{3} - x_{4} = d_{1} \\ b_{1}x_{1} + x_{2} + b_{3}x_{3} + b_{4}x_{4} = d_{2} \\ 2x_{1} + c_{2}x_{2} + c_{3}x_{3} + x_{4} = d_{2} \\ \end{cases}$ 的三个解 ，求该方程组的通解 $$\begin{array}{ll} & \bar{\boldsymbol{A}} = \left( \begin{matrix} 1 & a_{2} & 4 & -1 & d_{1}\\ b_{1} & 1 & b_{3} & b_{4} & d_{2} \\ 2 & c_{2} & c_{3} & 1 & d_{3} \\ \end{matrix} \right) \\ \because & \text{第1和第3行一定不成比例} \\ \therefore & r(\boldsymbol{A}) \ge 2 \\ \because & \zeta_{1} = \eta_{1} - \eta_{2} = \left( \begin{smallmatrix} 1 \\ -3 \\ 1 \\ 3 \end{smallmatrix} \right); \zeta_{2} = \eta_{1} - \eta_{3} = \left( \begin{smallmatrix} -1 \\ -3 \\ -2 \\ 4 \end{smallmatrix} \right)\\ \therefore & \zeta_{1}, \zeta_{2}\text{不成比例}\\ & \boldsymbol{AX} = \boldsymbol{0} \text{的通解由至少两个线性无关的解向量构成} \\ \therefore & 4 - r(\boldsymbol{A}) \ge 2 \Rightarrow r(\boldsymbol{A})\le 2\\ \therefore & r(\boldsymbol{A}) = 2 \\ \therefore & \text{通解 } \boldsymbol{x} = k_{1}\zeta_{1} + k_{2}\zeta_{2} + \eta_{1} \\ \end{array}$$
5. 设 $\boldsymbol{A} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2} & \boldsymbol{\alpha}_{3} & \boldsymbol{\alpha}_{4} & \boldsymbol{\alpha}_{5} \end{matrix} \right)$，其中 $\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4}$ 线性无关，$\boldsymbol{\alpha}_{2} = \boldsymbol{\alpha}_{1} + 2 \boldsymbol{\alpha}_{3}- \boldsymbol{\alpha}_{4}, \boldsymbol{\alpha}_{5} = \boldsymbol{\alpha}_{1}-\boldsymbol{\alpha}_{2} - 2 \boldsymbol{\alpha}_{3} + 3 \boldsymbol{\alpha}_{4}$，求方程组 $\boldsymbol{AX} = \boldsymbol{0}$ 的解 $$\begin{array}{ll} \because & \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{3},\boldsymbol{\alpha}_{4} \text{线性无关} \\ \therefore & r(\boldsymbol{A}) = 3 \\ \because & \boldsymbol{\alpha}_{1} - \boldsymbol{\alpha}_{2} + 2\boldsymbol{\alpha}_{3}- \boldsymbol{\alpha}_{4} + 0 \boldsymbol{\alpha}_{5}= \boldsymbol{0} \\ & \boldsymbol{\alpha}_{1} - \boldsymbol{\alpha}_{2} - 2 \boldsymbol{\alpha}_{3} +3 \boldsymbol{\alpha}_{4} - \boldsymbol{\alpha}_{5} = \boldsymbol{0} \\ \therefore & \boldsymbol{x} = k_{1} \left( \begin{smallmatrix} 1 \\ -1 \\ 2 \\ -1 \\ 0 \end{smallmatrix} \right) + k_{2} \left( \begin{smallmatrix} 1 \\ -1 \\ -2 \\ 3 \\ -1 \end{smallmatrix} \right) \\ \end{array}$$

型三 含参数方程组解的讨论

1. 设 $\boldsymbol{A} = \left( \begin{smallmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & t & t \\ 1 & t & 0 & 1 \\ \end{smallmatrix} \right)$，且 $\boldsymbol{AX}= \boldsymbol{0}$ 含两个线性无关的解向量，求 $\boldsymbol{AX}= \boldsymbol{0}$ 的通解 $$\begin{array}{ll} & \boldsymbol{A} = \left( \begin{smallmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & t & t \\ 1 & t & 0 & 1 \end{smallmatrix} \right) \rightarrow \left( \begin{smallmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & t & t \\ 0 & t-2 & -1 & -1 \end{smallmatrix} \right) \\ \because & \boldsymbol{AX}= \boldsymbol{0} \text{含两个线性无关的解向量} \\ \therefore & 4 -r(\boldsymbol{A}) = 2 \Rightarrow