[MIT18.06 Notes] Course 1 - The Geometry of Linear Equations

文章目录

  • Row picture & Column picture
  • 求解n linear equations,n unknowns
    • Example 1: n=2
    • Example 2: n=3

Row picture & Column picture

PS: Row pictureColumn picture 在国内这两个概念不多见,翻译过来为行图像和列图像。

  • Row picture:可以理解为线性方程的函数图像。
  • Column picture:可以理解为将矩阵乘以向量的形式分解为 x ⋅ u ⃗ + y ⋅ v ⃗ + . . . x \cdot \vec{u} + y \cdot \vec{v} + ... xu +yv +...形式的图像。
    下面会有详细的例子。

求解n linear equations,n unknowns

n equations, n unknowns直译过来就是n个线性方程的方程组,n个未知量。对于一个n equations, n unknowns的方程组,Ax=b。

Example 1: n=2

一个线性的二元一次方程组:
{ 2 x − y = 0 − x + 2 y = 3 \left\{ \begin{aligned} & 2x - y =0\\ & -x + 2y =3 \end{aligned} \right. {2xy=0x+2y=3
它的图像就称作Row Picture:
[MIT18.06 Notes] Course 1 - The Geometry of Linear Equations_第1张图片

解得
{ x = 1 y = 2 \left\{ \begin{aligned} x=1\\ y=2 \end{aligned} \right. {x=1y=2
将线性方程组转化为对应的矩阵形式:
[ 2 − 1 − 1 − 1 ] [ x y ] = [ 0 3 ] \begin{bmatrix} 2 & -1\\ -1 & -1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}= \begin{bmatrix} 0\\ 3 \end{bmatrix} [2111][xy]=[03]
其中 A = [ 2 − 1 − 1 − 1 ] , b = [ 0 3 ] A=\begin{bmatrix}2 & -1\\-1 & -1\end{bmatrix},b=\begin{bmatrix}0\\3\end{bmatrix} A=[2111],b=[03],然后我们将它拆分一下:
x [ 2 − 1 ] + y [ − 1 2 ] = [ 0 3 ] x \begin{bmatrix} 2\\ -1 \end{bmatrix} +y \begin{bmatrix} -1\\ 2 \end{bmatrix}= \begin{bmatrix} 0\\ 3 \end{bmatrix} x[21]+y[12]=[03]
Column picture:
[MIT18.06 Notes] Course 1 - The Geometry of Linear Equations_第2张图片

也可解得
{ x = 1 y = 2 \left\{ \begin{aligned} x=1\\ y=2 \end{aligned} \right. {x=1y=2

Example 2: n=3

{ x + 2 y + 3 z = 6 2 x + 5 y + 2 z = 9 6 x − 3 y + z = 4 \left\{ \begin{aligned} & x+2y+3z =6\\ & 2x+5y+2z =9\\ & 6x−3y+z= 4 \end{aligned} \right. x+2y+3z=62x+5y+2z=96x3y+z=4
有了上面的经验我们可以将其转换为矩阵乘以向量的形式:
[ 1 2 3 2 5 2 6 − 3 1 ] [ x y z ] = [ 6 9 4 ] \begin{bmatrix} 1 & 2 &3\\ 2 & 5 & 2\\ 6 & -3 & 1 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix}= \begin{bmatrix} 6\\ 9\\ 4 \end{bmatrix} 126253321xyz=694
拆分:
x [ 1 2 6 ] + y [ 2 5 − 3 ] + z [ 3 2 1 ] = [ 6 9 4 ] x \begin{bmatrix} 1\\ 2\\ 6 \end{bmatrix} +y \begin{bmatrix} 2\\ 5\\ -3 \end{bmatrix} +z \begin{bmatrix} 3\\ 2\\ 1 \end{bmatrix}= \begin{bmatrix} 6\\ 9\\ 4 \end{bmatrix} x126+y253+z321=694
解得
{ x = 1 y = 1 z = 1 \left\{ \begin{aligned} x=1\\ y=1\\ z=1 \end{aligned} \right. x=1y=1z=1
我们来看一下图像是不是这样的:
Row picture:
[MIT18.06 Notes] Course 1 - The Geometry of Linear Equations_第3张图片

Column picture:
[MIT18.06 Notes] Course 1 - The Geometry of Linear Equations_第4张图片
本节的解都是由图像得出,具体矩阵求解请参见下一节[MIT18.06 Notes] Course 2 - Elimination with Matrices(Part 1)

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