线性代数作业代写linear algebra代考|A classification algorithm

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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|Rotate the

Rotate the $\left(x_{1}, y_{1}\right)$ axes with the new positive $x_{2}$-axis in the direction of
$$\begin{gathered} {[(b-a} \ +4 h^{2} \end{gathered}$$
where $R=\sqrt{(a-b)^{2}+4 h^{2}}$. Then equation $7.12$ becomes
$$\lambda_{1} x_{2}^{2}+\lambda_{2} y_{2}^{2}=-\frac{\Delta}{C}$$
where
$$\lambda_{1}=(a+b-R) / 2, \lambda_{2}=(a+b+R) / 2$$
Here $\lambda_{1} \lambda_{2}=C$.
(a) $C<0$ : Hyperbola. Here $\lambda_{2}>0>\lambda_{1}$ and equation $7.13$ becomes
$$\frac{x_{2}^{2}}{u^{2}}-\frac{y_{2}^{2}}{v^{2}}=\frac{-\Delta}{|\Delta|},$$
where
$$u=\sqrt{\frac{|\Delta|}{C \lambda_{1}}}, v=\sqrt{\frac{|\Delta|}{-C \lambda_{2}}}$$
(b) $C>0$ and $a \Delta>0$ : Empty set.
(c) $C>0$ and $a \Delta<0$ : Ellipse.
Here $\lambda_{1}, \lambda_{2}, a, b$ have the same sign and $\lambda_{1} \neq \lambda_{2}$ and equation $7.13$ becomes
$$\frac{x_{2}^{2}}{u^{2}}+\frac{y_{2}^{2}}{v^{2}}=1$$
where
$$u=\sqrt{\frac{\Delta}{-C \lambda_{1}}}, v=\sqrt{\frac{\Delta}{-C \lambda_{2}}}$$

线性代数作业代写linear algebra代考|coefficients

Identify the curve
$$x^{2}+2 x y+y^{2}++2 x+2 y+1=0 .$$
Solution. Here
$$\Delta=\left|\begin{array}{lll} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{array}\right|=0 .$$
Let $x=x_{1}+\alpha, y=y_{1}+\beta$ and substitute in equation $7.16$ to get $\left(x_{1}+\alpha\right)^{2}+2\left(x_{1}+\alpha\right)\left(y_{1}+\beta\right)+\left(y_{1}+\beta\right)^{2}+2\left(x_{1}+\alpha\right)+2\left(y_{1}+\beta\right)+1=0 .$
Then equating the coefficients of $x_{1}$ and $y_{1}$ to 0 gives the same equation
$$2 \alpha+2 \beta+2=0 .$$
Take $\alpha=0, \beta=-1$. Then equation $7.17$ simplifies to
$$x_{1}^{2}+2 x_{1} y_{1}+y_{1}^{2}=0=\left(x_{1}+y_{1}\right)^{2},$$
and in terms of $x, y$ coordinates, equation $7.16$ becomes
$$(x+y+1)^{2}=0, \text { or } x+y+1=0 .$$

线性代数作业代写linear algebra代考|Rotate the

Rotate the (x1,y1) axes with the new positive x2-axis in the direction of
[(b−a +4h2
where R=(a−b)2+4h2. Then equation 7.12 becomes
λ1×22+λ2y22=−ΔC
where
λ1=(a+b−R)/2,λ2=(a+b+R)/2
Here λ1λ2=C.
(a) C<0 : Hyperbola. Here λ2>0>λ1 and equation 7.13 becomes
x22u2−y22v2=−Δ|Δ|,
where
u=|Δ|Cλ1,v=|Δ|−Cλ2
(b) C>0 and aΔ>0 : Empty set.
(c) C>0 and aΔ<0 : Ellipse.
Here λ1,λ2,a,b have the same sign and λ1≠λ2 and equation 7.13 becomes
x22u2+y22v2=1
where
u=Δ−Cλ1,v=Δ−Cλ2

线性代数作业代写linear algebra代考|coefficients

Identify the curve
x2+2xy+y2++2x+2y+1=0.
Solution. Here
Δ=|111 111 111|=0.
Let x=x1+α,y=y1+β and substitute in equation 7.16 to get (x1+α)2+2(x1+α)(y1+β)+(y1+β)2+2(x1+α)+2(y1+β)+1=0.
Then equating the coefficients of x1 and y1 to 0 gives the same equation
2α+2β+2=0.
Take α=0,β=−1. Then equation 7.17 simplifies to
x12+2x1y1+y12=0=(x1+y1)2,
and in terms of x,y coordinates, equation 7.16 becomes
(x+y+1)2=0, or x+y+1=0.

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