AI笔记: 数学基础之齐次与非齐次线性方程组解的结构定理

<>对称矩阵

* 元素以对角线为对称轴对应相等的矩阵就叫做对称矩阵
* 对称矩阵具有的特性：
* 对称矩阵中 a i j = a j i a_{ij} = a_{ji} aij​=aji​
* 对称矩阵一定是方阵, 并且对于任何的方阵A, A + A T A + A^T A+AT是对称矩阵
* 除对角线外的其他元素均为0的矩阵叫做对角矩阵
* 矩阵中的每个元素都是实数的对称矩阵叫做实对称矩阵
* A = { a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ ⋯ a n 1 a n 2 ⋯ a n n } A
=\left \{\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1n} \\a_{21} & a_{22}
& \cdots & a_{2n} \\\cdots & \cdots & \cdots & \cdots \\a_{n1} & a_{n2} &
\cdots & a_{nn}\end{array} \right \}A=⎩⎪⎪⎨⎪⎪⎧​a11​a21​⋯an1​​a12​a22​⋯an2​​⋯⋯⋯⋯​a
1n​a2n​⋯ann​​⎭⎪⎪⎬⎪⎪⎫​
* a 12 = a 21 a i j = a j i a 1 n = a n 1 a_{12} = a_{21} \\ a_{ij} = a_{ji}
\\ a_{1n} = a_{n1}a12​=a21​aij​=aji​a1n​=an1​
<>线性方程组

* 设有n个未知数m个方程的线性方程组 { a 11 x 1 + a 12 x 2 + . . . + a 1 n x n = b 1 a 21 x 1
+ a 22 x 2 + . . . + a 2 n x n = b 1 ⋯ a m 1 x 1 + a m 2 x 2 + . . . + a m n x
n = b m \left \{\begin{array}{cccc}a_{11}x_1 + a_{12}x_2 + ... + a_{1n}x_n =
b_1 \\a_{21}x_1 + a_{22}x_2 + ... + a_{2n}x_n = b_1 \\\cdots \\a_{m1}x_1 +
a_{m2}x_2 + ... + a_{mn}x_n = b_m\end{array} \right.⎩⎪⎪⎨⎪⎪⎧​a11​x1​+a12​x2​+...+
a1n​xn​=b1​a21​x1​+a22​x2​+...+a2n​xn​=b1​⋯am1​x1​+am2​x2​+...+amn​xn​=bm​​
* 可以写成以向量x为未知元的向量方程 A x = b Ax = b Ax=b
* 可将上述线性方程组和向量方程混同使用
定理1

* n元齐次线性方程组 A x = 0 Ax = 0 Ax=0 有非零解的充要条件是 R ( A ) < n R(A) < n R(A)<n
* 推论：当m < n时，齐次线性方程组 A m × n x = 0 A_{m×n} x = 0 Am×n​x=0 一定有非零解
定理2

* 对于n元线性方程组 A x = b Ax=b Ax=b
* 无解的充要条件是 R ( A ) < R ( A , b ) ; R(A) < R(A, b); R(A)<R(A,b);
* 有唯一解的充要条件是 R ( A ) = R ( A , b ) = n R(A) = R(A,b) = n R(A)=R(A,b)=n
* 有无穷多解的充要条件是 R ( A ) = R ( A , b ) < n R(A) = R(A,b) < n R(A)=R(A,b)<n
求解线性方程组的步骤

* (1) 对于非齐次线性方程组，把它的增广矩阵B化成行阶梯形，从中可同时看出 R ( A ) R(A) R(A)和 R ( B ) R(B) R(B).
若 R ( B ) < R ( B ) R(B) < R(B) R(B)<R(B), 则方程组无解.
* (2) 若 R ( A ) = R ( B ) R(A) = R(B) R(A)=R(B), 则进一步把B化成行最简形.
而对于齐次线性方程组，则把系数矩阵A化成行最简形.
* (3) 设 R ( A ) = R ( B ) = r R(A) = R(B) = r R(A)=R(B)=r,
把行最简形中r个非零行的非零首元所对应的未知量取作非自由未知量, 其余n-r个未知量取作自由未知量, 并令自由未知量分别等于 c 1 , c 2 , . .
. , c n − r c_1, c_2, ..., c_{n-r}c1​,c2​,...,cn−r​, 由B(或A)的行最简形，即可写出含n-r个参数的通解
<>齐次方程组解的结构定理

