返回

19 , 设 { β 1 = α 2 + α 3 + ⋯ α n β 2 = α 1 + α 3 + ⋯ α n ⋯ ⋯ ⋯ β n = α 1 + α 2 + ⋯ α n − 1 \left\{ \begin{array}{r} \begin{matrix} \beta_{1} = \alpha_{2} + \alpha_{3} + \cdots\alpha_{n} \\ \beta_{2} = \alpha_{1} + \alpha_{3} + \cdots\alpha_{n} \end{matrix} \\ \begin{matrix} \cdots\ \ \ \cdots\ \ \ \cdots \\ \beta_{n} = \alpha_{1} + \alpha_{2} + \cdots\alpha_{n - 1} \end{matrix} \end{array} \right.\

试证明向量组α1 , α2 , ⋯ , αn与向量组β1 , β2 , ⋯ , βn等价

证: 要证明向量组α1 , α2 , ⋯ , αn与向量组β1 , β2 , ⋯ , βn等价，

需证明两个向量组可以相互线性表示。

步骤一：证明向量组β1 , β2 , ⋯ , βn可由向量组α1 , α2 , ⋯ , αn线性表示

已知β1=α2 + α3 + ⋯ + αn

β2=α1 + α3 + ⋯ + αn

⋯⋯⋯⋯⋯

βn=α1 + α2 + ⋯ + αn-1

由上述表达式可知，β1 , β2 , ⋯ , βn的每一个向量

都可以直接写成α1 , α2 , ⋯ , αn的线性组合形式，

所以向量组β1 , β2 , ⋯ , βn可由向量组α1 , α2 , ⋯ , αn线性表示。

步骤二：证明向量组α1 , α2 , ⋯ , αn可由向量组β1 , β2 , ⋯ , βn线性表示

将β1 , β2 , ⋯ , βn的表达式相加可得：

β1 +β2 + ⋯ + βn=(n-1)(α1 + α2 + ⋯ + αn)

则α1 + α2 + ⋯ + αn= 1 n − 1 \frac{1}{n - 1} (β1 +β2 + ⋯ + βn)

(n≠1 , 当n=1时 , β1 =0 , α1与β1不构成等价向量组 , 这里默认n⩾2)

用α1 + α2 + ⋯ + αn= 1 n − 1 \frac{1}{n - 1} (β1 +β2 + ⋯ + βn)分别减去β1 , β2 , ⋯ , βn的表达式

α1 =(α1 + α2 + ⋯ + αn)-β1= 1 n − 1 \frac{1}{n - 1} (β1 +β2 + ⋯ + βn)-β1= − ( n − 2 ) n − 1 \frac{- (n - 2)}{n - 1} β1 + 1 n − 1 \frac{1}{n - 1} β2+⋯

α2 =(α1 + α2 + ⋯ + αn)-β2= 1 n − 1 \frac{1}{n - 1} (β1 +β2 + ⋯ + βn)-β2= 1 n − 1 \frac{1}{n - 1} β1 + − ( n − 2 ) n − 1 \frac{- (n - 2)}{n - 1} β2+⋯

⋯⋯⋯⋯⋯⋯⋯

αn =(α1 + α2 + ⋯ + αn)-βn= 1 n − 1 \frac{1}{n - 1} (β1 +β2 + ⋯ + βn)-βn= 1 n − 1 \frac{1}{n - 1} β1 + 1 n − 1 \frac{1}{n - 1} β2+⋯

由此可知，α1 , α2 , ⋯ , αn的每一个向量都可以写成β1 , β2 , ⋯ , βn的线性组合形式

所以向量组α1 , α2 , ⋯ , αn可由向量组β1 , β2 , ⋯ , βn线性表示。

步骤三：得出结论

因为向量组β1 , β2 , ⋯ , βn可由向量组α1 , α2 , ⋯ , αn线性表示，

且向量组α1 , α2 , ⋯ , αn可由向量组β1 , β2 , ⋯ , βn线性表示，

所以向量组α1 , α2 , ⋯ , αn与向量组β1 , β2 , ⋯ , βn等价。

综上，命题得证。

20 . 已知3阶矩阵A与3维列向量x满足等式A3x=3Ax-A2x

且向量组x , Ax , A2x线性无关

(1)记y=Ax , z=Ay , P=(x , y , z) , 求3阶矩阵B , 使AP=PB

(2)求|A|

解: (1)因矩阵P的列向量组线性无关 , 故P可逆 , 从而B=P-1AP，

AP=A(x , y , z)=(Ax , Ay , Az)