r(\boldsymbol{A}) = 2 \\ \therefore & t-2 = \frac{-1}{t} = \frac{-1}{t} \\ \therefore & t = 1 \\ & \boldsymbol{A} \rightarrow \left( \begin{smallmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{smallmatrix} \right) \rightarrow \left( \begin{smallmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{smallmatrix} \right) \\ \therefore & \boldsymbol{x} = k_{1} \left( \begin{smallmatrix} 1 \\ -1 \\ 1 \\ 0 \\ \end{smallmatrix} \right) + k_{2} \left( \begin{smallmatrix} 0 \\ -1 \\ 0 \\ 1 \\ \end{smallmatrix} \right)\\ \end{array}$$
2. 已知齐次线性方程组 $\begin{cases} x_{1} - x_{2} + 2x_{3} = 0 \\ 2x_{1} + (a-3)x_{2} + 5x_{3} = 0 \\ -x_{1} + x_{2} + (4-a)x_{3} = 0 \\ ax_{1} - ax_{2} + (2a +3)x_{3} = 0 \\ \end{cases}$ 有非零解，求常数 $a$，并求其通解 $$\begin{array}{ll} & \boldsymbol{A} = \left( \begin{smallmatrix} 1 & -1 & 2 \\ 2 & a-3 & 5 \\ -1 & 1 & 4-a \\ a & -a & 2a+3 \\ \end{smallmatrix} \right) \rightarrow \left( \begin{smallmatrix} 1 & -1 & 2 \\ 0 & a-1 & 1 \\ 0 & 0 & 6-a \\ 0 & 0 & 3 \\ \end{smallmatrix} \right) \\ \because & \boldsymbol{AX} = \boldsymbol{0}\text{有非零解} \\ \therefore & r(\boldsymbol{A}) < 3 \\ \therefore & a- 1 = 0\Rightarrow a = 1 \\ \therefore & \boldsymbol{A} = \left( \begin{smallmatrix} 1 & -1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{smallmatrix} \right)\rightarrow \left( \begin{smallmatrix} 1 & -1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{smallmatrix} \right)\\ \therefore & \boldsymbol{x} = k \left( \begin{smallmatrix} 1 \\ 1 \\ 0 \end{smallmatrix} \right) \\ \end{array}$$
3. 设方程组 $\left( \begin{smallmatrix} 1 & 2 & 1 \\ 2 & 3 & a+2 \\ 1 & a & -2 \\ \end{smallmatrix} \right)\left( \begin{smallmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{smallmatrix} \right) = \left( \begin{smallmatrix} 1 \\ 3 \\ 0 \\ \end{smallmatrix} \right)$，讨论 $a$ 的取值，使得方程组有唯一解，无解，无穷多个解，当方程组有无穷多解时，求出其通解 $$\begin{array}{ll} & \bar{\boldsymbol{A}} = \left( \begin{smallmatrix} 1 & 2 & 1 & 1 \\ 2 & 3 & a+2 & 3 \\ 1 & a & -2 & 0 \end{smallmatrix} \right) \rightarrow \left( \begin{smallmatrix} 1 & 2 & 1 & 1 \\ 0 & -1 & a & 1 \\ 0 & a-2 & -3 & -1 \end{smallmatrix} \right) \\ & \text{当} r(\boldsymbol{A}) = r(\bar{\boldsymbol{A}})=3 \text{时，方程组有唯一解} \\ \therefore & \frac{-1}{a-2} \ne \frac{a}{-3} \\ \therefore & a \ne 3 \text{且} a\ne -1 \text{时，方程组有唯一解} \\\\ & \text{当} r(\boldsymbol{A}) \ne r(\bar{\boldsymbol{A}}) \text{时，方程组无解} \\ \therefore & \frac{-1}{a-2} = \frac{a}{-3} \ne -1 \\ \therefore & a=-1 \text{时，方程组无解} \\ \\ & \text{当} r(\boldsymbol{A}) = r(\bar{\boldsymbol{A}})< 3 \text{时，方程组有无数解} \\ \therefore & \frac{-1}{a-2} = \frac{a}{-3} = -1 \\ \therefore & a = 3 \text{时，方程组有无数解} \\ \therefore & \bar{\boldsymbol{A}} = \left( \begin{smallmatrix} 1 & 2 & 1 & 1 \\ 0 & 1 & -3 & -1 \\ 0 & 0 & 0 & 0 \end{smallmatrix} \right)\rightarrow \left( \begin{smallmatrix} 1 & 0 & 7 & 3 \\ 0 & 1 & -3 & -1 \\ 0 & 0 & 0 & 0 \end{smallmatrix} \right)\\ \therefore & \boldsymbol{x} = k \left( \begin{smallmatrix} -7 \\ 3 \\ 1 \end{smallmatrix} \right) + \left( \begin{smallmatrix} 3 \\ -1 \\ 0 \end{smallmatrix} \right)\\ \end{array}$$
