### 数学代写|线性代数代写linear algebra代考|MAST10022

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|线性代数代写linear algebra代考|Techniques for Solving Systems of Linear Equations

In this section, we will describe two techniques for solving systems of equations. We use these two techniques to solve systems like the one presented in the previous section that arose from a question about images.
Method of elimination
In this section, we solve the system of equations in $2.4$ using the method of elimination. You may have used this method before, but we include it here to introduce some terminology to which we will refer in later sections. We will also give a parallel method later in this section.

Two systems of equations are said to be equivalent if they have the same solution set.
The idea behind the method of elimination is that we seek to manipulate the equations in a system so that the solution is easier to obtain. Specifically, in the new system, one or more of the equations will be of the form $x_{i}=c$. Since one of the equations tells us directly the value of one of the variables, we can substitute that value into the other equations and the remaining, smaller system has the same solution (together, of course, with $x_{i}=c$ ).

Before we solve the system in $(2.4)$, we provide the list of allowed operations for solving a system of equations, using the method of elimination.
Allowed operations for solving systems of equations
(1) Multiply both sides of an equation by a nonzero number.
(2) Change one equation by adding a nonzero multiple of another equation to it.
(3) Change the order of equations.
You may find these operations familiar from your earlier experience solving systems of equations; they do not change the solution set of a system. In other words, every time we change a system using one of these operations, we obtain an equivalent system of equations.

## 数学代写|线性代数代写linear algebra代考|echelon form

A matrix is in echelon form if the following three statements are true:
-All leading entries are 1 .
-Each leading entry is to the right of the leading entry in the row above it.
-Any row of zeros is below all rows that are not all zero.
The matrix $P$, above, is not in echelon form. Two of the three criteria above are not true. First, 2 is a leading entry so the first rule is not true. Second, the leading entry in the second row is to the left of the leading entry in the first row, violating the second rule.

However, if we start with $P$, then multiply row two by $1 / 2$ and then switch the first two rows, we get
$$Q=\left(\begin{array}{cccc|c} 1 & 0 & 1 / 2 & 5 / 2 & 1 \ 0 & 1 & 2 & 4 & 0 \ 0 & 0 & 0 & 1 & 1 \end{array}\right)$$
which is in echelon form. It is not just a coincidence that we could row reduce $P$ to echelon form; in fact, every matrix can be put into echelon form. It is indeed convenient that every matrix is row equivalent to a matrix that is in echelon form, and you should be able to see that an augmented matrix in echelon form corresponds to a simpler system to solve.

However, augmented matrices in echelon form are not the simplest to solve, and moreover, echelon form is not unique. For example, both the matrices $R$ and $S$ below are also in echelon form and are row equivalent to $Q$.

$$\begin{gathered} R=\left(\begin{array}{cccc|c} 1 & -1 & -3 / 2 & -1 & 3 / 2 \ 0 & 1 & 2 & 6 & 2 \ 0 & 0 & 0 & 1 & 1 \end{array}\right), \ S=\left(\begin{array}{cccc|c} 1 & 0 & 1 / 2 & 0 & -3 / 2 \ 0 & 1 & 2 & 0 & -4 \ 0 & 0 & 0 & 1 & 1 \end{array}\right) . \end{gathered}$$
Since the matrices $P, Q, R$, and $S$ are all row equivalent, they correspond to equivalent systems. Which would you rather work with? What makes your choice nicer to solve than the others? Likely, what you are discovering is that matrices corresponding to systems that are quite easy to solve are matrices in reduced echelon form.

## 数学代写|线性代数代写linear algebra代考|Elementary Matrix

In this section, we will briefly connect matrix reduction to a set of matrix products ${ }^{5}$. This connection will prove useful later. To begin, let us define an elementary matrix. We begin with the identity matrix.

An $n \times n$ elementary matrix $E$ is a matrix that can be obtained by performing one row operation on $I_{n}$.
Let us give a couple examples of elementary matrices before we give some results.
Example $2.2 .23$ The following are $3 \times 3$ elementary matrices:
$E_{1}=\left(\begin{array}{lll}1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0\end{array}\right)$ is obtained by changing the order of rows 2 and 3 of the identity matrix.
$E_{2}=\left(\begin{array}{lll}2 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right)$ is obtained by multiplying row 1 of $I_{3}$ by 2

• $E_{3}=\left(\begin{array}{lll}1 & 0 & 0 \ 3 & 1 & 0 \ 0 & 0 & 1\end{array}\right)$ is obtained by adding 3 times row 1 to row $2 .$
Since $M=\left(\begin{array}{ccc}2 & 0 & 0 \ -3 & 1 & 0 \ 0 & 0 & 1\end{array}\right)$ cannot be obtained by performing a single row operation on $I_{3}$, so is not an elementary matrix.
Let us now see how these are related to matrix reduction. Consider the following example:
Example 2.2.24 Let $M=\left(\begin{array}{ccc}2 & 3 & 5 \ 1 & 2 & 1 \ 3 & 4 & -1\end{array}\right)$. Let us see what happens when we multiply by each of the elementary matrices in Example 2.2.23.
$$E_{1} M=\left(\begin{array}{lll} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{array}\right)\left(\begin{array}{ccc} 2 & 3 & 5 \ 1 & 2 & 1 \ 3 & 4 & -1 \end{array}\right)=\left(\begin{array}{ccc} 2 & 3 & 5 \ 3 & 4 & -1 \ 1 & 2 & 1 \end{array}\right)$$
The matrix multiplication results in $M$ altered by changing rows 2 and 3 .
$$E_{2} M=\left(\begin{array}{lll} 2 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right)\left(\begin{array}{ccc} 2 & 3 & 5 \ 1 & 2 & 1 \ 3 & 4 & -1 \end{array}\right)=\left(\begin{array}{ccc} 4 & 6 & 10 \ 1 & 2 & 1 \ 3 & 4 & -1 \end{array}\right)$$
The matrix multiplication results in $M$ altered by multiplying row 1 by 2 .

## 数学代写|线性代数代写linear algebra代考|Techniques for Solving Systems of Linear Equations

(1) 将方程两边乘以一个非零数。
(2) 通过添加另一个方程的非零倍数来改变一个方程。
(3) 改变方程的顺序。

## 数学代写|线性代数代写linear algebra代考|echelon form

-所有前导条目均为 1 。
– 每个前导条目都位于其上方行中的前导条目的右侧。
– 任何零行低于所有非全零的行。

R=(1−1−3/2−13/2 01262 00011), 小号=(101/20−3/2 0120−4 00011).

## 数学代写|线性代数代写linear algebra代考|Elementary Matrix

• 和3=(100 310 001)通过将第 1 行与第 1 行相加 3 次获得2.
自从米=(200 −310 001)不能通过对上执行单行操作来获得我3，所以不是初等矩阵。
现在让我们看看这些与矩阵约简有何关系。考虑以下示例：
示例 2.2.24 让米=(235 121 34−1). 让我们看看当我们乘以示例 2.2.23 中的每个基本矩阵时会发生什么。
和1米=(100 001 010)(235 121 34−1)=(235 34−1 121)
矩阵乘法导致米通过更改第 2 行和第 3 行来更改。
和2米=(200 010 001)(235 121 34−1)=(4610 121 34−1)
矩阵乘法导致米通过将第 1 行乘以 2 来更改。

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。