13.求解下列齐次线性方程组:

(1) $\left\{ \begin{matrix} x_{1} + x_{2} + 2x_{3} - x_{4} = 0 \\ 2x_{1} + x_{2} + x_{3} - x_{4} = 0 \\ 2x_{1} + 2x_{2} + x_{3} + 2x_{4} = 0 \end{matrix} \right.\$ (2) $\left\{ \begin{matrix} x_{1} + 2x_{2} + x_{3} - x_{4} = 0 \\ 3x_{1} + 6x_{2} - x_{3} - 3x_{4} = 0 \\ 5x_{1} + 10x_{2} + x_{3} - 5x_{4} = 0 \end{matrix} \right.\$

(3) $\left\{ \begin{array}{r} \begin{matrix} 2x_{1} + 3x_{2} - x_{3} - 7x_{4} = 0 \\ 3x_{1} + x_{2} + 2x_{3} - 7x_{4} = 0 \end{matrix} \\ \begin{matrix} 4x_{1} + x_{2} - 3x_{3} + 6x_{4} = 0 \\ x_{1} - 2x_{2} + 5x_{3} - 5x_{4} = 0 \end{matrix} \end{array} \right.\$ (4) $\left\{ \begin{array}{r} \begin{matrix} 3x_{1} + 4x_{2} - 5x_{3} + 7x_{4} = 0 \\ 2x_{1} - 3x_{2} + 3x_{3} - 2x_{4} = 0 \end{matrix} \\ \begin{matrix} 4x_{1} + 11x_{2} - 13x_{3} + 16x_{4} = 0 \\ 7x_{1} - 2x_{2} + x_{3} + 3x_{4} = 0 \end{matrix} \end{array} \right.\$

解: 对系数矩阵A作初等行变换 , 化为行最简形

(1) A=$\left\lbrack \begin{matrix} 1 \\ 2 \\ 2 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 1 \\ 2 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ 1 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ - 1 \\ 2 \end{matrix} \right\rbrack\overset{\begin{matrix} r_{2} - 2r_{1} \\ r_{3} - 2r_{1} \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ - 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ - 3 \\ - 3 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 1 \\ 4 \end{matrix} \right\rbrack\overset{\begin{matrix} r_{2} \times ( - 1) \\ r_{3} \div ( - 3) \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ 3 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ - 1 \\ \frac{- 4}{3} \end{matrix} \right\rbrack$

$$\overset{r_{1} - r_{2}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 3 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ - 1 \\ \frac{- 4}{3} \end{matrix} \right\rbrack\overset{\begin{matrix} r_{1} + r_{3} \\ r_{2} - 3r_{3} \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} \frac{- 4}{3} \\ 3 \\ \frac{- 4}{3} \end{matrix} \right\rbrack$$

于是R(A)=3 , 故方程组有4-R(A)=1个自由未知数

得同解方程组：$\left\{ \begin{matrix} x_{1} - \frac{4}{3}x_{4} = 0 \\ x_{2} + 3x_{4} = 0 \\ x_{3} - \frac{4}{3}x_{4} = 0 \end{matrix} \right.\$ ,

得参数形式$\left\{ \begin{matrix} x_{1} = \frac{4}{3}x_{4} \\ x_{2} = - 3x_{4} \\ x_{3} = \frac{4}{3}x_{4} \end{matrix} \right.\$ ( x₄可任意取值)

得参数形式$\left\{ \begin{array}{r} \begin{matrix} x_{1} = \frac{4}{3}c \\ x_{2} = - 3c \end{matrix} \\ \begin{matrix} x_{3} = \frac{4}{3}c\ \ \\ x_{4} = c\ \ \ \ \end{matrix} \end{array} \right.\$ 令 x₄=c , 其中c为任意实数

得齐次通解$\left\lbrack \begin{array}{r} \begin{matrix} x_{1} \\ x_{2} \end{matrix} \\ \begin{matrix} x_{3} \\ x_{4} \end{matrix} \end{array} \right\rbrack$=$\left\lbrack \begin{array}{r} \begin{matrix} \frac{4}{3}c \\ - 3c \end{matrix} \\ \begin{matrix} \frac{4}{3}c\ \ \\ c\ \ \ \ \end{matrix} \end{array} \right\rbrack$=$c\left\lbrack \begin{array}{r} \begin{matrix} \frac{4}{3} \\ - 3 \end{matrix} \\ \begin{matrix} \frac{4}{3} \\ 1 \end{matrix} \end{array} \right\rbrack$ , c属于R

