【11408学习记录】考研数学核心突破:线性代数之线性方程组深度解析
线性方程组
- 数学
- 线性代数
- 线性方程组
- 齐次线性方程组
- 有解的条件
- 解的性质
- 基础解系和解的结构
- 求解方法和步骤
- 英语
- 每日一句
- 词汇
- 第一步:找谓语
- 第二步:断句
- 第三步:简化
- 主句
- 同位语从句
- 主句
- 定语从句
数学
线性代数
线性方程组
从本质上来说,方程组问题就是向量组问题,方程组和向量组是同一个问题的两种表现形式,其本质一样,所以解决方法也一样。
求解线性方程组,就是对增广矩阵作初等行变换,化成行阶梯形矩阵,然后求解。
解方程组所得到的解 x1,x2,⋯,xnx_1, x_2, \cdots, x_nx1,x2,⋯,xn 就是描述向量与向量之间关系的表示系数:
[x1,x2,⋯,xn]T={自由项β被a1,a2,⋯,an线性表示的组合系数(仅数学一)a1,a2,⋯,an在基下的坐标方程组的解[x_1, x_2, \cdots, x_n]^T = \begin{cases} 自由项 \beta 被 a_1, a_2, \cdots, a_n 线性表示的组合系数 \\ (仅数学一)a_1, a_2, \cdots, a_n 在基下的坐标 \\ 方程组的解 \end{cases} [x1,x2,⋯,xn]T=⎩⎨⎧自由项β被a1,a2,⋯,an线性表示的组合系数(仅数学一)a1,a2,⋯,an在基下的坐标方程组的解
齐次线性方程组
方程组:
{a11x1+a12x2+⋯+a1nxn=0,a21x1+a22x2+⋯+a2nxn=0,⋯am1x1+am2x2+⋯+amnxn=0\begin{cases} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = 0, \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = 0, \\ \cdots \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n = 0 \end{cases} ⎩⎨⎧a11x1+a12x2+⋯+a1nxn=0,a21x1+a22x2+⋯+a2nxn=0,⋯am1x1+am2x2+⋯+amnxn=0
称为 m 个方程,n 个未知量的齐次线性方程组。
其向量形式为:x1a1+x2a2+⋯+xnan=0x_1a_1 + x_2a_2 + \cdots + x_na_n = 0x1a1+x2a2+⋯+xnan=0 ,其中 aj=[a1ja2j⋮amj],j=1,2,⋯,na_j = \begin{bmatrix}a_{1j} \\ a_{2j} \\ \vdots \\ a_{mj} \end{bmatrix} , \ j = 1, 2, \cdots, naj=a1ja2j⋮amj, j=1,2,⋯,n.
其矩阵形式为:Am×nx=0A_{m×n}x = 0Am×nx=0 ,其中 Am×n=[a11a12⋯a1na21a22⋯a2n⋮⋮⋱⋮am1am2⋯amn],x=[x1x2⋮xn]A_{m×n} = \begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}, \ x = \begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}Am×n=a11a21⋮am1a12a22⋮am2⋯⋯⋱⋯a1na2n⋮amn, x=x1x2⋮xn ,这里的 Am×nA_{m×n}Am×n 称为系数矩阵,xxx 称为解向量
有解的条件
- 当 r(A)=nr(A) = nr(A)=n (a1,a2,⋯,ana_1, a_2, \cdots, a_na1,a2,⋯,an 线性无关) 时,方程组有唯一解
- 当 r(A)=r<nr(A) = r < nr(A)=r<n (a1,a2,⋯,ana_1, a_2, \cdots, a_na1,a2,⋯,an 线性相关)时,方程组有非零解(无穷多解),且有 n−rn-rn−r 个线性无关解.
解的性质
- 若 Aξ1=0,Aξ2=0A\xi_1 = 0, A\xi_2 = 0Aξ1=0,Aξ2=0 ,则 A(k1ξ1+k2ξ2)=0A(k_1\xi_1 + k_2\xi_2) = 0A(k1ξ1+k2ξ2)=0 ,其中 k1,k2k_1, k_2k1,k2 是任意常数.