* 齐次方程组 A m × n X = 0 A_{m×n}X = 0 Am×n​X=0的基础解系所含向量个数为 n − r     ( r = R (
A ) ) n-r \ \ \ (r=R(A))n−r   (r=R(A)) 设一个基础解系为： ξ 1 , ξ 2 , ⋯   , ξ n − r
\xi_1, \xi_2, \cdots, \xi_{n-r}ξ1​,ξ2​,⋯,ξn−r​, 则通解为： x = k 1 ξ 1 + k 2 ξ 2 + .
. . + k n − r ξ n − r     ( k i ∈ R ) x = k_1 \xi_1 + k_2\xi_2 + ... + k_{n-r}
\xi_{n-r} \ \ \ (k_i \in R)x=k1​ξ1​+k2​ξ2​+...+kn−r​ξn−r​   (ki​∈R)
例1

*
求 { x 1 + x 2 − x 3 − x 4 = 0 2 x 1 − 5 x 2 + 3 x 3 + 2 x 4 = 0 7 x 1 − 7 x 2
+ 3 x 3 + x 4 = 0 \left \{\begin{array}{cccc}x_1 + x_2 - x_3 - x_4 = 0 \\2x_1 -
5x_2 + 3x_3 + 2x_4 = 0 \\7x_1 - 7x_2 + 3x_3 + x_4 = 0\end{array} \right.⎩⎨⎧​x1​+
x2​−x3​−x4​=02x1​−5x2​+3x3​+2x4​=07x1​−7x2​+3x3​+x4​=0​ 基础解系和通解

*
分析

* 对系数矩阵A作初等行变换，变为行最简形矩阵， 有
* A = ( 1 1 − 1 − 1 2 − 5 3 2 7 − 7 3 1 ) ∼ ( 1 0 − 2 7 − 3 7 0 1 − 5 7 − 4
7 0 0 0 0 ) A = \left (\begin{array}{cccc} 1 & 1 & -1 & -1 \\ 2 & -5 & 3 & 2 \\
7 & -7 & 3 & 1 \end{array} \right ) \sim \left ( \begin{array}{cccc} 1 & 0 &
-\frac{2}{7} & -\frac{3}{7} \\ 0 & 1 & -\frac{5}{7} & -\frac{4}{7} \\ 0 & 0 & 0
& 0 \end{array} \right )A=⎝⎛​127​1−5−7​−133​−121​⎠⎞​∼⎝⎛​100​010​−72​−75​0​−73​−7
4​0​⎠⎞​
* 化为： { x 1 = 2 7 x 3 + 3 7 x 4 x 2 = 5 7 x 3 + 4 7 x 4 \left
\{\begin{array}{cccc}x_1 = \frac{2}{7} x_3 + \frac{3}{7} x_4 \\x_2 =
\frac{5}{7} x_3 + \frac{4}{7} x_4\end{array} \right.{x1​=72​x3​+73​x4​x2​=75​x3​
+74​x4​​
* 令 ( x 3 x 4 ) = ( 1 0 ) , ( 0 1 ) \left (\begin{array}{cccc}x_3
\\x_4\end{array} \right ) = \left (\begin{array}{cccc}1 \\0\end{array} \right
),\left (\begin{array}{cccc}0 \\1\end{array} \right )(x3​x4​​)=(10​),(01​) 则 (
x 1 x 2 ) = ( 2 7 5 7 ) , ( 3 7 4 7 ) \left (\begin{array}{cccc}x_1
\\x_2\end{array} \right ) = \left (\begin{array}{cccc}\frac{2}{7}
\\\frac{5}{7}\end{array} \right ),\left (\begin{array}{cccc}\frac{3}{7}
\\\frac{4}{7}\end{array} \right )(x1​x2​​)=(72​75​​),(73​74​​)，合起来得到基础解系
* 基础解系为： ξ 1 = ( 2 7 5 7 1 0 ) , ξ 2 = ( 3 7 4 7 0 1 ) \xi_1 =\left
(\begin{array}{cccc}\frac{2}{7} \\\frac{5}{7} \\1 \\0\end{array} \right ),\xi_2
=\left (\begin{array}{cccc}\frac{3}{7} \\\frac{4}{7} \\0 \\1\end{array} \right )
ξ1​=⎝⎜⎜⎛​72​75​10​⎠⎟⎟⎞​,ξ2​=⎝⎜⎜⎛​73​74​01​⎠⎟⎟⎞​
* 通解为： A = ( x 1 x 2 x 3 x 4 ) = c 1 ( 2 7 5 7 1 0 ) + c 2 ( 3 7 4 7 0 1 ) ,
    ( c 1 , c 2 ∈ R ) A =\left (\begin{array}{cccc}x_1 \\x_2 \\x_3
\\x_4\end{array} \right ) = c_1\left (\begin{array}{cccc}\frac{2}{7}
\\\frac{5}{7} \\1 \\0\end{array} \right ) + c_2\left
(\begin{array}{cccc}\frac{3}{7} \\\frac{4}{7} \\0 \\1\end{array} \right ), \ \
\ (c_1, c_2 \in R)A=⎝⎜⎜⎛​x1​x2​x3​x4​​⎠⎟⎟⎞​=c1​⎝⎜⎜⎛​72​75​10​⎠⎟⎟⎞​+c2​⎝⎜⎜⎛​73​74
​01​⎠⎟⎟⎞​,   (c1​,c2​∈R)
*
设 η ∗ \eta^* η∗是非齐次方程组 A m ✖ × n X = b A_{m✖×n} X = b Am✖×n​X=b
的一特解，则当非齐次线性方程组有无穷多解时其通解为： x = k 1 ξ 1 + k 2 ξ 2 + ⋯ + k n − r ξ n − r + η ∗ (
k i ∈ R ) x = k_1\xi_1 + k_2\xi_2 + \cdots + k_{n-r}\xi_{n-r} + \eta^* (k_i \in
R)x=k1​ξ1​+k2​ξ2​+⋯+kn−r​ξn−r​+η∗(ki​∈R)