(2) A=$\left\lbrack \begin{matrix} 1 \\ 3 \\ 5 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ 6 \\ 10 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ - 3 \\ - 5 \end{matrix} \right\rbrack\overset{\begin{matrix} r_{2} - 3r_{1} \\ r_{3} - 5r_{1} \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ - 4 \\ - 4 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 0 \\ 0 \end{matrix} \right\rbrack\overset{\begin{matrix} r_{2} \div ( - 4) \\ r_{1} - r_{2} \\ r_{3} + 4r_{2} \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 0 \\ 0 \end{matrix} \right\rbrack$

得同解方程组：$\left\{ \begin{matrix} x_{1} + 2x_{2} - x_{4} = 0 \\ x_{3} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} \right.\$

得参数形式：$\left\{ \begin{matrix} x_{1} = - 2x_{2} + x_{4} \\ x_{3} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} \right.\$ (x₂ , x₄可任意取值)

得参数形式：$\left\{ \begin{array}{r} \begin{matrix} x_{1} = 2c_{1} + c_{2} \\ x_{2} = c_{1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} \\ \begin{matrix} x_{3} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x_{4} = c_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} \end{array} \right.\$ 令x₂=c₁ , x₄=c₂ , 其中c₁ , c₂为任意实数

得齐次通解：$\left\lbrack \begin{array}{r} \begin{matrix} x_{1} \\ x_{2} \end{matrix} \\ \begin{matrix} x_{3} \\ x_{4} \end{matrix} \end{array} \right\rbrack$=$\left\lbrack \begin{array}{r} \begin{matrix} - 2c_{1} + c_{2} \\ c_{1} \end{matrix} \\ \begin{matrix} 0 \\ \ c_{2} \end{matrix} \end{array} \right\rbrack$=$c_{1}\left\lbrack \begin{array}{r} \begin{matrix} - 2 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack$+$c_{2}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack$ , 其中c₁ , c₂属于R

(3) A =$\left\lbrack \begin{array}{r} \begin{matrix} 2 \\ 3 \end{matrix} \\ \begin{matrix} 4 \\ 1 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 3 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 2 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 1 \\ 2 \end{matrix} \\ \begin{matrix} - 3 \\ 5 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 7 \\ - 7 \end{matrix} \\ \begin{matrix} 6 \\ - 5 \end{matrix} \end{array} \right\rbrack\overset{\begin{array}{r} \begin{matrix} r_{1} \leftrightarrow r_{4} \\ r_{2} - 3r_{1} \end{matrix} \\ \begin{matrix} r_{3} - 4r_{1} \\ r_{4} - 2r_{1} \end{matrix} \end{array}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 2 \\ 7 \end{matrix} \\ \begin{matrix} 9 \\ 7 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 5 \\ - 13 \end{matrix} \\ \begin{matrix} - 23 \\ - 11 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 5 \\ 8 \end{matrix} \\ \begin{matrix} 26 \\ 3 \end{matrix} \end{array} \right\rbrack\overset{\begin{matrix} r_{3} - r_{2} \\ r_{4} - r_{2} \end{matrix}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 2 \\ 7 \end{matrix} \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 5 \\ - 13 \end{matrix} \\ \begin{matrix} - 10 \\ 2 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 5 \\ 8 \end{matrix} \\ \begin{matrix} 18 \\ 5 \end{matrix} \end{array} \right\rbrack$

$$\overset{\begin{array}{r} \begin{matrix} r_{2} \leftrightarrow r_{3} \\ r_{2} \div 2 \end{matrix} \\ \begin{matrix} r_{1} + 2r_{2} \\ r_{3} - 7r_{2} \end{matrix} \end{array}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 5 \\ - 5 \end{matrix} \\ \begin{matrix} 22 \\ 2 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 13 \\ 9 \end{matrix} \\ \begin{matrix} - 55 \\ - 5 \end{matrix} \end{array} \right\rbrack\overset{\begin{array}{r} \begin{matrix} r_{3} \div 22 \\ r_{1} + 5r_{3} \end{matrix} \\ \begin{matrix} r_{2} + 5r_{3} \\ r_{4} - 2r_{3} \end{matrix} \end{array}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} \frac{1}{2} \\ \frac{- 7}{2} \end{matrix} \\ \begin{matrix} \frac{- 5}{2} \\ 0 \end{matrix} \end{array} \right\rbrack$$

