# 数学代写|代数学代写Algebra代考|Math4120

## 数学代写|代数学代写Algebra代考|Matrix Equations

One of the primary uses of matrices is that they give us a way of working with linear systems much more compactly and cleanly. In particular, any system of linear equations
\begin{aligned} a_{1,1} x_1+a_{1,2} x_2+\cdots+a_{1, n} x_n &=b_1 \ a_{2,1} x_1+a_{2,2} x_2+\cdots+a_{2, n} x_n &=b_2 \ & \vdots \ a_{m, 1} x_1+a_{m, 2} x_2+\cdots+a_{m, n} x_n &=b_m \end{aligned}
can be rewritten as the single matrix equation $A \mathbf{x}=\mathbf{b}$, where $A \in \mathcal{M}{m, n}$ is the coefficient matrix whose $(i, j)$-entry is $a{i, j}, \mathbf{b}=\left(b_1, b_2, \ldots, b_m\right) \in \mathbb{R}^m$ is a vector containing the constants from the right-hand side, and $\mathbf{x}=\left(x_1, x_2, \ldots, x_n\right) \in \mathbb{R}^n$ is a vector containing the variables.

The advantage of writing linear systems in this way (beyond the fact that it requires less writing) is that we can now make use of the various properties of matrices and matrix multiplication that we already know to help us understand linear systems a bit better. For example, we can now prove the observation that we made earlier: every linear system has either zero, one, or infinitely many solutions.
Every system of linear equations has either
a) no solutions,
b) exactly one solution, or
c) infinitely many solutions.
Proof. Another way of phrasing this theorem is as follows: if a system of linear equations has at least two solutions then it must have infinitely many solutions. With this in mind, we start by assuming that there are two distinct solutions to the linear system.

If $A \mathbf{x}=\mathbf{b}$ is the matrix form of the linear system (where $A \in \mathcal{M}_{m, n}, \mathbf{x} \in \mathbb{R}^n$, and $\mathbf{b} \in \mathbb{R}^m$ ), then there existing two distinct solutions of the linear system means that there exist vectors $\mathbf{x}_1 \neq \mathbf{x}_2 \in \mathbb{R}^n$ such that $A \mathbf{x}_1=\mathbf{b}$ and $A \mathbf{x}_2=\mathbf{b}$. Then for any scalar $c \in \mathbb{R}$, it is the case that
$$A\left((1-c) \mathbf{x}_1+c \mathbf{x}_2\right)=(1-c) A \mathbf{x}_1+c A \mathbf{x}_2=(1-c) \mathbf{b}+c \mathbf{b}=\mathbf{b},$$
so every vector of the form $(1-c) \mathbf{x}_1+c \mathbf{x}_2$ is a solution of the linear system. Since there are infinitely many such vectors (one for each choice of $c \in \mathbb{R}$ ), the proof is complete.

## 数学代写|代数学代写Algebra代考|The Inverse of a Matrix

One of the nice features of the elementary matrices is that they are invertible: multiplication by them can be undone by multiplying by some other matrix. For example, if
$$E_1=\left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ -3 & 0 & 1 \end{array}\right] \text { and } E_2=\left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 3 & 0 & 1 \end{array}\right] \text {, }$$
then it is straightforward to check that $E_1 E_2=I$ and $E_2 E_1=I$. We thus say that $E_1$ and $E_2$ are inverses of each other, which makes sense in this case because $E_1$ is the elementary matrix corresponding to the elementary row operation $R_3-3 R_1$, and $E_2$ is the elementary matrix corresponding to the elementary row operation $R_3+3 R_1$, and performing these row operations one after another has the same effect as not doing anything at all.

It is also useful to talk about invertibility of (not necessarily elementary) matrices in general, and the idea is exactly the same – two matrices are inverses of each other if multiplying by one of them “undoes” the multiplication by the other.

A square matrix $A \in \mathcal{M}_n$ is called invertible if there exists a matrix, which we denote by $A^{-1}$ and call the inverse of $A$, such that
$$A A^{-1}=A^{-1} A=I .$$
In this definition, we referred to $A^{-1}$ as the inverse of $A$ (as opposed to an inverse of $A$ ). To justify this terminology, we should show that every matrix has at most one inverse. To see this, suppose for a moment that a matrix $A \in \mathcal{M}_n$ had two inverses $B, C \in \mathcal{M}_n$ (i.e., $A B=B A=I$ and $A C=C A=I$ ). It would then follow that
$$B=I B=(C A) B=C(A B)=C I=C,$$
so in fact these two inverses must be the same as each other. It follows that inverses (when they exist) are indeed unique.

Given a pair of matrices, it is straightforward to check whether or not they are inverses of each other-just multiply them together and see if we get the identity matrix.

# 代数学代考

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## 数学代写|代数学代写Algebra代考|Systems of Linear equation

.

$n$变量$x_1, x_2, \ldots, x_n$中的线性方程可以写成
$$a_1 x_1+a_2 x_2+\cdots+a_n x_n=b,$$
，其中$a_1, a_2, \ldots, a_n$和$b$是称为线性方程系数的常数。

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