*
其中 k 1 ξ 1 + ⋯ + k n − r ξ n − r k_1\xi_1 + \cdots + k_{n-r} \xi_{n-r} k1​ξ1​+⋯
+kn−r​ξn−r​为对应齐次线性方程组的通解

例2

* 求解方程组 { x 1 − x 2 − x 3 + x 4 = 0 , x 1 − x 2 + x 3 − 3 x 4 = 1 , x 1 − x
2 − 2 x 3 + 3 x 4 = − 1 2 , \left \{\begin{array}{cccc}x_1 - x_2 - x_3 + x_4 =
0,x_1 - x_2 + x_3 - 3x_4 = 1,x_1 - x_2 - 2x_3 + 3x_4 = -
\frac{1}{2},\end{array} \right.{x1​−x2​−x3​+x4​=0,x1​−x2​+x3​−3x4​=1,x1​−x2​−2x3
​+3x4​=−21​,​
* 分析
* B = ( 1 − 1 − 1 1 1 − 1 1 − 3 1 − 1 − 2 3 ∣ 0 1 − 1 2 ) → ( 1 − 1 0 − 1 0
0 1 − 2 0 0 0 0 ∣ 1 2 1 2 0 ) B = \left (\begin{array}{cccc}1 & -1 & -1 & 1 \\1
& -1 & 1 & -3 \\1 & -1 & -2 & 3\end{array} \right |\left.\begin{array}{cccc}0
\\1 \\-\frac{1}{2}\end{array} \right ) \to\left (\begin{array}{cccc}1 & -1 & 0
& -1 \\0 & 0 & 1 & -2 \\0 & 0 & 0 & 0\end{array} \right
|\left.\begin{array}{cccc}\frac{1}{2} \\\frac{1}{2} \\0\end{array} \right )B=⎝⎛​
111​−1−1−1​−11−2​1−33​∣∣∣∣∣∣​01−21​​⎠⎞​→⎝⎛​100​−100​010​−1−20​∣∣∣∣∣∣​21​21​0​⎠⎞​
* 可见 R ( A ) = R ( B ) = 2 < 4 R(A) = R(B) = 2 < 4 R(A)=R(B)=2<4, 故方程组有无穷多解
* { x 1 = x 2 + x 4 + 1 2 x 3 = 2 x 4 + 1 2 ⇒ { x 1 = x 2 + x 4 + 1 2 x 2 =
x 2 x 3 = 2 x 4 + 1 2 x 4 = x 4 ⇒ ( x 1 x 2 x 3 x 4 ) = C 1 ( 1 1 0 0 ) + C 2 (
1 0 2 1 ) + ( 1 2 0 1 2 0 )     ( C 1 , C 2 ∈ R ) . \left
\{\begin{array}{cccc}x_1 = x_2 + x_4 + \frac{1}{2} \\x_3 = 2x_4 +
\frac{1}{2}\end{array} \right. \Rightarrow\left \{\begin{array}{cccc}x_1 = x_2
+ x_4 + \frac{1}{2} \\x_2 = x_2 \\x_3 = 2x_4 + \frac{1}{2} \\x_4 =
x_4\end{array} \right. \Rightarrow \left ( \begin{array}{cccc} x_1 \\ x_2 \\
x_3 \\ x_4 \end{array} \right ) = C_1 \left ( \begin{array}{cccc} 1 \\ 1 \\ 0
\\ 0 \end{array} \right ) + C_2 \left ( \begin{array}{cccc} 1 \\ 0 \\ 2 \\ 1
\end{array} \right ) + \left ( \begin{array}{cccc} \frac{1}{2} \\ 0 \\
\frac{1}{2} \\ 0 \end{array} \right ) \ \ \ (C_1, C_2 \in R).{x1​=x2​+x4​+21​x3​