得同解方程组：$\left\{ \begin{array}{r} x_{1} + \frac{1}{2}x_{4} = 0 \\ x_{2} - \frac{7}{2}x_{4} = 0 \\ x_{3} - \frac{5}{2}x_{4} = 0 \end{array} \right.\$ ，得参数形式：$\left\{ \begin{array}{r} x_{1} = \frac{- 1}{2}x_{4} \\ x_{2} = \frac{7}{2}x_{4} \\ x_{3} = \frac{5}{2}x_{4} \end{array} \right.\$ ( x₄可任意取值)

得参数形式：$\left\{ \begin{array}{r} \begin{matrix} x_{1} = \frac{- 1}{2}c \\ x_{2} = \frac{7}{2}c \end{matrix} \\ \begin{matrix} x_{3} = \frac{5}{2}c\ \ \ \\ x_{4} = c\ \ \end{matrix} \end{array} \right.\$ 令 x₄=c , 其中 c为任意实数

得齐次通解：$\left\lbrack \begin{array}{r} \begin{matrix} x_{1} \\ x_{2} \end{matrix} \\ \begin{matrix} x_{3} \\ x_{4} \end{matrix} \end{array} \right\rbrack$=$\left\lbrack \begin{array}{r} \begin{matrix} \frac{- 1}{2}c \\ \frac{7}{2}c \end{matrix} \\ \begin{matrix} \frac{5}{2}c \\ \ c \end{matrix} \end{array} \right\rbrack$=$c\left\lbrack \begin{array}{r} \begin{matrix} \frac{- 1}{2} \\ \frac{7}{2} \end{matrix} \\ \begin{matrix} \frac{5}{2} \\ 1 \end{matrix} \end{array} \right\rbrack$ , 其中c属于R

(4) A =$\left\lbrack \begin{array}{r} \begin{matrix} 3 \\ 2 \end{matrix} \\ \begin{matrix} 4 \\ 7 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 4 \\ - 3 \end{matrix} \\ \begin{matrix} 11 \\ - 2 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 5 \\ 3 \end{matrix} \\ \begin{matrix} - 13 \\ 1 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 7 \\ - 2 \end{matrix} \\ \begin{matrix} 16 \\ 3 \end{matrix} \end{array} \right\rbrack\overset{r_{1} - r_{2}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 2 \end{matrix} \\ \begin{matrix} 4 \\ 7 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 7 \\ - 3 \end{matrix} \\ \begin{matrix} 11 \\ - 2 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 8 \\ 3 \end{matrix} \\ \begin{matrix} - 13 \\ 1 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 9 \\ - 2 \end{matrix} \\ \begin{matrix} 16 \\ 3 \end{matrix} \end{array} \right\rbrack$

$$\overset{\begin{matrix} r_{2} - 2r_{1} \\ r_{3} - 4r_{1} \\ r_{4} - 7r_{1} \end{matrix}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 7 \\ - 17 \end{matrix} \\ \begin{matrix} - 17 \\ - 51 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 8 \\ 19 \end{matrix} \\ \begin{matrix} 19 \\ 57 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 9 \\ - 20 \end{matrix} \\ \begin{matrix} - 20 \\ - 60 \end{matrix} \end{array} \right\rbrack\overset{\begin{matrix} r_{3} - r_{2} \\ r_{4} - 3r_{2} \\ r_{2} \div ( - 17) \end{matrix}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 7 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 8 \\ \frac{- 19}{17} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 9 \\ \frac{20}{17} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack\overset{r_{1} - 7r_{2}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} \frac{- 3}{17} \\ \frac{- 19}{17} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} \frac{13}{17} \\ \frac{20}{17} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack$$

得同解方程组：$\left\{ \begin{matrix} x_{1} - \frac{3}{17}x_{3} + \frac{13}{17}x_{4} = 0 \\ x_{2} - \frac{19}{17}x_{3} + \frac{20}{17}x_{4} = 0 \end{matrix} \right.\$

