数学代写代考| Gaussian Elimination 离散数学代写 - 代考代写：100%准时可靠您的作业代写专家

Gaussian elimination with backward substitution is an important method used in solving a set of linear equations. A matrix is used to represent the set of linear equations, and Gaussian elimination reduces the matrix to a triangular or reduced form, which may then be solved by backward substitution.

This allows the set of $n$ linear equations $\left(\mathrm{E}{1}\right.$ to $\left.\mathrm{E}{n}\right)$ defined below to be solved by applying operations to the equations to reduce the matrix to triangular form. This reduced form is easier to solve, and it provides exactly the original set of equations. The set of equations is defined as
$$
\begin{array}{cccccccc}
E_{1}: & a_{11} x_{1} & + & a_{12} x_{2} & + & \cdots & + & a_{1 n} x_{n} & = & b_{1} \
E_{2}: & a_{21} x_{1} & + & a_{22} x_{2} & + & \cdots & + & a_{2 n} x_{n} & = & b_{1} \
\vdots & \vdots & & \vdots & & & & \vdots & & \vdots \
E_{n}: & a_{n 1} x_{1} & + & a_{n 2} x_{2} & + & \cdots & + & a_{n n} x_{n} & = & b_{n}
\end{array}
$$
Three operations are permitted on the equations, and these operations transform the linear system into a reduced form. They are
(a) Any equation may be multiplied by a non-zero constant.
(b) An equation $E_{i}$ may be multiplied by a constant and added to another equation $E_{j}$, with the resulting equation replacing $E_{j}$.
(c) Equations $E_{i}$ and $E_{j}$ may be transposed with $E_{j}$ replacing $E_{i}$ and vice versa.
This method for solving a set of linear equations is best illustrated by an example, and we consider an example taken from [1]. Then, the solution to a set of linear equations with four unknowns may be determined as follows:
$8.6$ Gaussian Elimination
143
$E_{1}: \quad x_{1}+x_{2}+3 x_{4}=4$
$E_{2}: \quad 2 x_{1}+x_{2}-x_{3}+x_{4}=1$
$E_{3}: \quad 3 x_{1}-x_{2}-x_{3}+2 x_{4}=-3$
$E_{4}: \quad-x_{1}+2 x_{2}+3 x_{3}-x_{4}=4$.
First, the unknown $x_{1}$ is eliminated from $E_{2}, E_{3}$ and $E_{4}$ and this is done by replacing $E_{2}$ with $E_{2}-2 E_{1}$; replacing $E_{3}$ with $E_{3}-3 E_{1}$; and replacing $E_{4}$ with $E_{4}+E_{1}$. The resulting system is
$\begin{aligned}&E_{1}: \quad x_{1}+x_{2}+3 x_{4}=4 \&E_{2}: \quad-x_{2}-x_{3}-5 x_{4}=-7 \&E_{3}: \quad-4 x_{2}-x_{3}-7 x_{4}=-15 \&E_{4}: & 3 x_{2}+3 x_{3}+2 x_{4}=8 .\end{aligned}$ with $E_{3}-4 E_{2}$ and replacing $E_{4}$ with $E_{4}+3 E_{2}$. The resulting system is now in triangular form and the unknown variable may be solved easily by backward substitution. That is, we first use equation $E_{4}$ to find the solution to $x_{4}$ and then we use equation $E_{3}$ to find the the solutions to $x_{2}$ and $x_{1} .$
$$
\begin{aligned}
E_{1}: & & x_{1}+x_{2} &+3 x_{4} &=4 \
E_{2}: & & &-x_{2}-x_{3}-5 x_{4} &=-7 \
E_{3}: & & 3 x_{3}+13 x_{4} &=13 \
E_{4}: & & &-13 x_{4} &=-13 .
\end{aligned}
$$

