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# 数学代写|数值分析代写Numerical analysis代考|AMATH352 Compact Variants of Gaussian Elimination

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## 数学代写|数值分析代写Numerical analysis代考|Compact Variants of Gaussian Elimination

Rather than constructing $L$ and $U$ by using the elimination steps (8.22) to (8.25) of Section $8.2$, it is possible to solve directly for these matrices. The number of operations will be the same as with Gaussian elimination, but there are some advantages to such compact variants of Gaussian elimination.

To illustrate the direct computation of $L$ and $U$, consider the $n=3$ case. Write $A=L U$ as
$$\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \ m_{21} & 1 & 0 \ m_{31} & m_{32} & 1 \end{array}\right]\left[\begin{array}{ccc} u_{11} & u_{12} & u_{13} \ 0 & u_{22} & u_{23} \ 0 & 0 & u_{33} \end{array}\right]$$
Multiply $L$ and $U$, and match the elements of the product with the corresponding elements in $A$. Doing this for the first row and column of the product yields
$a_{11}=u_{11} \quad a_{12}=u_{12} \quad a_{13}=u_{13}$
$a_{21}=m_{21} u_{11} \quad a_{31}=m_{31} u_{11}$
This gives the first column of $L$ and the first row of $U$. Next, multiply row 2 of $L$ times columns 2 and 3 of $U$, to obtain

$$a_{22}=m_{21} u_{12}+u_{22} \quad a_{23}=m_{21} u_{13}+u_{23}$$
These can be solved for $u_{22}$ and $u_{23}$. Next multiply row 3 of $L$ to obtain
\begin{aligned} &m_{31} u_{12}+m_{32} u_{22}=a_{32} \ &m_{31} u_{13}+m_{32} u_{23}+u_{33}=a_{33} \end{aligned}
These equations yield values for $m_{32}$ and $u_{33}$, completing the construction of $L$ and $U$. In this process, we must have $u_{11} \neq 0, u_{22} \neq 0$ in order to solve for $L$. There are modifications of the method to avoid this assumption, using pivoting, but we will simply consider those cases where the necessary elements will be nonzero.

## 数学代写|数值分析代写Numerical analysis代考|Tridiagonal Systems

When the coefficient matrix $A$ has a special form, it is often possible to simplify Gaussian elimination. We will do this for tridiagonal systems of linear equations.
A system $A x=f$ is called tridiagonal if its matrix has the form
$$A=\left[\begin{array}{cccccccc} b_1 & c_1 & 0 & 0 & \cdots & 0 \ a_2 & b_2 & c_2 & 0 & & & 0 \ 0 & a_3 & b_3 & c_3 & & & 0 \ \vdots & & \ddots & & & \vdots \ 0 & & \cdots & & 0 & a_{n-1} & b_{n-1} & c_{n-1} \ 0 & & & & 0 \end{array}\right]$$
This is called a tridiagonal matrix. Tridiagonal systems occur commonly in solving problems in many different areas of numerical analysis. In Chapter 5, spline function interpolation led to the tridiagonal system in (5.69).

When Gaussian elimination is applied to $A x=f$, most of the multipliers $m_{i k}=0$ and most of the elements of $U$ are also zero. With this in mind, it can be shown that the $L U$ factorization will have the following general form, provided that pivoting is not used.

$$L U=\left[\begin{array}{lllll} 1 & 0 & 0 & \cdots & 0 \ a_2 & 1 & 0 & & 0 \ 0 & a_3 & 1 & & \ \vdots & & & \ddots & \vdots \ 0 & \cdots & a_n & 1 \end{array}\right] \cdot\left[\begin{array}{cccccc} \beta_1 & c_1 & 0 & & \cdots & 0 \ 0 & \beta_2 & c_2 & 0 & & 0 \ \vdots & & \ddots & & \vdots \ 0 & & \cdots & & 0 & \beta_n \end{array}\right]$$
Multiply in succession each row of $L$ times the various columns of $U$, and then equate the results to the corresponding elements of $A$. This yields
$\beta_1=b_1 \quad: \quad$ row 1 of $L U$
$a_2 \beta_1=a_2, \quad a_2 c_1+\beta_2=b_2 \quad: \quad$ row 2 of $L U$
$a_j \beta_{j-1}=a_j, \quad a_j c_{j-1}+\beta_j=b_j \quad: \quad$ row $j$ of $L U$
for $j=3, \ldots, n$. The reader should check these equations for the first few rows of A.

## 数学代写|数值分析代写数值分析代考|高斯消去的紧致变式

$$\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \ m_{21} & 1 & 0 \ m_{31} & m_{32} & 1 \end{array}\right]\left[\begin{array}{ccc} u_{11} & u_{12} & u_{13} \ 0 & u_{22} & u_{23} \ 0 & 0 & u_{33} \end{array}\right]$$
$L$与$U$相乘，并将乘积的元素与$A$中相应的元素匹配。对product的第一行和第一列执行此操作将得到
$a_{11}=u_{11} \quad a_{12}=u_{12} \quad a_{13}=u_{13}$
$a_{21}=m_{21} u_{11} \quad a_{31}=m_{31} u_{11}$

$$a_{22}=m_{21} u_{12}+u_{22} \quad a_{23}=m_{21} u_{13}+u_{23}$$

\begin{aligned} &m_{31} u_{12}+m_{32} u_{22}=a_{32} \ &m_{31} u_{13}+m_{32} u_{23}+u_{33}=a_{33} \end{aligned}

## 数学代写|数值分析代写数值分析代考|三对角系统

$$A=\left[\begin{array}{cccccccc} b_1 & c_1 & 0 & 0 & \cdots & 0 \ a_2 & b_2 & c_2 & 0 & & & 0 \ 0 & a_3 & b_3 & c_3 & & & 0 \ \vdots & & \ddots & & & \vdots \ 0 & & \cdots & & 0 & a_{n-1} & b_{n-1} & c_{n-1} \ 0 & & & & 0 \end{array}\right]$$

$$L U=\left[\begin{array}{lllll} 1 & 0 & 0 & \cdots & 0 \ a_2 & 1 & 0 & & 0 \ 0 & a_3 & 1 & & \ \vdots & & & \ddots & \vdots \ 0 & \cdots & a_n & 1 \end{array}\right] \cdot\left[\begin{array}{cccccc} \beta_1 & c_1 & 0 & & \cdots & 0 \ 0 & \beta_2 & c_2 & 0 & & 0 \ \vdots & & \ddots & & \vdots \ 0 & & \cdots & & 0 & \beta_n \end{array}\right]$$

$\beta_1=b_1 \quad: \quad$第1行$L U$
$a_2 \beta_1=a_2, \quad a_2 c_1+\beta_2=b_2 \quad: \quad$第2行$L U$
$a_j \beta_{j-1}=a_j, \quad a_j c_{j-1}+\beta_j=b_j \quad: \quad$行$j$$L U$
$j=3, \ldots, n$。读者应该检查a的前几行方程式

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