- 类消去律:若 Am×n,r(A)=n,AB=ACA_{m×n} , r(A) = n, AB = ACAm×n,r(A)=n,AB=AC ,则 B=CB = CB=C
基础解系和解的结构
-
基础解系:设 ξ1,ξ2,⋯,ξn−r\xi_1, \xi_2, \cdots, \xi_{n-r}ξ1,ξ2,⋯,ξn−r 满足:{是方程组Ax=0的解线性无关方程组Ax=0的任意解均可由ξ1,ξ2,⋯,ξn−r线性表示\begin{cases}是方程组 Ax = 0 的解 \\ 线性无关 \\ 方程组 Ax = 0 的任意解均可由 \xi_1, \xi_2, \cdots, \xi_{n-r} 线性表示\end{cases}⎩⎨⎧是方程组Ax=0的解线性无关方程组Ax=0的任意解均可由ξ1,ξ2,⋯,ξn−r线性表示
则称 ξ1,ξ2,⋯,ξn−r\xi_1, \xi_2, \cdots, \xi_{n-r}ξ1,ξ2,⋯,ξn−r 为 Ax=0Ax = 0Ax=0 的基础解系. -
通解: 设 ξ1,ξ2,⋯,ξn−r\xi_1, \xi_2, \cdots, \xi_{n-r}ξ1,ξ2,⋯,ξn−r 为 Ax=0Ax = 0Ax=0 的基础解系,则 k1ξ1+k2ξ2+⋯+kn−rξn−rk_1\xi_1 + k_2\xi_2 + \cdots + k_{n-r}\xi_{n-r}k1ξ1+k2ξ2+⋯+kn−rξn−r 是方程组 Ax=0Ax= 0Ax=0 的通解,其中 k1,k2,⋯,kn−rk_1, k_2, \cdots, k_{n-r}k1,k2,⋯,kn−r 是任意常数.
求解方法和步骤
- 将系数矩阵 AAA 作初等行变换化成行阶梯型矩阵 BBB (或行最简阶梯形矩阵 BBB),(初等行变换将方程组化为同解方程组,故 Ax=0Ax = 0Ax=0 和 Bx=0Bx = 0Bx=0 同解,只需解 Bx=0Bx = 0Bx=0 即可.)阶梯数为 rrr ,并记 r(A)=rr(A) = rr(A)=r ,
A→初等行变换B=[c11c12⋯c1r⋯c1n0c22⋯c2r⋯c2n⋮⋮⋱⋮⋱⋮00⋯crr⋯crn00⋯0⋯0⋮⋮⋱⋮⋱⋮00⋯0⋯0]m×nA \xrightarrow{初等行变换} B = \begin{bmatrix} c_{11} & c_{12} & \cdots & c_{1r} & \cdots & c_{1n} \\ 0 & c_{22} & \cdots & c_{2r} & \cdots & c_{2n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & c_{rr} & \cdots & c_{rn} \\ 0 & 0 & \cdots & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & \cdots & 0 \end{bmatrix}_{m × n} A初等行变换B=c110⋮00⋮0c12c22⋮00⋮0⋯⋯⋱⋯⋯⋱⋯c1rc2r⋮crr0⋮0⋯⋯⋱⋯⋯⋱⋯c1nc2n⋮crn0⋮0m×n
其中,m 是原方程组中方程的个数,n是未知量个数
- 按列找出一个秩为 rrr 的子矩阵,剩余列位置的未知数设为自由变量
- 按基础解系定义求出 ξ1,ξ2,⋯,ξn−r\xi_1, \xi_2, \cdots, \xi_{n-r}ξ1,ξ2,⋯,ξn−r 并写出通解
英语
每日一句
Their analysis ruled out the possibility that it was firms’ political influence, rather than their CSR stand, that accounted for the leniency: Companies that contributed more to political campaigns did not receive lower fines.(2016, Reading Comprehension, Part A Text 3)
词汇
rule out the possibility: 排除可能性
political influence: 政治影响力
stand: n. 立场
account for…: 导致…,说明…;占……比例
political campaign: 政治运动
fine: n. 罚款,罚金
第一步:找谓语
Their analysis ruled out the possibility that it was firms’ political influence, rather than their CSR stand, that accounted for the leniency: Companies that contributed more to political campaigns did not receive lower fines.