=2x4​+21​​⇒⎩⎪⎪⎨⎪⎪⎧​x1​=x2​+x4​+21​x2​=x2​x3​=2x4​+21​x4​=x4​​⇒⎝⎜⎜⎛​x1​x2​x3​x4​​
⎠⎟⎟⎞​=C1​⎝⎜⎜⎛​1100​⎠⎟⎟⎞​+C2​⎝⎜⎜⎛​1021​⎠⎟⎟⎞​+⎝⎜⎜⎛​21​021​0​⎠⎟⎟⎞​   (C1​,C2​∈R).
* 在 { x 1 = x 2 + x 4 + 1 2 x 3 = 2 x 4 + 1 2 \left \{\begin{array}{cccc}x_1
= x_2 + x_4 + \frac{1}{2} \\x_3 = 2x_4 + \frac{1}{2}\end{array} \right.{x1​=x2​+
x4​+21​x3​=2x4​+21​​
* 取 x 2 = x 4 = 0 x_2 = x_4 = 0 x2​=x4​=0, 则 x 1 = x 3 = 1 2 x_1 = x_3 =
\frac{1}{2}x1​=x3​=21​, 即得方程组的一个解 η ∗ = ( 1 2 0 1 2 0 ) \eta^* =\left
(\begin{array}{cccc}\frac{1}{2} \\ 0 \\ \frac{1}{2} \\ 0\end{array} \right )η∗=⎝
⎜⎜⎛​21​021​0​⎠⎟⎟⎞​
* 在对应的齐次线性方程组 { x 1 = x 2 + x 4 , x 3 = 2 x 4 \left \{
\begin{array}{cccc}x_1 = x_2 + x_4, \\ x_3 = 2x_4\end{array} \right.{x1​=x2​+x4​
,x3​=2x4​​ 中取
* ( x 2 x 4 ) = ( 1 0 ) \left (\begin{array}{cccc}x_2 \\x_4\end{array}
\right ) = \left (\begin{array}{cccc}1 \\0\end{array} \right )(x2​x4​​)=(10​) 及
( 0 1 ) \left (\begin{array}{cccc}0 \\1\end{array} \right )(01​)，则 ( x 1 x 3 )
= ( 1 0 ) \left (\begin{array}{cccc}x_1 \\x_3\end{array} \right ) =\left
(\begin{array}{cccc}1 \\0\end{array} \right )(x1​x3​​)=(10​) 及 ( 1 2 ) \left
(\begin{array}{cccc}1 \\2\end{array} \right )(12​)
* 即得对应的齐次线性方程组的基础解系 ξ 1 = ( 1 1 0 0 ) , ξ 2 = ( 1 0 2 1 ) \xi_1 =\left
(\begin{array}{cccc}1 \\1 \\0 \\0 \end{array} \right ), \xi_2 = \left (
\begin{array}{cccc}1 \\0 \\2 \\1\end{array} \right )ξ1​=⎝⎜⎜⎛​1100​⎠⎟⎟⎞​,ξ2​=⎝⎜⎜⎛
​1021​⎠⎟⎟⎞​
* 于是所求通解为： ( x 1 x 2 x 3 x 4 ) = c 1 ( 1 1 0 0 ) + c 2 ( 1 0 2 1 ) + ( 1 2 0
1 2 0 ) ,     ( c 1 , c 2 ∈ R ) . \left (\begin{array}{cccc}x_1 \\x_2 \\x_3
\\x_4\end{array} \right ) = c_1 \left (\begin{array}{cccc}1 \\1 \\0
\\0\end{array} \right ) + c_2 \left (\begin{array}{cccc}1 \\0 \\2
\\1\end{array} \right ) + \left (\begin{array}{cccc}\frac{1}{2} \\0
\\\frac{1}{2} \\0\end{array} \right ), \ \ \ (c_1, c_2 \in R).⎝⎜⎜⎛​x1​x2​x3​x4​​
⎠⎟⎟⎞​=c1​⎝⎜⎜⎛​1100​⎠⎟⎟⎞​+c2​⎝⎜⎜⎛​1021​⎠⎟⎟⎞​+⎝⎜⎜⎛​21​021​0​⎠⎟⎟⎞​,   (c1​,c2​∈R).
例3

