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# 数学代考|线性代数代考LINEAR ALGEBRA代考|MATH270 APPLICATIONS OF LINEAR SYSTEMS

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## 数学代考|线性代数代考LINEAR ALGEBRA代考|A Homogeneous System in Econo

The system of 500 equations in 500 variables, mentioned in this chapter’s introduction, is now known as a Leontief “input-output” (or “production”) model. ${ }^{1}$ Section $2.6$ will examine this model in more detail, when more theory and better notation are available. For now, we look at a simpler “exchange model,” also due to Leontief.

Suppose a nation’s economy is divided into many sectors, such as various manufacturing, communication, entertainment, and service industries. Suppose that for each sector we know its total output for one year and we know exactly how this output is divided or “exchanged” among the other sectors of the economy. Let the total dollar value of a sector’s output be called the price of that output. Leontief proved the following result.
There exist equilibrium prices that can be assigned to the total outputs of the various sectors in such a way that the income of each sector exactly balances its expenses.
The following example shows how to find the equilibrium prices.

## 数学代考|线性代数代考LINEAR ALGEBRA代考|Balancing Chemical Equations

Chemical equations describe the quantities of substances consumed and produced by chemical reactions. For instance, when propane gas burns, the propane $\left(\mathrm{C}{3} \mathrm{H}{8}\right)$ combines with oxygen $\left(\mathrm{O}{2}\right)$ to form carbon dioxide $\left(\mathrm{CO}{2}\right)$ and water $\left(\mathrm{H}{2} \mathrm{O}\right)$, according to an equation of the form $$\left(x{1}\right) \mathrm{C}{3} \mathrm{H}{8}+\left(x_{2}\right) \mathrm{O}{2} \rightarrow\left(x{3}\right) \mathrm{CO}{2}+\left(x{4}\right) \mathrm{H}{2} \mathrm{O}$$ To “balance” this equation, a chemist must find whole numbers $x{1}, \ldots, x_{4}$ such that the total numbers of carbon $(C)$, hydrogen $(H)$, and oxygen $(O)$ atoms on the left match the corresponding numbers of atoms on the right (because atoms are neither destroyed nor created in the reaction).

A systematic method for balancing chemical equations is to set up a vector equation that describes the numbers of atoms of each type present in a reaction. Since equation (4) involves three types of atoms (carbon, hydrogen, and oxygen), construct a vector in $\mathbb{R}^{3}$ for each reactant and product in (4) that lists the numbers of “atoms per molecule,” as follows:
\mathrm{C}{3} \mathrm{H}{8}:\left[\begin{array}{l} 3 \ 8 \ 0 \end{array}\right], \mathrm{O}{2}:\left[\begin{array}{l} 0 \ 0 \ 2 \end{array}\right], \mathrm{CO}{2}:\left[\begin{array}{l} 1 \ 0 \ 2 \end{array}\right], \mathrm{H}{2} \mathrm{O}:\left[\begin{array}{l} 0 \ 2 \ 1 \end{array}\right] \begin{aligned} &+\text { Carbon } \ &\leftarrow \text { Hydrogen } \end{aligned} To balance equation (4), the coefficients $x{1}, \ldots, x_{4}$ must satisfy
$$x_{1}\left[\begin{array}{l} 3 \ 8 \ 0 \end{array}\right]+x_{2}\left[\begin{array}{l} 0 \ 0 \ 2 \end{array}\right]=x_{3}\left[\begin{array}{l} 1 \ 0 \ 2 \end{array}\right]+x_{4}\left[\begin{array}{l} 0 \ 2 \ 1 \end{array}\right]$$
To solve, move all the terms to the left (changing the signs in the third and fourth vectors):
$$x_{1}\left[\begin{array}{l} 3 \ 8 \ 0 \end{array}\right]+x_{2}\left[\begin{array}{l} 0 \ 0 \ 2 \end{array}\right]+x_{3}\left[\begin{array}{r} -1 \ 0 \ -2 \end{array}\right]+x_{4}\left[\begin{array}{r} 0 \ -2 \ -1 \end{array}\right]=\left[\begin{array}{l} 0 \ 0 \ 0 \end{array}\right]$$
Row reduction of the augmented matrix for this equation leads to the general solution
$$x_{1}=\frac{1}{4} x_{4}, x_{2}=\frac{5}{4} x_{4}, x_{3}=\frac{3}{4} x_{4} \text {, with } x_{4} \text { free }$$
Since the coefficients in a chemical equation must be integers, take $x_{4}=4$, in which case $x_{1}=1, x_{2}=5$, and $x_{3}=3$. The balanced equation is
$$\mathrm{C}{3} \mathrm{H}{8}+5 \mathrm{O}{2} \rightarrow 3 \mathrm{CO}{2}+4 \mathrm{H}_{2} \mathrm{O}$$
The equation would also be balanced if, for example, each coefficient were doubled. For most purposes, however, chemists prefer to use a balanced equation whose coefficients are the smallest possible whole numbers.

## 数学代考线性代数代考LINEAR ALGEBRA代考|Balancing Chemical Equations

$$(x 1) \mathrm{C} 3 \mathrm{H} 8+\left(x_{2}\right) \mathrm{O} 2 \rightarrow(x 3) \mathrm{CO} 2+(x 4) \mathrm{H} 2 \mathrm{O}$$

C3H8 : [3 80$], \mathrm{O} 2:[002], \mathrm{CO} 2:[102], \mathrm{H} 2 \mathrm{O}:[021]+$ Carbon 5 Hydrogen 为了平衡等式 $(4)$ ，系数 $x 1, \ldots, x_{4}$ 必须满足
$$x_{1}\left[\begin{array}{lll} 3 & 8 & 0 \end{array}\right]+x_{2}\left[\begin{array}{lll} 0 & 0 & 2 \end{array}\right]=x_{3}\left[\begin{array}{lll} 1 & 0 & 2 \end{array}\right]+x_{4}\left[\begin{array}{lll} 0 & 2 & 1 \end{array}\right]$$

$$x_{1}\left[\begin{array}{lll} 3 & 8 & 0 \end{array}\right]+x_{2}\left[\begin{array}{lll} 0 & 0 & 2 \end{array}\right]+x_{3}\left[\begin{array}{lll} -1 & 0 & -2 \end{array}\right]+x_{4}[0-2-1]=\left[\begin{array}{lll} 0 & 0 & 0 \end{array}\right]$$

$$x_{1}=\frac{1}{4} x_{4}, x_{2}=\frac{5}{4} x_{4}, x_{3}=\frac{3}{4} x_{4}, \text { with } x_{4} \text { free }$$

$$\mathrm{C} 3 \mathrm{H} 8+5 \mathrm{O} 2 \rightarrow 3 \mathrm{CO} 2+4 \mathrm{H}_{2} \mathrm{O}$$

## 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。