第二步:断句
原句中存在5处谓语,包含5件事,其谓语分别位于以下分句中:
- ruled 为主句谓语
- was 为 that 引导的同位语从句谓语
- accounted 为 that 引导的同位语从句谓语
- contributed 为 that 引导的定语从句谓语
- did not receive 为冒号后的主句谓语
按照标点、引导词以及谓语,可以将原句断开为以下分句:
- Their analysis ruled out the possibility —— 主句
- that it was firms’ political influence, rather than their CSR stand, that accounted for the leniency: —— 同位语从句
- Companies …… did not receive lower fines. —— 主句
- that contributed more to political campaigns —— 定语从句
第三步:简化
主句
Their analysis ruled out the possibility that it was firms’ political influence, rather than their CSR stand, that accounted for the leniency:
- 主句主语部分:Their analysis
- 限定词:their 修饰名词:analysis
- 名词:analysis 为主句主语核心词 - 主句谓语部分:ruled out
- 副词:out 修饰动词:ruled
- 动词:ruled 为及物动词,与副词:out 组成及物动词短语,后接宾语
- 主句宾语部分:the possibility that it was firms’ political influence, rather than their CSR stand, that accounted for the leniency:
- 定冠词:the 修饰名词:possibility
- 名词:possibility 为动词短语:ruled out 的宾语核心词
- 同位语从句部分:that it was firms’ political influence, rather than their CSR stand, that accounted for the leniency: 为 that 引导的同位语从句,用于解释说明抽象名词:possibility
去掉主句扩展部分,就得到了主句核心:
- …… analysis ruled …… possibility …… —— …… 分析排除了…… 可能性
同位语从句
that it was firms’ political influence, rather than their CSR stand, that accounted for the leniency:
- 从句引导词:that 引导同位语从句,解释说明抽象名词:possibility
- 引导词:that 在从句中不充当任何成分
- 从句主语部分:it was firms’ political influence, rather than their CSR stand, that
- 强调句式:it was …… that 突出主语的重要性
- 名词所有格: firms’ 与形容词:political 共同修饰名词:influence
- 名词:influence 为从句主语核心词
- 比较连词:rather than 连接名词词组:firms’ political influenc 与名词词组:their CSR stand 构成并列结构
- 并列结构:A rather than B 表示的含义为:“是 A 而不是 B”
- 名词词组:their CSR stand 为被排除的选项与 firms’ political influenc 通过 rather than 构成并列结构
- 限定词:their 与 CSR 共同修饰名词:stand
- 从句谓语部分:accounted 为不及物动词
- 从句状语部分:for the leniency 修饰动词:accounted ,表示动作发生的对象
去掉从句扩展部分,就得到了从句核心:
- that it was …… influence, rather than …… stand, that accounted …… —— 正是 …… 影响,而不是 …… 立场,导致了……
主句
Companies that contributed more to political campaigns did not receive lower fines.
- 主句主语部分:Companies that contributed more to political campaigns
- 名词:Companies 为主句主语核心词
- 定语从句:that contributed more to political campaigns 修饰主语 companies ,进一步解释说明 companies
- 主句谓语部分:did not receive 为及物动词的否定形式,后接宾语
- 主句宾语部分:lower fines
- 形容词比较级:lower 修饰名词:fines
- 名词:fines 为主句宾语核心词
去掉主句扩展部分,就得到了主句核心:
- Companies …… did not receive …… fines. —— 公司 …… 没有收到 …… 罚款
定语从句
that contributed more to political campaigns
- 从句引导词:that 引导定语从句,修饰先行词:companies
- 引导词:that 在从句中充当主语,代替先行词:companies
- 从句谓语部分:contributed 为及物动词,后接宾语
- 从句宾语部分:more
- 从句状语部分:to political campaigns 为目的状语,修饰动词:contributed 表示动作发生的对象
- 形容词:political 修饰名词:campaigns
- 名词:campaigns 为介词:to 的宾语
去掉从句扩展部分,就得到了从句核心:
- that contributed more …… —— 公司贡献了更多……