因Ax=y , Ay=z , Az=AAy=A3x=3Ax -A2x=3y-z

故AP =(y , z , 3y-z)=(x , y , z) [ 0 0 0 1 0 3 0 1 − 1 ] \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 3 \\ 0 & 1 & - 1 \end{bmatrix} =P [ 0 0 0 1 0 3 0 1 − 1 ] \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 3 \\ 0 & 1 & - 1 \end{bmatrix}

于是B=P-1AP= [ 0 0 0 1 0 3 0 1 − 1 ] \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 3 \\ 0 & 1 & - 1 \end{bmatrix}

(其实矩阵B就是向量组Ax , Ay , Az由向量组x , y , z线性表示的系数矩阵)

(2)由B=P-1AP , 两边取行列式 , 便有|A|=|B|=0

21 , 求下列齐次线性方程组的基础解系

(1) { x 1 − 8 x 2 + 10 x 3 + 2 x 4 = 0 2 x 1 + 4 x 2 + 5 x 3 − x 4 = 0 3 x 1 + 8 x 2 + 6 x 3 − 2 x 4 = 0 \left\{ \begin{matrix} x_{1} - 8x_{2} + 10x_{3} + 2x_{4} = 0 \\ 2x_{1} + 4x_{2} + 5x_{3} - x_{4} = 0 \\ 3x_{1} + 8x_{2} + 6x_{3} - 2x_{4} = 0 \end{matrix} \right.\ (2) { 2 x 1 − 3 x 2 − 2 x 3 + x 4 = 0 3 x 1 + 5 x 2 + 4 x 3 − 2 x 4 = 0 8 x 1 + 7 x 2 + 6 x 3 − 3 x 4 = 0 \left\{ \begin{matrix} 2x_{1} - 3x_{2} - 2x_{3} + x_{4} = 0 \\ 3x_{1} + 5x_{2} + 4x_{3} - 2x_{4} = 0 \\ 8x_{1} + 7x_{2} + 6x_{3} - 3x_{4} = 0 \end{matrix} \right.\

(3) nx1+(n-1)x2+⋯+2xn-1+xn=0

解: (1) A= [ 1 2 3 − 8 4 8 10 5 6 2 − 1 − 2 ] ⇒ r 2 − 2 r 1 r 3 − 3 r 1 [ 1 0 0 − 8 20 32 10 − 15 − 24 2 − 5 − 8 ] \left\lbrack \begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\ \ \ \ \begin{matrix} - 8 \\ 4 \\ 8 \end{matrix}\ \ \ \ \begin{matrix} 10 \\ 5 \\ 6 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ - 1 \\ - 2 \end{matrix} \right\rbrack\overset{\begin{matrix} r_{2} - 2r_{1} \\ r_{3} - 3r_{1} \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 8 \\ 20 \\ 32 \end{matrix}\ \ \ \ \begin{matrix} 10 \\ - 15 \\ - 24 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ - 5 \\ - 8 \end{matrix} \right\rbrack

⇒ r 2 ÷ ( − 5 ) r 3 + 8 r 2 [ 1 0 0 − 8 − 4 0 10 3 0 2 1 0 ] ⇒ r 1 − 2 r 2 [ 1 0 0 0 − 4 0 4 3 0 0 1 0 ] \overset{\begin{matrix} r_{2} \div ( - 5) \\ r_{3} + 8r_{2} \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 8 \\ - 4 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 10 \\ 3 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ 1 \\ 0 \end{matrix} \right\rbrack\overset{r_{1} - 2r_{2}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ - 4 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 4 \\ 3 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right\rbrack

