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# 数学代写代考| Two  Two Matrices 离散数学

## 数学代写| Two Two Matrices 代考

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## 离散数学代写

Matrices arose in practice as a means of solving a set of linear equations. One of the earliest examples of their use is in a Chinese text dating from between $300 \mathrm{BC}$ and $200 \mathrm{AD}$. The Chinese text showed how matrices could be employed to solve simultaneous equations. Consider the set of equations:
\begin{aligned} &a x+b y=r \ &c x+d y=s \end{aligned}
Matrices $\quad 135$
Then, the coefficients of the linear equations in $x$ and $y$ above may be represented by the matrix $A$, where $A$ is given by
$$\mathrm{A}=\left(\begin{array}{ll} a & b \ c & d \end{array}\right)$$
The linear equations may be represented as the multiplication of the matrix $\mathrm{A}$ and a vector $x$ resulting in a vector $v$ :
$$A x=v$$
The matrix representation of the linear equations and its solution are as follows:
$$\left(\begin{array}{ll} a & b \ c & d \end{array}\right)\left(\begin{array}{l} x \ y \end{array}\right)=\left(\begin{array}{l} r \ s \end{array}\right)$$
The vector $x$ may be calculated by determining the inverse of the matrix $\mathrm{A}$ (provided that its inverse exists). The vector $x$ is then given by
$$x=A^{-1} v$$
The solution to the set of linear equations is then given by
$$\left(\begin{array}{l} x \ y \end{array}\right)=\left(\begin{array}{ll} a & b \ c & d \end{array}\right)^{-1}\left(\begin{array}{l} r \ s \end{array}\right)$$
The inverse of a matrix $\mathrm{A}$ exists if and only if its determinant is non-zero, and if this is the case the vector $x$ is given by
$$\left(\begin{array}{l} x \ y \end{array}\right)=\frac{1}{\operatorname{det} \mathrm{A}}\left(\begin{array}{cc} d & -b \ -c & a \end{array}\right)\left(\begin{array}{l} r \ s \end{array}\right)$$
The determinant of a $2 \times 2$ matrix $A$ is given by
$$\operatorname{det} \mathrm{A}=a d-c b$$
The determinant of a $2 \times 2$ matrix is denoted by
$$\left|\begin{array}{ll} a & b \ c & d \end{array}\right|$$
A key property of determinants is that
$$\operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B)$$
and columns, and is given by
$$\mathrm{A}^{T}=\left(\begin{array}{ll} a & c \ b & d \end{array}\right)$$
The inverse of the matrix $A$ (denoted by $A^{-1}$ ) is given by
$$\mathrm{A}^{-1}=\frac{1}{\operatorname{det} \mathrm{A}}\left(\begin{array}{cc} d & -b \ -c & a \end{array}\right)$$
Further, $\mathrm{A} \cdot \mathrm{A}^{-1}=\mathrm{A}^{-1} \cdot \mathrm{A}=\mathrm{I}$ where $\mathrm{I}$ is the identity matrix of the $2 \times 2$ matrices under multiplication. That is,
$$\mathrm{AA}^{-1}=\mathrm{A}^{-1} \mathrm{~A}=\left(\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}\right)$$
The addition of two $2 \times 2$ matrices $\mathrm{A}$ and $\mathrm{B}$ is given by a matrix wh are the addition of the individual components of $\mathrm{A}$ and $\mathrm{B}$. The addit matrices is commutative, and we have
$$\mathrm{A}+\mathrm{B}=\mathrm{B}+\mathrm{A}=\left(\begin{array}{ll} a+p & b+1 \ c+r & d+s \end{array}\right)$$
where $A, B$ are given by
$$\mathrm{A}=\left(\begin{array}{ll} a & b \ c & d \end{array}\right) \quad \mathrm{B}=\left(\begin{array}{ll} p & q \ r & s \end{array}\right)$$
The identity matrix under addition is given by the matrix whose entrie and it has the property that $\mathrm{A}+0=0+\mathrm{A}=\mathrm{A}$.
$$\left(\begin{array}{ll} 0 & 0 \ 0 & 0 \end{array}\right)$$
The multiplication of two $2 \times 2$ matrices is given by
$$\mathrm{AB}=\left(\begin{array}{ll} a p+b r & a q+b s \ c p+d r & c q+d s \end{array}\right)$$
The multiplication of matrices is not commutative: i.e. $A B \neq B A$. plicative identity matrix I has the property that $\mathrm{A} \cdot \mathrm{I}=\mathrm{I} \cdot \mathrm{A}=\mathrm{A}$, and it

$$\开始{对齐} &a x+b y=r \ &c x+d y=s \end{对齐}$$

$$\mathrm{A}=\left(\begin{数组}{ll} a & b \ 开发 \end{数组}\右）$$

$$一个\下划线{x}=v$$

$$\left(\begin{数组}{ll} a & b \ 开发 \end{array}\right)\left(\begin{array}{l} X \ 是的 \end{array}\right)=\left(\begin{array}{l} r \ s \end{数组}\右）$$

$$\下划线{x}=A^{-1} \下划线{v}$$

$$\left(\begin{数组}{l} X \ 是的 \end{array}\right)=\left(\begin{array}{ll} a & b \ 开发 \end{array}\right)^{-1}\left(\begin{array}{l} r \ s \end{数组}\右）$$

$$\left(\begin{数组}{l} X \ 是的 \end{array}\right)=\frac{1}{\operatorname{det} \mathrm{A}}\left(\begin{array}{cc} D b \ -c & a \end{array}\right)\left(\begin{array}{l} r \ s \end{数组}\右）$$
$2 \times 2$ 矩阵 $A$ 的行列式由下式给出
$$\operatorname{det} \mathrm{A}=a d-c b$$
$2 \times 2$ 矩阵的行列式表示为
$$\left|\begin{数组}{ll} a & b \ 开发 \end{数组}\right|$$

$$\operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B)$$

$$\mathrm{A}^{T}=\left(\begin{array}{ll} 一个 & c \ b&d \end{数组}\右）$$

$$\mathrm{A}^{-1}=\frac{1}{\operatorname{det} \mathrm{A}}\left(\begin{array}{cc} D b \ -c & a \end{数组}\右）$$

$$\mathrm{AA}^{-1}=\mathrm{A}^{-1} \mathrm{~A}=\left(\begin{array}{ll} 1 & 0 \ 0 & 1 \end{数组}\右）$$

$$\mathrm{A}+\mathrm{B}=\mathrm{B}+\mathrm{A}=\left(\begin{array}{ll} a+p & b+1 \ c+r & d+s \end{数组}\右）$$

$$\mathrm{A}=\left(\begin{数组}{ll} a & b \ 开发 \end{数组}\right) \quad \mathrm{B}=\left(\begin{array}{ll} p&q \ r&s \end{数组}\右）$$

$$\left(\begin{数组}{ll} 0 & 0 \ 0 & 0 \end{数组}\右）$$

$$\mathrm{AB}=\left(\begin{数组}{ll} a p+b r & a q+b s \ c p+d r & c q+d s \end{数组}\右）$$

## 密码学代考

• 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

1. 数据压缩（或信源编码
2. 前向错误更正（或信道编码
3. 加密编码
4. 线路码