* 求解齐次线性方程组 { x 1 + 2 x 2 + 2 x 3 + x 4 = 0 2 x 1 + x 2 − 2 x 3 − 2 x 4 = 0
x 1 − x 2 − 4 x 3 − 3 x 4 = 0 \left \{\begin{array}{cccc}x_1 + 2x_2 + 2x_3 +
x_4 = 0 \\2x_1 + x_2 - 2x_3 - 2x_4 = 0 \\x_1 - x_2 - 4x_3 - 3x_4 = 0\end{array}
\right.⎩⎨⎧​x1​+2x2​+2x3​+x4​=02x1​+x2​−2x3​−2x4​=0x1​−x2​−4x3​−3x4​=0​
* 分析：
* 对系数矩阵 A = ( 1 2 2 1 2 1 − 2 − 2 1 − 1 − 4 − 3 ) A =\left
(\begin{array}{cccc}1 & 2 & 2 & 1 \\2 & 1 & -2 & -2 \\1 & -1 & -4 & -3
\\\end{array} \right )A=⎝⎛​121​21−1​2−2−4​1−2−3​⎠⎞​施行初等行变换化为最简阶梯形
* r 2 − 2 r 1 , r 3 − r 1 r_2 - 2r_1, r_3 - r_1 r2​−2r1​,r3​−r1​
* ( 1 2 2 1 0 − 3 − 6 − 4 0 − 3 − 6 − 4 ) \left (\begin{array}{cccc}1 & 2 &
2 & 1 \\0 & -3 & -6 & -4 \\0 & -3 & -6 & -4\end{array} \right )⎝⎛​100​2−3−3​2−6−
6​1−4−4​⎠⎞​
* r 3 − r 2 r_3 - r_2 r3​−r2​
* ( 1 2 2 1 0 − 3 − 6 − 4 0 0 0 0 ) \left (\begin{array}{cccc}1 & 2 & 2 & 1
\\0 & -3 & -6 & -4 \\0 & 0 & 0 & 0\end{array} \right )⎝⎛​100​2−30​2−60​1−40​⎠⎞​
* − 1 3 r 2 - \frac{1}{3} r_2 −31​r2​
* ( 1 2 2 1 0 1 2 4 3 0 0 0 0 ) \left (\begin{array}{cccc}1 & 2 & 2 & 1 \\0
& 1 & 2 & \frac{4}{3} \\0 & 0 & 0 & 0\end{array} \right )⎝⎛​100​210​220​134​0​⎠⎞
​
* r 1 − 2 r 2 r_1 - 2r_2 r1​−2r2​
* ( 1 0 − 2 − 5 3 0 1 2 − 4 3 0 0 0 0 ) \left (\begin{array}{cccc}1 & 0 & -2
& -\frac{5}{3} \\0 & 1 & 2 & -\frac{4}{3} \\0 & 0 & 0 & 0\end{array} \right )⎝⎛​
100​010​−220​−35​−34​0​⎠⎞​
* 等价式: { x 1 = 2 x 3 + 5 3 x 4 x 2 = − 2 x 3 + 4 3 x 4 \left
\{\begin{array}{cccc}x_1 = 2x_3 + \frac{5}{3}x_4 \\x_ 2= -2x_3 +
\frac{4}{3}x_4\end{array}\right.{x1​=2x3​+35​x4​x2​=−2x3​+34​x4​​
* 令： x 3 = c 1 , x 4 = c 2 x_3 = c_1, x_4 = c_2 x3​=c1​,x4​=c2​
* 写出参数形式的通解，再改写为向量形式
* 通解： ( x 1 = 2 c 1 + 5 3 c 2 x 2 = − 2 c 1 − 4 3 c 2 x 3 = c 1 x 4 = c 2 )
\left (\begin{array}{cccc}x_1 = 2c_1 + \frac{5}{3}c_2 \\x_2 = -2c_1 -
\frac{4}{3}c_2 \\x_3 = c_1 \\x_4 = c_2\end{array} \right )⎝⎜⎜⎛​x1​=2c1​+35​c2​x2
​=−2c1​−34​c2​x3​=c1​x4​=c2​​⎠⎟⎟⎞​
* 即： ( x 1 x 2 x 3 x 4 ) = c 1 ( 2 − 2 1 0 ) + c 2 ( 5 3 − 4 3 0 1 ) \left
(\begin{array}{cccc}x_1 \\x_2 \\x_3 \\x_4\end{array} \right ) = c_1\left
(\begin{array}{cccc}2 \\-2 \\1 \\0\end{array} \right ) + c_2\left
(\begin{array}{cccc}\frac{5}{3} \\- \frac{4}{3} \\ 0 \\ 1 \end{array} \right )⎝⎜
⎜⎛​x1​x2​x3​x4​​⎠⎟⎟⎞​=c1​⎝⎜⎜⎛​2−210​⎠⎟⎟⎞​+c2​⎝⎜⎜⎛​35​−34​01​⎠⎟⎟⎞​, 其中 c 1 , c
2 c_1, c_2c1​,c2​为任意实数
例4