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

得参数形式： { x 1 = − 4 x 3 x 2 = 3 4 x 3 + 1 4 x 4 \left\{ \begin{array}{r} x_{1} = - 4x_{3}\ \ \ \ \ \ \ \ \ \\ x_{2} = \frac{3}{4}x_{3} + \frac{1}{4}x_{4} \end{array} \right.\ (x3 , x4可任意取值)

得参数形式： { x 1 = − 4 c 1 x 2 = 3 4 c 1 + 1 4 c 2 x 3 = c 1 x 4 = c 2 \left\{ \begin{array}{r} \begin{matrix} x_{1} = - 4c_{1}\ \ \ \ \ \ \ \ \ \ \ \ \\ x_{2} = \frac{3}{4}c_{1} + \frac{1}{4}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 )

得具体通解： [ x 1 x 2 x 3 x 4 ] \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 = [ − 4 c 1 3 4 c 1 + 1 4 c 2 c 1 c 2 ] \left\lbrack \begin{array}{r} \begin{matrix} - 4c_{1} \\ \frac{3}{4}c_{1} + \frac{1}{4}c_{2} \end{matrix} \\ \begin{matrix} c_{1} \\ \ c_{2} \end{matrix} \end{array} \right\rbrack = c 1 [ − 4 3 4 1 0 ] c_{1}\left\lbrack \begin{array}{r} \begin{matrix} - 4 \\ \frac{3}{4} \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack + c 2 [ 0 1 4 0 1 ] c_{2}\left\lbrack \begin{array}{r} \begin{matrix} 0 \\ \frac{1}{4} \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack (c1 , c2为任意实数)

得基础解系 ξ1= [ − 4 3 4 1 0 ] \left\lbrack \begin{array}{r} \begin{matrix} - 4 \\ \frac{3}{4} \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack , ξ2= [ 0 1 4 0 1 ] \left\lbrack \begin{array}{r} \begin{matrix} 0 \\ \frac{1}{4} \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack

(2) A= [ 2 3 8 − 3 5 7 − 2 4 6 1 − 2 − 3 ] ⇒ 1 2 r 1 [ 1 3 8 − 3 2 5 7 − 1 4 6 1 2 − 2 − 3 ] ⇒ r 2 + ( − 3 ) r 1 r 3 + ( − 8 ) r 1 [ 1 0 0 − 3 2 19 2 19 − 1 7 14 1 2 − 7 2 − 7 ] \left\lbrack \begin{matrix} 2 \\ 3 \\ 8 \end{matrix}\ \ \ \ \begin{matrix} - 3 \\ 5 \\ 7 \end{matrix}\ \ \ \ \begin{matrix} - 2 \\ 4 \\ 6 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ - 2 \\ - 3 \end{matrix} \right\rbrack\overset{\frac{1}{2}r_{1}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 3 \\ 8 \end{matrix}\ \ \ \ \begin{matrix} \frac{- 3}{2} \\ 5 \\ 7 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 4 \\ 6 \end{matrix}\ \ \ \ \begin{matrix} \frac{1}{2} \\ - 2 \\ - 3 \end{matrix} \right\rbrack\overset{\begin{matrix} r_{2} + ( - 3)r_{1} \\ r_{3} + ( - 8)r_{1} \end{matrix}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{- 3}{2} \\ \frac{19}{2} \\ 19 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 7 \\ 14 \end{matrix}\ \ \ \ \begin{matrix} \frac{1}{2} \\ \frac{- 7}{2} \\ - 7 \end{matrix} \right\rbrack

⇒ 2 r 2 [ 1 0 0 − 3 2 19 19 − 1 14 14 1 2 − 7 − 7 ] ⇒ r 3 + ( − 19 ) r 2 [ 1 0 0 − 3 2 19 0 − 1 14 0 1 2 − 7 0 ] ⇒ 1 2 r 2 [ 1 0 0 − 3 2 1 0 − 1 14 19 0 1 2 − 7 19 0 ] \overset{2r_{2}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{- 3}{2} \\ 19 \\ 19 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 14 \\ 14 \end{matrix}\ \ \ \ \begin{matrix} \frac{1}{2} \\ - 7 \\ - 7 \end{matrix} \right\rbrack\overset{r_{3} + ( - 19)r_{2}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{- 3}{2} \\ 19 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ 14 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{1}{2} \\ - 7 \\ 0 \end{matrix} \right\rbrack\overset{\frac{1}{2}r_{2}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{- 3}{2} \\ 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} - 1 \\ \frac{14}{19} \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{1}{2} \\ \frac{- 7}{19} \\ 0 \end{matrix} \right\rbrack

