# 线性代数作业代写linear algebra代考| Least squares solution of equations

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• 数值分析
• 高等线性代数
• 矩阵论
• 优化理论
• 线性规划
• 逼近论

## 线性代数作业代写linear algebra代考|equations represented

Suppose $A X=B$ represents a system of linear equations with real coefficients which may be inconsistent, because of the possibility of experimental errors in determining $A$ or $B$. For example, the system
\begin{aligned} x &=1 \ y &=2 \ x+y &=3.001 \end{aligned}
is inconsistent.
It can be proved that the associated system $A^{t} A X=A^{t} B$ is always consistent and that any solution of this system minimizes the sum $r_{1}^{2}+\ldots+$ $r_{m}^{2}$, where $r_{1}, \ldots, r_{m}$ (the residuals) are defined by
$$r_{i}=a_{i 1} x_{1}+\ldots+a_{i n} x_{n}-b_{i},$$
for $i=1, \ldots, m$. The equations represented by $A^{t} A X=A^{t} B$ are called the normal equations corresponding to the system $A X=B$ and any solution of the system of normal equations is called a least squares solution of the original system.

## 线性代数作业代写linear algebra代考|The points have to satisfy

Points $\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)$ are experimentally determined and should lie on a line $y=m x+c$. Find a least squares solution to the problem.
Solution. The points have to satisfy
\begin{aligned} m x_{1}+c=& y_{1} \ & \vdots \ m x_{n}+c=& y_{n}, \end{aligned}
or $A x=B$, where
$$A=\left[\begin{array}{cc} x_{1} & 1 \ \vdots & \vdots \ x_{n} & 1 \end{array}\right], X=\left[\begin{array}{c} m \ c \end{array}\right], B=\left[\begin{array}{c} y_{1} \ \vdots \ y_{n} \end{array}\right]$$
The normal equations are given by $\left(A^{t} A\right) X=A^{t} B$. Here
$A^{t} A=\left[\begin{array}{ccc}x_{1} & \ldots & x_{n} \ 1 & \ldots & 1\end{array}\right]\left[\begin{array}{cc}x_{1} & 1 \ \vdots & \vdots \ x_{n} & 1\end{array}\right]=\left[\begin{array}{cc}x_{1}^{2}+\ldots+x_{n}^{2} & x_{1}+\ldots+x_{n} \ x_{1}+\ldots+x_{n} & n\end{array}\right]$
$$A^{t} A=\left[\begin{array}{ccc} x_{1} & \ldots & x_{n} \ 1 & \ldots & 1 \end{array}\right]\left[\begin{array}{cc} x_{1} & 1 \ \vdots & \vdots \ x_{n} & 1 \end{array}\right]=\left[\begin{array}{cc} x_{1}^{2}+\ldots+x_{n}^{2} & x_{1}+\ldots+x_{n} \ x_{1}+\ldots+x_{n} & n \end{array}\right]$$
It is not difficult to prove that
prove that
$$\Delta=\operatorname{det}\left(A^{t} A\right)=\sum_{1 \leq i<j \leq n}\left(x_{i}-x_{j}\right)^{2},$$
which is positive unless $x_{1}=\ldots=x_{n}$. Hence if not all of $x_{1}, \ldots, x_{n}$ are equal, $A^{t} A$ is non-singular and the normal equations have a unique solution. This can be shown to be
$$m=\frac{1}{\Delta} \sum_{1 \leq i<j \leq n}\left(x_{i}-x_{j}\right)\left(y_{i}-y_{j}\right), c=\frac{1}{\Delta} \sum_{1 \leq i<j \leq n}\left(x_{i} y_{j}-x_{j} y_{i}\right)\left(x_{i}-x_{j}\right) .$$

## 线性代数作业代写linear algebra代考|equations represented

X=1 和=2 X+和=3.001

r一世=一种一世1X1+…+一种一世nXn−b一世,

## 线性代数作业代写linear algebra代考|The points have to satisfy

Δ=这⁡(一种吨一种)=∑1≤一世<j≤n(X一世−Xj)2,

# 计量经济学代写

## 在这种情况下，如何学好线性代数？如何保证线性代数能获得高分呢？

1.1 mark on book

【重点的误解】划重点不是书上粗体，更不是每个定义，线代概念这么多，很多朋友强迫症似的把每个定义整整齐齐用荧光笔标出来，然后整本书都是重点，那期末怎么复习呀。我认为需要标出的重点为

A. 不懂，或是生涩，或是不熟悉的部分。这点很重要，有的定义浅显，但证明方法很奇怪。我会将晦涩的定义，证明方法标出。在看书时，所有例题将答案遮住，自己做，卡住了就说明不熟悉这个例题的方法，也标出。

B. 老师课上总结或强调的部分。这个没啥好讲的，跟着老师走就对了

C. 你自己做题过程中，发现模糊的知识点

1.2 take note

1.3 understand the relation between definitions