### 数学代写|线性代数代写linear algebra代考|MTH2106

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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|ROW REDUCTION AND ECHELON FORMS

This section refines the method of Section $1.1$ into a row reduction algorithm that will enable us to analyze any system of linear equations. ${ }^1$ By using only the first part of the algorithm, we will be able to answer the fundamental existence and uniqueness questions posed in Section 1.1.

The algorithm applies to any matrix, whether or not the matrix is viewed as an augmented matrix for a linear system. So the first part of this section concerns an arbitrary rectangular matrix and begins by introducing two important classes of matrices that include the “triangular” matrices of Section 1.1. In the definitions that follow, a nonzero row or column in a matrix means a row or column that contains at least one nonzero entry; a leading entry of a row refers to the leftmost nonzero entry (in a nonzero row).

An echelon matrix (respectively, reduced echelon matrix) is one that is in echelon form (respectively, reduced echelon form). Property 2 says that the leading entries form an echelon (“steplike”) pattern that moves down and to the right through the matrix. Property 3 is a simple consequence of property 2 , but we include it for emphasis.
The “triangular” matrices of Section 1.1, such as
$$\left[\begin{array}{rrrc} 2 & -3 & 2 & 1 \ 0 & 1 & -4 & 8 \ 0 & 0 & 0 & 5 / 2 \end{array}\right] \text { and }\left[\begin{array}{lllr} 1 & 0 & 0 & 29 \ 0 & 1 & 0 & 16 \ 0 & 0 & 1 & 3 \end{array}\right]$$
are in echelon form. In fact, the second matrix is in reduced echelon form. Here are additional examples.

EXAMPLE 1 The following matrices are in echelon form. The leading entries ( $\boldsymbol{)}$ ) may have any nonzero value; the starred entries $(*)$ may have any value (including zero).

## 数学代写|线性代数代写linear algebra代考|Solutions of Linear Systems

The row reduction algorithm leads directly to an explicit description of the solution set of a linear system when the algorithm is applied to the augmented matrix of the system.
Suppose, for example, that the augmented matrix of a linear system has been changed into the equivalent reduced echelon form
$$\left[\begin{array}{rrrr} 1 & 0 & -5 & 1 \ 0 & 1 & 1 & 4 \ 0 & 0 & 0 & 0 \end{array}\right]$$
There are three variables because the augmented matrix has four columns. The associated system of equations is
$$\begin{array}{r} x_1-5 x_3=1 \ x_2+x_3=4 \ 0=0 \end{array}$$
The variables $x_1$ and $x_2$ corresponding to pivot columns in the matrix are called basic variables. ${ }^2$ The other variable, $x_3$, is called a free variable.

Whenever a system is consistent, as in (4), the solution set can be described explicitly by solving the reduced system of equations for the basic variables in terms of the free variables. This operation is possible because the reduced echelon form places each basic variable in one and only one equation. In (4), solve the first equation for $x_1$ and the second for $x_2$. (Ignore the third equation; it offers no restriction on the variables.)
$$\left{\begin{array}{l} x_1=1+5 x_3 \ x_2=4-x_3 \ x_3 \text { is free } \end{array}\right.$$
The statement ” $x_3$ is free” means that you are free to choose any value for $x_3$. Once that is done, the formulas in (5) determine the values for $x_1$ and $x_2$. For instance, when $x_3=0$, the solution is $(1,4,0)$; when $x_3=1$, the solution is $(6,3,1)$. Each different choice of $x_3$ determines a (different) solution of the system, and every solution of the system is determined by a choice of $x_3$.

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|ROW REDUCTION AND ECHELON FORMS

$1.1$ 节的“三角”矩阵，如

## 数学代写|线性代数代写linear algebra代考|Solutions of Linear Systems

$$\left[\begin{array}{llllllllllll} 1 & 0 & -5 & 1 & 0 & 1 & 1 & 4 & 0 & 0 & 0 & 0 \end{array}\right]$$

$$x_1-5 x_3=1 x_2+x_3=40=0$$

x_1=1+5 x_3 x_2=4-x_3 x_3 \text { is free }
$$正确的。 \ \$$

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## MATLAB代写

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