反向代入高斯消元法是求解一组线性方程组的重要方法。矩阵用于表示线性方程组，高斯消元将矩阵简化为三角形或简化形式，然后可以通过反向替换来求解。

这允许求解下面定义的 $n$ 线性方程组 $\left(\mathrm{E}{1}\right.$ to $\left.\mathrm{E}{n}\right)$通过对方程应用运算以将矩阵简化为三角形形式。这种简化形式更容易求解，而且它提供的正是原始方程组。方程组定义为
$$
\开始{数组}{cccccccc}
E_{1}: & a_{11} x_{1} & + & a_{12} x_{2} & + & \cdots & + & a_{1 n} x_{n} & = & b_{1} \ \
E_{2}: & a_{21} x_{1} & + & a_{22} x_{2} & + & \cdots & + & a_{2 n} x_{n} & = & b_{1} \ \
\vdots & \vdots & & \vdots & & & & \vdots & & \vdots \
E_{n}: & a_{n 1} x_{1} & + & a_{n 2} x_{2} & + & \cdots & + & a_{n n} x_{n} & = & b_{n}
\结束{数组}
$$
允许对方程进行三个操作，这些操作将线性系统转换为简化形式。他们是
(a) 任何方程都可以乘以一个非零常数。
(b) 方程$E_{i}$ 可以乘以一个常数并添加到另一个方程$E_{j}$，结果方程代替$E_{j}$。
(c) 方程 $E_{i}$ 和 $E_{j}$ 可以用 $E_{j}$ 代替 $E_{i}$ 进行转置，反之亦然。
这种求解一组线性方程的方法最好用一个例子来说明，我们考虑一个取自 [1] 的例子。然后，一组具有四个未知数的线性方程组的解可以确定如下：
$8.6$ 高斯消除
143
$E_{1}: \quad x_{1}+x_{2}+3 x_{4}=4$
$E_{2}: \quad 2 x_{1}+x_{2}-x_{3}+x_{4}=1$
$E_{3}: \quad 3 x_{1}-x_{2}-x_{3}+2 x_{4}=-3$
$E_{4}: \quad-x_{1}+2 x_{2}+3 x_{3}-x_{4}=4$。
首先，未知的 $x_{1}$ 从 $E_{2}、E_{3}$ 和 $E_{4}$ 中消除，这是通过将 $E_{2}$ 替换为 $E_{2}- 来完成的 – 2 E_{1}$;将 $E_{3}$ 替换为 $E_{3}-3 E_{1}$;并将 $E_{4}$ 替换为 $E_{4}+E_{1}$。结果系统是
$\begin{对齐}&E_{1}: \quad x_{1}+x_{2}+3 x_{4}=4 \&E_{2}: \quad-x_{2}-x_{3}- 5 x_{4}=-7 \&E_{3}: \quad-4 x_{2}-x_{3}-7 x_{4}=-15 \&E_{4}: & 3 x_{2} +3 x_{3}+2 x_{4}=8 .\end{aligned}$ 与 $E_{3}-4 E_{2}$ 并用 $E_{4}+3 替换 $E_{4}$ E_{2}$。得到的系统现在是三角形的，未知变量可以很容易地通过向后替换来解决。也就是说，我们首先使用方程 $E_{4}$ 找到 $x_{4}$ 的解，然后我们使用方程 $E_{3}$ 找到 $x_{2}$ 和 $x_ 的解{1} .$
$$
\开始{对齐}
E_{1}: & & x_{1}+x_{2} &+3 x_{4} &=4 \
E_{2}: & & &-x_{2}-x_{3}-5 x_{4} &=-7 \
E_{3}: & & 3 x_{3}+13 x_{4} &=13 \
E_{4}: & & &-13 x_{4} &=-13 。
\end{对齐}
$$

图论代考

自然数 $\mathbb{N}$ 由数字 $\{1,2,3, \ldots\}$ 组成。整数 $\mathbb{Z}$ 由 $\{\ldots-2,-1,0,1,2, \ldots\}$ 组成。有理数 $\mathbb{Q}$ 由 $\left\{{ }^{p} /_{q}\right.$ 形式的所有数字组成，其中 $p$ 和 $q$ 是整数，$ \left.q \neq 0\right\}$。实数 $\mathbb{R}$ 被定义为有理数收敛序列的集合，它们是有理数的超集。它们包含有理数和无理数。复数 $\mathbb{C}$ 由 $\{a+bi$ 形式的所有数字组成，其中 $a, b \in \mathbb{R}$ 和 $i=\sqrt{-} 1\}美元。毕达哥拉斯三元组（图 3.2）是满足毕达哥拉斯方程 $x^{2}+y^{2}=z^{2}$ 的三个整数的组合。有无数个这样的三元组，这种三元组的一个例子是 $3,4,5$，因为 $3^{2}+4^{2}=5^{2}$。毕达哥拉斯学派发现了音乐和数字之间的数学关系，他们的哲学是数字隐藏在从音乐到科学和自然的一切事物中。这导致了他们的哲学，即“一切都是数字”。

数学代写| DISCRETE MATHEMATICS代考请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。

Cryptosystem
A system that describes how to encrypt or decrypt messages
Plaintext
Message in its original form
Ciphertext
Message in its encrypted form
Cryptographer
Invents encryption algorithms
Cryptanalyst
Breaks encryption algorithms or implementations

编码理论代写

编码理论（英语：Coding theory）是研究编码的性质以及它们在具体应用中的性能的理论。编码用于数据压缩、加密、纠错，最近也用于网络编码中。不同学科（如信息论、电机工程学、数学、语言学以及计算机科学）都研究编码是为了设计出高效、可靠的数据传输方法。这通常需要去除冗余并校正（或检测）数据传输中的错误。

编码共分四类：^[1]