⇒ r 1 + 3 2 r 2 [ 1 0 0 0 1 0 2 19 14 19 0 − 1 19 − 7 19 0 ] \overset{r_{1} + \frac{3}{2}r_{2}}{\Rightarrow}\left\lbrack \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{2}{19} \\ \frac{14}{19} \\ 0 \end{matrix}\ \ \ \ \begin{matrix} \frac{- 1}{19} \\ \frac{- 7}{19} \\ 0 \end{matrix} \right\rbrack

得同解方程组： { x 1 + 2 19 x 3 − 1 19 x 4 = 0 x 2 + 14 19 x 3 − 7 19 x 4 = 0 \left\{ \begin{matrix} x_{1} + \frac{2}{19}x_{3} - \frac{1}{19}x_{4} = 0\ \ \ \ \ \ \ \ \ \ \\ x_{2} + \frac{14}{19}x_{3} - \frac{7}{19}x_{4} = 0\ \ \ \ \ \ \ \ \ \ \end{matrix} \right.\

得参数形式： { x 1 = − 2 19 x 3 + 1 19 x 4 x 2 = − 14 19 x 3 + 7 19 x 4 \ \left\{ \begin{matrix} x_{1} = \frac{- 2}{19}x_{3} + \frac{1}{19}x_{4}\ \ \ \ \ \ \ \ \ \ \\ x_{2} = \frac{- 14}{19}x_{3} + \frac{7}{19}x_{4}\ \ \ \ \ \ \ \ \ \ \end{matrix} \right.\ (x3 , x4可任意取值)

得参数形式： { x 1 = − 2 19 c 1 + 1 19 c 2 x 2 = − 14 19 c 1 + 7 19 c 2 x 3 = c 1 x 4 = c 2 \left\{ \begin{array}{r} \begin{matrix} x_{1} = \frac{- 2}{19}c_{1} + \frac{1}{19}c_{2} \\ x_{2} = \frac{- 14}{19}c_{1} + \frac{7}{19}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 )

得具体通解： [ x 1 x 2 x 3 x 4 ] \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 = [ − 2 19 c 1 + 1 19 c 2 − 14 19 c 1 + 7 19 c 2 c 1 c 2 ] \left\lbrack \begin{array}{r} \begin{matrix} \frac{- 2}{19}c_{1} + \frac{1}{19}c_{2} \\ \frac{- 14}{19}c_{1} + \frac{7}{19}c_{2} \end{matrix} \\ \begin{matrix} c_{1} \\ \ c_{2} \end{matrix} \end{array} \right\rbrack = c 1 [ − 2 19 − 14 19 1 0 ] c_{1}\left\lbrack \begin{array}{r} \begin{matrix} \frac{- 2}{19} \\ \frac{- 14}{19} \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack + c 2 [ 1 19 7 19 0 1 ] c_{2}\left\lbrack \begin{array}{r} \begin{matrix} \frac{1}{19} \\ \frac{7}{19} \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack (c1 , c2为任意实数)

得基础解系 ξ1= [ − 2 19 − 14 19 1 0 ] \left\lbrack \begin{array}{r} \begin{matrix} \frac{- 2}{19} \\ \frac{- 14}{19} \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack , ξ2= [ 1 19 7 19 0 1 ] \left\lbrack \begin{array}{r} \begin{matrix} \frac{1}{19} \\ \frac{7}{19} \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack

(3)

步骤1：选择自由变量

选择 x 1 , x 2 , … , x n − 1 x_{1},x_{2},\ldots,x_{n - 1} 为自由变量，用它们表示 x n x_{n} ：

x n = − n x 1 − ( n − 1 ) x 2 − ⋯ − 2 x n − 1 x_{n} = - nx_{1} - (n - 1)x_{2} - \cdots - 2x_{n - 1}

