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aij=ajia_{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+ATA + A^TA+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={a11a12⋯a1na21a22⋯a2n⋯⋯⋯⋯an1an2⋯ann}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​​⋯⋯⋯⋯​a1n​a2n​⋯ann​​⎭⎪⎪⎬⎪⎪⎫​a12=a21aij=ajia1n=an1a_{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 {a11x1+a12x2+...+a1nxn=b1a21x1+a22x2+...+a2nxn=b1⋯am1x1+am2x2+...+amnxn=bm\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 Ax=bAx = bAx=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 Ax=0Ax = 0Ax=0 有非零解的充要条件是R(A)<nR(A) < nR(A)<n推论：当m < n时，齐次线性方程组 Am×nx=0A_{m×n} x = 0Am×n​x=0 一定有非零解

定理2

对于n元线性方程组 Ax=bAx=bAx=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)=nR(A) = R(A,b) = nR(A)=R(A,b)=n有无穷多解的充要条件是 R(A)=R(A,b)<nR(A) = R(A,b) < nR(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)=rR(A) = R(B) = rR(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c1,c2,...,cn−rc_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 Am×nX=0A_{m×n}X = 0Am×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：ξ1,ξ2,⋯,ξn−r\xi_1, \xi_2, \cdots, \xi_{n-r}ξ1​,ξ2​,⋯,ξn−r​, 则通解为：x=k1ξ1+k2ξ2+...+kn−rξn−r(ki∈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 {x1+x2−x3−x4=02x1−5x2+3x3+2x4=07x1−7x2+3x3+x4=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=(11−1−12−5327−731)∼(10−27−3701−57−470000)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​−74​0​⎠⎞​化为：{x1=27x3+37x4x2=57x3+47x4\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 (x3x4)=(10),(01)\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 (x1x2)=(2757),(3747)\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：ξ1=(275710),ξ2=(374701)\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=(x1x2x3x4)=c1(275710)+c2(374701),(c1,c2∈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Am✖×nX=bA_{m✖×n} X = bAm✖×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=k1ξ1+k2ξ2+⋯+kn−rξn−r+η∗(ki∈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k1ξ1+⋯+kn−rξn−rk_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 {x1−x2−x3+x4=0,x1−x2+x3−3x4=1,x1−x2−2x3+3x4=−12,\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=(1−1−111−11−31−1−23∣01−12)→(1−10−1001−20000∣12120)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<4R(A) = R(B) = 2 < 4R(A)=R(B)=2<4, 故方程组有无穷多解{x1=x2+x4+12x3=2x4+12⇒{x1=x2+x4+12x2=x2x3=2x4+12x4=x4⇒(x1x2x3x4)=C1(1100)+C2(1021)+(10)(C1,C2∈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{x1=x2+x4+12x3=2x4+12\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x2=x4=0x_2 = x_4 = 0x2​=x4​=0, 则 x1=x3=12x_1 = x_3 = \frac{1}{2}x1​=x3​=21​, 即得方程组的一个解 η∗=(10)\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 {x1=x2+x4,x3=2x4\left \{ \begin{array}{cccc}x_1 = x_2 + x_4, \\ x_3 = 2x_4\end{array} \right.{x1​=x2​+x4​,x3​=2x4​​ 中取(x2x4)=(10)\left (\begin{array}{cccc}x_2 \\x_4\end{array} \right ) = \left (\begin{array}{cccc}1 \\0\end{array} \right )(x2​x4​​)=(10​) 及 (01)\left (\begin{array}{cccc}0 \\1\end{array} \right )(01​)，则(x1x3)=(10)\left (\begin{array}{cccc}x_1 \\x_3\end{array} \right ) =\left (\begin{array}{cccc}1 \\0\end{array} \right )(x1​x3​​)=(10​) 及 (12)\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 ξ1=(1100),ξ2=(1021)\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：(x1x2x3x4)=c1(1100)+c2(1021)+(10),(c1,c2∈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 {x1+2x2+2x3+x4=02x1+x2−2x3−2x4=0x1−x2−4x3−3x4=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=(122121−2−21−1−4−3)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r2−2r1,r3−r1r_2 - 2r_1, r_3 - r_1r2​−2r1​,r3​−r1​ (12210−3−6−40−3−6−4)\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​⎠⎞​ r3−r2r_3 - r_2r3​−r2​ (12210−3−6−40000)\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​⎠⎞​ −13r2- \frac{1}{3} r_2−31​r2​ (1221012430000)\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​⎠⎞​ r1−2r2r_1 - 2r_2r1​−2r2​ (10−2−53012−430000)\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: {x1=2x3+53x4x2=−2x3+43x4\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：x3=c1,x4=c2x_3 = c_1, x_4 = c_2x3​=c1​,x4​=c2​写出参数形式的通解，再改写为向量形式通解：(x1=2c1+53c2x2=−2c1−43c2x3=c1x4=c2)\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：(x1x2x3x4)=c1(2−210)+c2(53−4301)\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 c1,c2c_1, c_2c1​,c2​为任意实数

例4

求解非齐次线性方程组 {x1−2x2+3x3−x4=13x1−x2+5x3−3x4=22x1+x2+2x3−2x4=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=[1−23−13−15−3212−2∣123]→r[1−23−105−400000∣1−12]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=20 = 20=2, 所以无解

例5

解方程组 {x1+2x2+x4=3x1+2x2+x3−3x4=82x1+4x2+2x4=6x1+2x2−x3+5x4=−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=[121−3240212−15∣386−2]→r[101−400000000∣3500]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[101−4∣35]→{x1=−2x2−x4+3x3=4x4+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。{x1=−2x2−x4+3x3=4x4+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 x2=k1,x4=k2x_2 = k_1, x_4 = k_2x2​=k1​,x4​=k2​, 通解：{x1=−2k1−k2+3x2=k1x3=4k2+5x4=k2\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=k1[−2100]+k2[−1041]+[3050]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​⎦⎥⎥⎤​k1,k2k_1, k_2k1​,k2​ 为任意常数