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13.求解下列齐次线性方程组:

(1) {x1+x2+2x3−x4=02x1+x2+x3−x4=02x1+2x2+x3+2x4=0\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) {x1+2x2+x3−x4=03x1+6x2−x3−3x4=05x1+10x2+x3−5x4=0\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) {2x1+3x2−x3−7x4=03x1+x2+2x3−7x4=04x1+x2−3x3+6x4=0x1−2x2+5x3−5x4=0\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) {3x1+4x2−5x3+7x4=02x1−3x2+3x3−2x4=04x1+11x2−13x3+16x4=07x1−2x2+x3+3x4=0\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=[122112211−1−12]⇒r2−2r1r3−2r1[1001−102−3−3−114]⇒r2×(−1)r3÷(−3)[100110231−1−1−43]\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

⇒r1−r2[100010−1310−1−43]⇒r1+r3r2−3r3[100010001−433−43]\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：{x1−43x4=0x2+3x4=0x3−43x4=0\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{x1=43x4x2=−3x4x3=43x4\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{x1=43cx2=−3cx3=43cx4=c\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[x1x2x3x4]\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=[43c−3c43cc]\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[43−3431]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=[13526101−11−1−3−5]⇒r2−3r1r3−5r1[1002001−4−4−100]⇒r2÷(−4)r1−r2r3+4r2[100200010−100]\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：{x1+2x2−x4=0x3=0\left\{ \begin{matrix} x_{1} + 2x_{2} - x_{4} = 0 \\ x_{3} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} \right.\

得参数形式：{x1=−2x2+x4x3=0\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：{x1=2c1+c2x2=c1x3=0x4=c2\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：[x1x2x3x4]\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=[−2c1+c2c10c2]\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=c1[−2100]c_{1}\left\lbrack \begin{array}{r} \begin{matrix} - 2 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack+c2[1001]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 =[23413112−12−35−7−76−5]⇒r1↔r4r2−3r1r3−4r1r4−2r1[1000−27975−13−23−11−58263]⇒r3−r2r4−r2[1000−27205−13−102−58185]\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

⇒r2↔r3r2÷2r1+2r2r3−7r2[10000100−5−5222139−55−5]⇒r3÷22r1+5r3r2+5r3r4−2r3[10000100001012−72−520]\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：{x1+12x4=0x2−72x4=0x3−52x4=0\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：{x1=−12x4x2=72x4x3=52x4\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：{x1=−12cx2=72cx3=52cx4=c\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：[x1x2x3x4]\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=[−12c72c52cc]\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[−1272521]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 =[32474−311−2−53−1317−2163]⇒r1−r2[12477−311−2−83−1319−2163]\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

⇒r2−2r1r3−4r1r4−7r1[10007−17−17−51−81919579−20−20−60]⇒r3−r2r4−3r2r2÷(−17)[10007100−8−1917009201700]⇒r1−7r2[10000100−317−1917001317201700]\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：{x1−317x3+1317x4=0x2−1917x3+2017x4=0\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：{x1=317x3−1317x4x2=1917x3−2017x4\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：{x1=317c1−1317c2x2=1917c1−2017c2x3=c1x4=c2\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：[x1x2x3x4]\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=[317c1+−1317c21917c1−2017c2c1c2]\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=c1[317191710]+c2[−1317−201701]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) {4x1+2x2−x3=23x1−x2+2x3=1011x1+3x2=8\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) {2x+3y+z=4x−2y+4z=−53x+8y−2z=134x−y+9z=−6\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) {2x+y−z+w=14x+2y−2z+w=22x+y−z−w=1\left\{ \begin{array}{r} 2x + y - z + w = 1 \\ 4x + 2y - 2z + w = 2 \\ 2x + y - z - w = 1 \end{array} \right.\ (4) {2x+y−z+w=13x−2y+z−3w=4x+4y−3z+5w=−2\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=[43112−13−1202108]⇒r1−r2[13113−13−320−8108]\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

⇒r2−3r1r3−11r1[1003−10−30−31133−83496]⇒r3−3r2[1003−100−3110−834−6]\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=[21343−28−114−294−513−6]⇒r2↔r2[1234−238−141−29−5413−6]⇒r2−2r1r3−3r1r4−4r1[1000−271474−7−14−7−5142814]\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

⇒r2÷7r3−14r2r4−7r2[1000−21004−100−5200]⇒r1+2r2[100001002−100−1200]\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：{x+2z=−1y−z=2\left\{ \begin{matrix} x + 2z = - 1 \\ y - z = 2\ \ \ \ \ \end{matrix} \right.\

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

得参数形式：{x=−2c−1y=c+2z=c\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：[xyz]\begin{bmatrix} x \\ y \\ z \end{bmatrix}=[−2c−1c+2c]\begin{bmatrix} - 2c - 1 \\ c + 2 \\ c \end{bmatrix}=c[−211]\begin{bmatrix} - 2 \\ 1 \\ 1 \end{bmatrix}+[−120]\begin{bmatrix} - 1 \\ 2 \\ 0 \end{bmatrix} (c属于R)

(3) B=[242121−1−2−111−1121]⇒r2−2r1r3−r1[200100−1001−1−2100]\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

⇒r1+r2r3−2r2[200100−1000−10100]⇒r1÷2r2×(−1)[1001200−12000−101200]\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：{x+12y−12z=12w=0\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：{x=−12y+12z+12w=0\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：{x=−12c1+12c2+12y=c1z=c2w=0\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：[xyzw]\left\lbrack \begin{array}{r} \begin{matrix} x \\ y \end{matrix} \\ \begin{matrix} z \\ w \end{matrix} \end{array} \right\rbrack=[−12c1+12c2+12c1+0+00+c2+00]\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=c1[−12100]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+c2[12010]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+[12000]\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=[2311−24−11−31−3514−2]⇒r1↔r3[1324−21−31−15−31−241]\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

⇒r2−3r1r3−2r1[1004−14−7−31055−18−9−2105]⇒r3−12r2r2÷(−14)[100410−3−5705970−2−570]\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

⇒r1−4r2[100010−17−570−1797067−570]\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：{x−17z−17w=67y−57z+97w=−57\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：{x=17z+17w+67y=57z−97w−57\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：{x=17c1+17c2+67y=57c1−97c2−57z=c1w=c2\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：[xyzw]\left\lbrack \begin{array}{r} \begin{matrix} x \\ y \end{matrix} \\ \begin{matrix} z \\ w \end{matrix} \end{array} \right\rbrack=[17c1+17c2+6757c1−97c2−57c1c2]\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=c1[175710]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+c2[17−9701]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+[67−5700]\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)