得参数形式：$\left\{ \begin{matrix} x_{1} = \frac{3}{17}x_{3} - \frac{13}{17}x_{4} \\ x_{2} = \frac{19}{17}x_{3} - \frac{20}{17}x_{4} \end{matrix} \right.\$ (x₃ , x₄可任意取值)

得参数形式：$\left\{ \begin{array}{r} \begin{matrix} x_{1} = \frac{3}{17}c_{1} - \frac{13}{17}c_{2} \\ x_{2} = \frac{19}{17}c_{1} - \frac{20}{17}c_{2} \end{matrix} \\ \begin{matrix} x_{3} = c_{1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x_{4} = c_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} \end{array} \right.\$ 其中x₃=c₁ , x₄=c₂ , c₁ , c₂为任意实数

得齐次通解：$\left\lbrack \begin{array}{r} \begin{matrix} x_{1} \\ x_{2} \end{matrix} \\ \begin{matrix} x_{3} \\ x_{4} \end{matrix} \end{array} \right\rbrack$=$\left\lbrack \begin{array}{r} \begin{matrix} \frac{3}{17}c_{1} + \frac{- 13}{17}c_{2} \\ \frac{19}{17}c_{1} - \frac{20}{17}c_{2} \end{matrix} \\ \begin{matrix} c_{1} \\ \ c_{2} \end{matrix} \end{array} \right\rbrack$=$c_{1}\left\lbrack \begin{array}{r} \begin{matrix} \frac{3}{17} \\ \frac{19}{17} \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack + c_{2}\left\lbrack \begin{array}{r} \begin{matrix} \frac{- 13}{17} \\ \frac{- 20}{17} \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack$ , 其中c₁ , c₂属于R

14. 求解下列非齐次线性方程组:

(1) $\left\{ \begin{array}{r} 4x_{1} + 2x_{2} - x_{3} = 2 \\ 3x_{1} - x_{2} + 2x_{3} = 10 \\ 11x_{1} + 3x_{2} = 8\ \ \ \ \ \ \ \ \ \ \end{array} \right.\$ (2) $\left\{ \begin{array}{r} \begin{matrix} 2x + 3y + z = 4 \\ x - 2y + 4z = - 5 \end{matrix} \\ \begin{matrix} 3x + 8y - 2z = 13 \\ 4x - y + 9z = - 6 \end{matrix} \end{array} \right.\$

(3) $\left\{ \begin{array}{r} 2x + y - z + w = 1 \\ 4x + 2y - 2z + w = 2 \\ 2x + y - z - w = 1 \end{array} \right.\$ (4) $\left\{ \begin{array}{r} 2x + y - z + w = 1 \\ 3x - 2y + z - 3w = 4 \\ x + 4y - 3z + 5w = - 2 \end{array} \right.\$

解: 本题中分别以A和B表示方程组的系数矩阵和增广矩阵

(1)B=$\left\lbrack \begin{matrix} 4 \\ 3 \\ 11 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ - 1 \\ 3 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 2 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ 10 \\ 8 \end{matrix} \right\rbrack\overset{r_{1} - r_{2}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 3 \\ 11 \end{matrix}\ \ \ \ \begin{matrix} 3 \\ - 1 \\ 3 \end{matrix}\ \ \ \ \begin{matrix} - 3 \\ 2 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 8 \\ 10 \\ 8 \end{matrix} \right\rbrack$

$$\overset{\begin{matrix} r_{2} - 3r_{1} \\ r_{3} - 11r_{1} \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 3 \\ - 10 \\ - 30 \end{matrix}\ \ \ \ \begin{matrix} - 3 \\ 11 \\ 33 \end{matrix}\ \ \ \ \begin{matrix} - 8 \\ 34 \\ 96 \end{matrix} \right\rbrack\overset{r_{3} - 3r_{2}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 3 \\ - 10 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 3 \\ 11 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 8 \\ 34 \\ - 6 \end{matrix} \right\rbrack$$

