# 线性代数作业代写linear algebra代考|Lines

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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|Use the formulae

If $A=(5,0,7)$ and $B=(2,-3,6)$, find the points $P$ on the line $A B$ which satisfy $A P / P B=3$. Solution. Use the formulae $$\mathbf{P}=\mathbf{A}+t \overrightarrow{A B} \quad \text { and } \quad\left|\frac{t}{1-t}\right|=\frac{A P}{P B}=3$$ Then $$\frac{t}{1-t}=3 \text { or }-3$$ so $t=\frac{3}{4}$ or $t=\frac{3}{2}$. The corresponding points are $\left(\frac{11}{4}, \frac{9}{4}, \frac{25}{4}\right)$ and $\left(\frac{1}{2}, \frac{9}{2}, \frac{11}{2}\right)$. DEFINITION 8.4.2 Let $X$ and $Y$ be non-zero vectors. Then $X$ is parallel or proportional to $Y$ if $X=t Y$ for some $t \in \mathbb{R}$. We write $X | Y$ if $X$ is parallel to $Y$. If $X=t Y$, we say that $X$ and $Y$ have the same or opposite direction, according as $t>0$ or $t<0$.

DEFINITION 8.4.3 if $A$ and $B$ are distinct points on a line $\mathcal{L}$, the nonzero vector $\overrightarrow{A B}$ is called a direction vector for $\mathcal{L}$. It is easy to prove that any two direction vectors for a line are parallel. DEFINITION 8.4.4 Let $\mathcal{L}$ and $\mathcal{M}$ be lines having direction vectors $X$ and $Y$, respectively. Then $\mathcal{L}$ is parallel to $\mathcal{M}$ if $X$ is parallel to $Y$. Clearly any line is parallel to itself.

It is easy to prove that the line through a given point $A$ and parallel to a given line $C D$ has an equation $\mathbf{P}=\mathbf{A}+t \overrightarrow{C D}$.

## 线性代数作业代写linear algebra代考|Suppose A, B, C, D

Suppose $A, B, C, D$ are distinct points such that no three are collinear. Then $\overrightarrow{A B}=\overrightarrow{C D}$ if and only if $\overrightarrow{A B} | \overrightarrow{C D}$ and $\overrightarrow{A C} | \overrightarrow{B D}$ (See Figure 8.1.) Proof. If $\overrightarrow{A B}=\overrightarrow{C D}$ then \begin{aligned} &\mathbf{B}-\mathbf{A}=\mathbf{D}-\mathbf{C} \ &\mathbf{C}-\mathbf{A}=\mathbf{D}-\mathbf{B} \end{aligned} and so $\overrightarrow{A C}=\overrightarrow{B D}$. Hence $\overrightarrow{A B} | \overrightarrow{C D}$ and $\overrightarrow{A C} | \overrightarrow{B D}$. Conversely, suppose that $\overrightarrow{A B} | \overrightarrow{C D}$ and $\overrightarrow{A C} | \overrightarrow{B D}$. Then $$\overrightarrow{A B}=s \overrightarrow{C D} \quad \text { and } \quad \overrightarrow{A C}=t \overrightarrow{B D},$$ or $$\mathbf{B}-\mathbf{A}=s(\mathbf{D}-\mathbf{C}) \quad \text { and } \quad \mathbf{C}-\mathbf{A}=t \mathbf{D}-\mathbf{B}$$ We have to prove $s=1$ or equivalently, $t=1$. Now subtracting the second equation above from the first, gives $$\mathbf{B}-\mathbf{C}=s(\mathbf{D}-\mathbf{C})-t(\mathbf{D}-\mathbf{B})$$ So $$(1-t) \mathbf{B}=(1-s) \mathbf{C}+(s-t) \mathbf{D}$$ If $t \neq 1$, then $$\mathbf{B}=\frac{1-s}{1-t} \mathbf{C}+\frac{s-t}{1-t} \mathbf{D}$$ and $B$ would lie on the line $C D$. Hence $t=1$.

## 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