标签：extract result sum ESS dia isp ror factor lin
\((1\times 1)\) \(ax_1=b\) \(\rightarrow\) \(x_1=\frac{b}{a}\)
\((2 \times 2)\) \(\begin{cases}a_{11}x_1+a_{12}x_2=b_1 \\ a_{21}x_1+a_{22}x_2=b_2\end{cases}\)
Ex1:
\[ \begin{cases}x_1+2x_2=5 \3x_1+4x_2=6\end{cases} \rightarrow x_1=-4,x_2=4.5\]
\((n \times n)\)
\[ \begin{cases}a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n=b_1 \a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n=b_2\\vdots\a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n=b_n\\end{cases}\]
Matix
\[A=\begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n}\a_{21} & a_{22} & \cdots & a_{2n}\\vdots\a_{11} & a_{12} & \cdots & a_{nn}
\end{bmatrix}, x=\begin{bmatrix}x_1 \cdots x_n\end{bmatrix}, b=\begin{bmatrix}b_1 \cdots b_n\end{bmatrix}, Ax=b\]
Matrix form extracts the must essential information .
Matrix form of Ex1:
\[\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix}5\\6\end{bmatrix}\]
Augment matrix
\[\begin{bmatrix}1 & 2 & 5 \\3 & 4 & 6\end{bmatrix} \rightarrow \begin{bmatrix}1 & 2 & 5 \\0 & -2 & -9\end{bmatrix} \rightarrow \begin{bmatrix}1 & 2 & 5 \\0 & 1 & 4.5\end{bmatrix} \rightarrow \begin{bmatrix}1 & 0 & -4 \\0 & 1 & 4.5\end{bmatrix} \]
Systematic method of Solving for Ex1:
\[ \begin{cases}x_1+2x_2=5 \3x_1+4x_2=6\end{cases} \rightarrow \begin{cases}x_1+2x_2=5 \0-2x_2=-9\end{cases} \rightarrow \begin{cases}x_1+2x_2=5 \x_2=4.5\end{cases}\]
Know as Gauss elimination Method：
Matlab: To solve \(Ax=b\), using \(x=A\backslash b\)
>> A = [1 2;3 4]
A =
1 2
3 4
>> b = [5 6]'
b =
5
6
>> x = A\b
x =
-4.0000
4.5000
>> format long
>> x = A\b
x =
-3.999999999999999
4.499999999999999Bring to an easily solved form:
For upper triangular (back substitution):
\[a_{nn}x_n=b_n \rightarrow x_n = \frac{b_n}{a_{nn}}\] \[a_{n-1,n-1}x_{n-1}+ a_{n-1,n}x_n=b_{n-1}\] repeat... \[x_i=\frac{b_i-\sum^n_{j=i+1}a_{ij}x_j}{a_{ii}}\]
For lower triangular (forward substitution):
\[a_{11}x_1=b_1 \rightarrow x_1 = \frac{b_1}{a_{11}}\] \[a_{2,1}x_{1}+ a_{2,2}x_2=b_{2}\] repeat... \[x_i=\frac{b_i-\sum^{i-1}_{j=1}a_{ij}x_j}{a_{ii}}\]
General Methods:
Starting from 1st, a suitable multiple of the current row (pivot row) is subtracted form all subsequent rows, the result is an upper triangular
Ex2: \[\begin{cases}2x_1+x_2-x_3=1\\x_1+2x_2+x_3=8\\-x_1+x_2-x_3=-5\end{cases}\]
matrix:
\[\begin{bmatrix}2 & 1 & -1 & 1\\ 1 &2 & 1 & 8\\ -1 & 1 & -1 & -5 \end{bmatrix} \rightarrow \begin{bmatrix}2 & 1 & -1 & 1\\ 0 &1.5 & 1.5 & 7.5\\ 0 & 1.5 & -1.5 & -4.5 \end{bmatrix}\]
\[\rightarrow \begin{bmatrix}2 & 1 & -1 & 1\\ 0 &1.5 & 1.5 & 7.5\\ 0 & 0 & -3 & -12\end{bmatrix} \rightarrow \begin{cases}-3x_2 = -12\\x_3=4\end{cases} \rightarrow \begin{cases}x_1 =2 \\x_2=1\\x_3=4\end{cases}\]
It is a problem if this element is zero. If a pivot is small number, this may lead to loss of accuracy.
Exchange rows when necessary:
\[\begin{bmatrix}
1 & 2 & 5\\ 3 & 4& 5
\end{bmatrix}, \] \[\begin{bmatrix}
0 & 2 & 5\\ 3 & 4& 5
\end{bmatrix}, \]
Pivoting is the procedure of rearranying rows to make sure Pivot is not zero and as large as possible in magnitude.