4. 设 $\boldsymbol{A} = \left( \begin{smallmatrix} 1 & a & 0 & 0 \\ 0 & 1 & a & 0 \\ 0 & 0 & 1 & a \\ a & 0 & 0 & 1 \\ \end{smallmatrix} \right), \boldsymbol{\beta} = \left( \begin{smallmatrix} 1 \\ -1 \\ 0 \\ 0 \end{smallmatrix} \right)$
   1. 计算 $\det(\boldsymbol{A})$
   2. 谈论 $a$ 取何值时，方程组 $\boldsymbol{AX} = \boldsymbol{\beta}$ 有无数个解，并求出其通解 $$\begin{array}{ll} & \det(\boldsymbol{A}) = \left\vert \begin{smallmatrix} 1 & a & 0 & 0 \\ 0 & 1 & a & 0 \\ 0 & 0 & 1 & a \\ a & 0 & 0 & 1 \end{smallmatrix} \right\vert = 1 - a \left\vert \begin{smallmatrix} 0 & a & 0 \\ 0 & 1 & a \\ a & 0 &1 \end{smallmatrix} \right\vert = 1-a^{4} \\\\ \because & \boldsymbol{AX} = \boldsymbol{\beta}\text{有无数个解} \\ \therefore & \det(\boldsymbol{A}) = 0 \\ \therefore & a = 1 \text{ 或 } a=-1 \\ & \text{当} a = 1 \text{时} \\ & \bar{\boldsymbol{A}} = \left( \begin{smallmatrix} 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ \end{smallmatrix} \right) \rightarrow \left( \begin{smallmatrix} 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & -1 & 0 & 1 & -1 \\ \end{smallmatrix} \right)\rightarrow \left( \begin{smallmatrix} 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & -2 \\ \end{smallmatrix} \right) \\ & \rightarrow \left( \begin{smallmatrix} 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & -2 \\ \end{smallmatrix} \right) \\ \therefore & r(\boldsymbol{A}) \ne r(\bar{\boldsymbol{A}}) \\ \therefore & \text{当}a=1 \text{时，此方程组无解} \\ & \text{当} a = -1 \text{时} \\ & \bar{\boldsymbol{A}} = \left( \begin{smallmatrix} 1 & -1 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 & -1 \\ 0 & 0 & 1 & -1 & 0 \\ -1 & 0 & 0 & 1 & 0 \\ \end{smallmatrix} \right) \rightarrow \left( \begin{smallmatrix} 1 & -1 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 & -1 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & -1 & 0 & 1 & 1 \\ \end{smallmatrix} \right)\rightarrow \left( \begin{smallmatrix} 1 & -1 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 & -1 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ \end{smallmatrix} \right) \\ & \rightarrow \left( \begin{smallmatrix} 1 & -1 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 & -1 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{smallmatrix} \right) \rightarrow \left( \begin{smallmatrix} 1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{smallmatrix} \right)\\ \therefore & r(\boldsymbol{A}) = r(\bar{\boldsymbol{A}}) < 4 \\ \therefore & \text{当}a=1 \text{时，此方程组有无数解} \\ \therefore & \boldsymbol{x} = k \left( \begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix} \right) + \left( \begin{smallmatrix} 0 \\ -1 \\ 0 \\ 0 \end{smallmatrix} \right) \\ \end{array}$$