因R(A)=2 , R(B)=3 , R(A)≠R(B) , 知方程组无解

(2) B=$\left\lbrack \begin{array}{r} \begin{matrix} 2 \\ 1 \end{matrix} \\ \begin{matrix} 3 \\ 4 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 3 \\ - 2 \end{matrix} \\ \begin{matrix} 8 \\ - 1 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 4 \end{matrix} \\ \begin{matrix} - 2 \\ 9 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 4 \\ - 5 \end{matrix} \\ \begin{matrix} 13 \\ - 6 \end{matrix} \end{array} \right\rbrack\overset{r_{2} \leftrightarrow r_{2}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 2 \end{matrix} \\ \begin{matrix} 3 \\ 4 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 2 \\ 3 \end{matrix} \\ \begin{matrix} 8 \\ - 1 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 4 \\ 1 \end{matrix} \\ \begin{matrix} - 2 \\ 9 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 5 \\ 4 \end{matrix} \\ \begin{matrix} 13 \\ - 6 \end{matrix} \end{array} \right\rbrack\overset{\begin{matrix} r_{2} - 2r_{1} \\ r_{3} - 3r_{1} \\ r_{4} - 4r_{1} \end{matrix}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 2 \\ 7 \end{matrix} \\ \begin{matrix} 14 \\ 7 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 4 \\ - 7 \end{matrix} \\ \begin{matrix} - 14 \\ - 7 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 5 \\ 14 \end{matrix} \\ \begin{matrix} 28 \\ 14 \end{matrix} \end{array} \right\rbrack$

$$\overset{\begin{matrix} r_{2} \div 7 \\ r_{3} - 14r_{2} \\ r_{4} - 7r_{2} \end{matrix}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 2 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 4 \\ - 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 5 \\ 2 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack\overset{r_{1} + 2r_{2}}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 2 \\ - 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 1 \\ 2 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack$$

因R(A)=R(B)=2<3 , 故方程组有无限多解 , 并且有3- R(A)=1个自由未知数

得同解方程组：$\left\{ \begin{matrix} x + 2z = - 1 \\ y - z = 2\ \ \ \ \ \end{matrix} \right.\$

得参数形式：$\left\{ \begin{array}{r} x = - 2z - 1 \\ y = z + 2\ \ \ \ \ \end{array} \right.\$ (z可任意取值)

得参数形式：$\left\{ \begin{matrix} x = - 2c - 1 \\ y = c + 2 \\ z = c \end{matrix} \right.\$ 其中z=c , c为任意实数

得非齐次通解：$\begin{bmatrix} x \\ y \\ z \end{bmatrix}$=$\begin{bmatrix} - 2c - 1 \\ c + 2 \\ c \end{bmatrix}$=c$\begin{bmatrix} - 2 \\ 1 \\ 1 \end{bmatrix}$+$\begin{bmatrix} - 1 \\ 2 \\ 0 \end{bmatrix}$ (c属于R)

(3) B=$\left\lbrack \begin{matrix} 2 \\ 4 \\ 2 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 2 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ - 2 \\ - 1 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 1 \\ - 1 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right\rbrack\overset{\begin{matrix} r_{2} - 2r_{1} \\ r_{3} - r_{1} \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 2 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ - 1 \\ - 2 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right\rbrack$

$$\overset{\begin{matrix} r_{1} + r_{2} \\ r_{3} - 2r_{2} \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 2 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ - 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right\rbrack\overset{\begin{matrix} r_{1} \div 2 \\ r_{2} \times ( - 1) \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{1}{2} \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{- 1}{2} \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ - 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{1}{2} \\ 0 \\ 0 \end{matrix} \right\rbrack$$

得同解方程组：$\left\{ \begin{matrix} x + \frac{1}{2}y - \frac{1}{2}z = \frac{1}{2} \\ w = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} \right.\$

得参数形式：$\left\{ \begin{array}{r} x = \frac{- 1}{2}y + \frac{1}{2}z + \frac{1}{2} \\ w = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array} \right.\$ (y , z可任意取值)

得参数形式：$\left\{ \begin{array}{r} \begin{matrix} x = \frac{- 1}{2}c_{1} + \frac{1}{2}c_{2} + \frac{1}{2} \\ y = c_{1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} \\ \begin{matrix} z = c_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ w = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} \end{array} \right.\$ 其中y=c₁ , z=c₂ , c₁ , c₂为任意实数