Ex3: \[ \begin{cases}0.0003x_1 + 12.34x_2=12.343 \\0.4321 x_1+x_2=5.321\end{cases}\] Using Gauss:
\[\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n}& b_1\\ \vdots & & & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} & b_n\end{bmatrix}\]
normalized and pivot row by dividing by the
normalize by the pivot; subtract the pivot row to elimate all non-zero in the column.
Ex1. \[\begin{bmatrix}4 & 1 & 2 & 21\\ 2 & -2 & 2 & 8\\ 1 & -2 & 4 & 16\end{bmatrix} \rightarrow \begin{bmatrix}1 & -2 & 4 & 16\\ 2 & -2 & 2 & 8 \\4 & 1 & 2 & 21 \end{bmatrix} \rightarrow \begin{bmatrix}1 & -2 & 4 & 16\\ 0 & 2 & -6 & -24 \\0 & 9 & -14 & -43 \end{bmatrix}\] \[ \rightarrow \begin{bmatrix}1 & -2 & 4 & 16\\ 0 & 1 & -3 & -12 \\0 & 9 & -14 & -43 \end{bmatrix} \rightarrow \begin{bmatrix}1 & 0 & -2 & -8\\ 0 & 1 & -3 & -12 \\0 & 0 & 13 & 65 \end{bmatrix} \rightarrow \begin{bmatrix}1 & 0 & -2 & -8\\ 0 & 1 & -3 & -12 \\0 & 0 & 1 & 5 \end{bmatrix}\] \[ \rightarrow \begin{bmatrix}1 & 0 & 0 & 2\\ 0 & 1 & 0 & 3 \\0 & 0 & 1 & 5 \end{bmatrix}\]
\(AB=I\), \(A\begin{bmatrix}b_1 & b_2 & b_3 \end{bmatrix} \rightarrow \begin{bmatrix}a_{11} & a_{12} & a_{13} & 1 & 0 & 0\\ a_{21} & a_{22} & a_{23} & 0 & 1 & 0\\ a_{31} & a_{32} & a_{33} & 0 & 0 & 1\end{bmatrix} \rightarrow \begin{bmatrix}1 & 0 & 0 &c_{11} & c_{12} & c_{13} & \\ 0 & 1 & 0 & c_{21} & c_{22} & c_{23} &\\ 0 & 0 & 1& c_{31} & c_{32} & c_{33} \end{bmatrix}\)
\(A=LU\), where \(L\) means lower triangular, \(U\) means upper triangular.
If this is achieved, then, solve \(Ax=b\) is equivalent to solve \(LUx=b\)
Work in 2 steps: \(Ux=y,Ly=b\), solve \(Ly=b\) then \(Ux=y\).
Two approaches:
\[\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\a_{21} & a_{22} & \cdots & a_{2n}\\vdots\a_{11} & a_{12} & \cdots & a_{nn} \end{bmatrix}=\begin{bmatrix}1 & & & & \\ m_{21} & 1 & & & \\ m_{31} & m_{32} & 1 & &\end{bmatrix} \begin{bmatrix}a'_{11} & a'_{12} & \cdots & a'_{1n} \\ & a'_{22} & \cdots & a'_{2n}\end{bmatrix}\]
general formulation: \(m_{ij}=\frac{a'_{ij}}{a'_{jj}}\)
Ex4: \(\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}=\begin{bmatrix}1 & 0\\3 & 1\end{bmatrix}\begin{bmatrix}1 & 2\\0 & -2\end{bmatrix}\)
\[\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\a_{21} & a_{22} & \cdots & a_{2n}\\vdots\a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}=\begin{bmatrix}L_{11} & & & & \\ L_{21} & L_{22} & & & \\ \vdots & \vdots & \ddots & \end{bmatrix} \begin{bmatrix}1 & U_{12} & U_{13} & \cdots & U_{1n} \\ & 1 & U_{23} & \cdots &U_{2n}\\ & & \ddots & & \vdots \\ & & & & 1\end{bmatrix} \]
Ex5:
\[\begin{bmatrix} a_{11} & a_{12} & a_{13}\a_{21} & a_{22} & a_{23}\a_{31} & a_{32} & a_{33} \end{bmatrix}=\begin{bmatrix} L_{11} & 0 & 0\L_{21} & L_{22} & 0\L_{31} & L_{32} & L_{33} \end{bmatrix}\begin{bmatrix} 1 & U_{12} & U_{13}\0 &1 & U_{23}\0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} L_{11} & L_{11}U_{12} & L_{11}U_{13}\L_{21} & L_{21}U_{12}+L_{22} & L_{21}U_{13}+ L_{22}U_{23}\L_{31} & L_{31}U_{12}+L_{32} & L_{33}+ L_{31}U_{13}+ L_{32}U_{23} \end{bmatrix}\]
标签：extract result sum ESS dia isp ror factor lin
原文地址：https://www.cnblogs.com/SweetZxl/p/11161730.html