步骤2：引入参数

令： x 1 = c 1 , x 2 = c 2 , … , x n − 1 = c n − 1 x_{1} = c_{1},\quad x_{2} = c_{2},\quad\ldots,\quad x_{n - 1} = c_{n - 1}

则： x n = − n c 1 − ( n − 1 ) c 2 − ⋯ − 2 c n − 1 x_{n} = - nc_{1} - (n - 1)c_{2} - \cdots - 2c_{n - 1}

步骤3：写出参数形式通解

[ x 1 x 2 ⋮ x n − 1 x n ] = [ c 1 c 2 ⋮ c n − 1 − n c 1 − ( n − 1 ) c 2 − ⋯ − 2 c n − 1 ] = c 1 [ 1 0 ⋮ 0 − n ] + c 2 [ 0 1 ⋮ 0 − ( n − 1 ) ] + ⋯ + c n − 1 [ 0 0 ⋮ 1 − 2 ] \begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n - 1} \\ x_{n} \end{bmatrix} = \begin{bmatrix} c_{1} \\ c_{2} \\ \vdots \\ c_{n - 1} \\ - nc_{1} - (n - 1)c_{2} - \cdots - 2c_{n - 1} \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \\ - n \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 1 \\ \vdots \\ 0 \\ - (n - 1) \end{bmatrix} + \cdots + c_{n - 1}\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ - 2 \end{bmatrix}

步骤4：写出基础解系

ξ 1 = [ 1 0 ⋮ 0 − n ] , ξ 2 = [ 0 1 ⋮ 0 − ( n − 1 ) ] , … , ξ n − 1 = [ 0 0 ⋮ 1 − 2 ] \xi_{1} = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \\ - n \end{bmatrix},\quad\xi_{2} = \begin{bmatrix} 0 \\ 1 \\ \vdots \\ 0 \\ - (n - 1) \end{bmatrix},\quad\ldots,\quad\xi_{n - 1} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ - 2 \end{bmatrix}

求方程组4x1+3x2+2x3+x4=0的基础解系

得参数形式：x4=-4x1-3x2-2x3

得参数形式： { x 1 = c 1 x 2 = c 2 x 3 = c 3 x 4 = − 4 c 1 − 3 c 2 − 2 c 3 \left\{ \begin{array}{r} \begin{matrix} x_{1} = c_{1} \\ x_{2} = c_{2} \end{matrix} \\ \begin{matrix} x_{3} = c_{3}\ \\ x_{4} = - 4c_{1} - 3c_{2} - 2c_{3}\ \ \end{matrix} \end{array} \right.\ (其中x1=c1 , x2=c2 , x3=c3 )

得齐次通解： [ x 1 x 2 x 3 x 4 ] \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 = [ c 1 c 2 c 3 − 4 c 1 − 3 c 2 − 2 c 3 ] \left\lbrack \begin{array}{r} \begin{matrix} c_{1} \\ c_{2} \end{matrix} \\ \begin{matrix} c_{3} \\ \ - 4c_{1} - 3c_{2} - 2c_{3} \end{matrix} \end{array} \right\rbrack = c 1 [ 1 0 0 − 4 ] + c 2 [ 0 1 0 − 3 ] + c 3 [ 0 0 1 − 2 ] c_{1}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ - 4 \end{matrix} \end{array} \right\rbrack + c_{2}\left\lbrack \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ - 3 \end{matrix} \end{array} \right\rbrack + c_{3}\left\lbrack \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 1 \\ - 2 \end{matrix} \end{array} \right\rbrack

(c1 , c2, c3为任意实数)

得基础解系 ξ1= [ 1 0 0 − 4 ] \left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ - 4 \end{matrix} \end{array} \right\rbrack , ξ2= [ 0 1 0 − 3 ] \left\lbrack \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ - 3 \end{matrix} \end{array} \right\rbrack , ξ3= [ 0 0 1 − 2 ] \left\lbrack \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 1 \\ - 2 \end{matrix} \end{array} \right\rbrack