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。

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