得非齐次通解：$\left\lbrack \begin{array}{r} \begin{matrix} x \\ y \end{matrix} \\ \begin{matrix} z \\ w \end{matrix} \end{array} \right\rbrack$=$\left\lbrack \begin{array}{r} \begin{matrix} \frac{- 1}{2}c_{1} + \frac{1}{2}c_{2} + \frac{1}{2} \\ c_{1} + 0 + 0 \end{matrix} \\ \begin{matrix} 0 + c_{2} + 0 \\ \ 0 \end{matrix} \end{array} \right\rbrack$=$c_{1}\left\lbrack \begin{array}{r} \begin{matrix} \frac{- 1}{2} \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack$+$c_{2}\left\lbrack \begin{array}{r} \begin{matrix} \frac{1}{2} \\ 0 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack$+$\left\lbrack \begin{array}{r} \begin{matrix} \frac{1}{2} \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack$ (c₁ , c₂属于R)

(4) B=$\left\lbrack \begin{matrix} 2 \\ 3 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ - 2 \\ 4 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 1 \\ - 3 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ - 3 \\ 5 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 4 \\ - 2 \end{matrix} \right\rbrack\overset{r_{1} \leftrightarrow r_{3}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 3 \\ 2 \end{matrix}\ \ \ \ \begin{matrix} 4 \\ - 2 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} - 3 \\ 1 \\ - 1 \end{matrix}\ \ \ \ \begin{matrix} 5 \\ - 3 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} - 2 \\ 4 \\ 1 \end{matrix} \right\rbrack$

$$\overset{\begin{matrix} r_{2} - 3r_{1} \\ r_{3} - 2r_{1} \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 4 \\ - 14 \\ - 7 \end{matrix}\ \ \ \ \begin{matrix} - 3 \\ 10 \\ 5 \end{matrix}\ \ \ \ \begin{matrix} 5 \\ - 18 \\ - 9 \end{matrix}\ \ \ \ \begin{matrix} - 2 \\ 10 \\ 5 \end{matrix} \right\rbrack\overset{\begin{matrix} r_{3} - {\frac{1}{2}r_{2}} \\ r_{2} \div ( - 14) \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 4 \\ 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 3 \\ \frac{- 5}{7} \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 5 \\ \frac{9}{7} \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 2 \\ \frac{- 5}{7} \\ 0 \end{matrix} \right\rbrack$$

$$\overset{r_{1} - 4r_{2}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{- 1}{7} \\ \frac{- 5}{7} \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{- 1}{7} \\ \frac{9}{7} \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{6}{7} \\ \frac{- 5}{7} \\ 0 \end{matrix} \right\rbrack$$

得同解方程组：$\left\{ \begin{matrix} x - \frac{1}{7}z - \frac{1}{7}w = \frac{6}{7} \\ y - \frac{5}{7}z + \frac{9}{7}w = - \frac{5}{7} \end{matrix} \right.\$

得参数形式：$\left\{ \begin{array}{r} x = \frac{1}{7}z + \frac{1}{7}w + \frac{6}{7} \\ y = \frac{5}{7}z - \frac{9}{7}w - \frac{5}{7} \end{array} \right.\$ (z , w可任意取值)

得参数形式：$\left\{ \begin{array}{r} \begin{matrix} x = \frac{1}{7}c_{1} + \frac{1}{7}c_{2} + \frac{6}{7} \\ y = \frac{5}{7}c_{1} - \frac{9}{7}c_{2} - \frac{5}{7} \end{matrix} \\ \begin{matrix} z = c_{1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ w = c_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} \end{array} \right.\$ 其中z=c₁ ,w=c₂ , c₁ , c₂为任意实数

得非齐次通解：$\left\lbrack \begin{array}{r} \begin{matrix} x \\ y \end{matrix} \\ \begin{matrix} z \\ w \end{matrix} \end{array} \right\rbrack$=$\left\lbrack \begin{array}{r} \begin{matrix} \frac{1}{7}c_{1} + \frac{1}{7}c_{2} + \frac{6}{7} \\ \frac{5}{7}c_{1} - \frac{9}{7}c_{2} - \frac{5}{7} \end{matrix} \\ \begin{matrix} c_{1} \\ \ c_{2} \end{matrix} \end{array} \right\rbrack$=$c_{1}\left\lbrack \begin{array}{r} \begin{matrix} \frac{1}{7} \\ \frac{5}{7} \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack$+$c_{2}\left\lbrack \begin{array}{r} \begin{matrix} \frac{1}{7} \\ \frac{- 9}{7} \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack$+$\left\lbrack \begin{array}{r} \begin{matrix} \frac{6}{7} \\ \frac{- 5}{7} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack$ (c₁ , c₂属于R)