* 求解非齐次线性方程组 { x 1 − 2 x 2 + 3 x 3 − x 4 = 1 3 x 1 − x 2 + 5 x 3 − 3 x 4 = 2
2 x 1 + x 2 + 2 x 3 − 2 x 4 = 3 \left \{\begin{array}{cccc}x_1 - 2x_2 + 3x_3 -
x_4 = 1 \\3x_1 - x_2 + 5x_3 - 3x_4 = 2 \\2x_1 + x_2 + 2x_3 - 2x_4 =
3\end{array} \right.⎩⎨⎧​x1​−2x2​+3x3​−x4​=13x1​−x2​+5x3​−3x4​=22x1​+x2​+2x3​−2x4
​=3​
* 分析：
* 对增广矩阵只用行变换化阶梯形
* B = [ 1 − 2 3 − 1 3 − 1 5 − 3 2 1 2 − 2 ∣ 1 2 3 ] → r [ 1 − 2 3 − 1 0 5 −
4 0 0 0 0 0 ∣ 1 − 1 2 ] B =\left [\begin{array}{cccc}1 & -2 & 3 & -1 \\3 & -1 &
5 & -3 \\2 & 1 & 2 & -2\end{array} \right |\left.\begin{array}{cccc}1 \\2
\\3\end{array} \right ] \overset{\text{r}}{\to}\left [\begin{array}{cccc}1 & -2
& 3 & -1 \\0 & 5 & -4 & 0 \\ 0 & 0 & 0 & 0\end{array} \right |
\left.\begin{array}{cccc} 1 \\ -1 \\2 \end{array} \right ]B=⎣⎡​132​−2−11​352​−1−
3−2​∣∣∣∣∣∣​123​⎦⎤​→r⎣⎡​100​−250​3−40​−100​∣∣∣∣∣∣​1−12​⎦⎤​
* 最后一行对应的方程是 0 = 2 0 = 2 0=2, 所以无解
例5

* 解方程组 { x 1 + 2 x 2 + x 4 = 3 x 1 + 2 x 2 + x 3 − 3 x 4 = 8 2 x 1 + 4 x 2 +
2 x 4 = 6 x 1 + 2 x 2 − x 3 + 5 x 4 = − 2 \left \{\begin{array}{cccc}x_1 + 2x_2
+ x_4 = 3 \\x_1 + 2x_2 + x_3 - 3x_4 = 8 \\2x_1 + 4x_2 + 2x_4 = 6 \\x_1 + 2x_2 -
x_3 + 5x_4 = -2\end{array} \right.⎩⎪⎪⎨⎪⎪⎧​x1​+2x2​+x4​=3x1​+2x2​+x3​−3x4​=82x1​+
4x2​+2x4​=6x1​+2x2​−x3​+5x4​=−2​
* 分析
* 第一步：把增广矩阵用行变换化阶梯形，如果 R ( A ) ≠ = R ( B ) R(A) \neq = R(B) R(A)​==R(B), 则无解
* 如果 R ( A ) = R ( B ) R(A) = R(B) R(A)=R(B), 则继续化为最简阶梯形
* B = [ 1 2 0 1 1 2 1 − 3 2 4 0 2 1 2 − 1 5 ∣ 3 8 6 − 2 ] → r [ 1 2 0 1 0 0
1 − 4 0 0 0 0 0 0 0 0 ∣ 3 5 0 0 ] B =\left [\begin{array}{cccc}1 & 2 & 0 & 1
\\1 & 2 & 1 & -3 \\2 & 4 & 0 & 2 \\1 & 2 & -1 & 5 \\\end{array} \right
|\left.\begin{array}{cccc}3 \\8 \\6 \\-2\end{array} \right ]
\overset{\text{r}}{\to} \left [ \begin{array}{cccc} 1 & 2 & 0 & 1 \\ 0 & 0 & 1
& -4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right | \left.
\begin{array}{cccc} 3 \\ 5 \\ 0 \\ 0 \end{array} \right ]B=⎣⎢⎢⎡​1121​2242​010−1​
1−325​∣∣∣∣∣∣∣∣​386−2​⎦⎥⎥⎤​→r⎣⎢⎢⎡​1000​2000​0100​1−400​∣∣∣∣∣∣∣∣​3500​⎦⎥⎥⎤​
* 第二步：写出等价的(独立的)方程组，保留第一个未知数在左边其余的移到右边，移到右边的称为自由变量
* [ 1 2 0 1 0 0 1 − 4 ∣ 3 5 ] → { x 1 = − 2 x 2 − x 4 + 3 x 3 = 4 x 4 + 5
\left [\begin{array}{cccc}1 & 2 & 0 & 1\\0 & 0 & 1 & -4\\\end{array} \right
|\left.\begin{array}{cccc}3 \\5\end{array} \right ] \to\left
\{\begin{array}{cccc}x_1 = -2x_2 - x_4 + 3 \\x_3 = 4x_4 + 5 \end{array} \right.[
10​20​01​1−4​∣∣∣∣​35​]→{x1​=−2x2​−x4​+3x3​=4x4​+5​
* 第三步：令自由变量为任意实数，写出通解。再改写为向量形式。 { x 1 = − 2 x 2 − x 4 + 3 x 3 = 4 x 4 + 5
\left \{\begin{array}{cccc}x_1 = -2x_2 - x_4 + 3 \\x_3 = 4x_4 + 5\end{array}
\right.{x1​=−2x2​−x4​+3x3​=4x4​+5​
* 令 x 2 = k 1 , x 4 = k 2 x_2 = k_1, x_4 = k_2 x2​=k1​,x4​=k2​, 通解： { x 1 =
− 2 k 1 − k 2 + 3 x 2 = k 1 x 3 = 4 k 2 + 5 x 4 = k 2 \left
\{\begin{array}{cccc}x_1 = -2k_1 - k_2 + 3 \\x_2 = k_1 \\x_3 = 4k_2 + 5 \\x_4 =
k_2\end{array} \right.⎩⎪⎪⎨⎪⎪⎧​x1​=−2k1​−k2​+3x2​=k1​x3​=4k2​+5x4​=k2​​
* 即： x = k 1 [ − 2 1 0 0 ] + k 2 [ − 1 0 4 1 ] + [ 3 0 5 0 ] x = k_1\left
[\begin{array}{cccc}-2 \\1 \\0 \\0\end{array} \right ] + k_2\left
[\begin{array}{cccc}-1 \\0 \\4 \\1\end{array} \right ] + \left
[\begin{array}{cccc}3 \\0 \\5 \\0\end{array} \right ]x=k1​⎣⎢⎢⎡​−2100​⎦⎥⎥⎤​+k2​⎣⎢
⎢⎡​−1041​⎦⎥⎥⎤​+⎣⎢⎢⎡​3050​⎦⎥⎥⎤​
* k 1 , k 2 k_1, k_2 k1​,k2​ 为任意常数

• « 上一篇：SpringBoot 配置 Redis 连接池
• » 下一篇：计算机考证含金量排行榜