## 张祖锦2023年数学专业真题分类70天之第70天
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1588、 9、 (16 分) 设 $\displaystyle V$ 是 $\displaystyle n$ 维欧氏空间, $\displaystyle V\_1,V\_2$ 是 $\displaystyle V$ 的子空间, 且 $\displaystyle \dim V\_1 < \dim V\_2$. 证明: 存在 $\displaystyle 0\neq \alpha\in V\_2$ 使得 $\displaystyle \alpha\perp V\_1$. (中南大学2023年高等代数考研试题) [欧氏空间 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} \dim (V\_2\cap V\_1^\perp)&=\dim V\_2+\dim V\_1^\perp-\dim (V\_2\cap V\_1^\perp)\\\\ &=\dim V\_2+(n-\dim V\_1)-\dim (V\_2\cap V\_1^\perp)\\\\ &=(\dim V\_2-\dim V\_1)+\left\[n-\dim (V\_2\cap V\_1^\perp)\right\]\\\\ &\geq \dim V\_2-\dim V\_1 > 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \exists\ 0\neq \alpha\in V\_2\cap V\_1^\perp$. 此 $\displaystyle \alpha$ 即满足题意.跟锦数学微信公众号. [在线资料](
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1589、 10、 (15 分)设 $\displaystyle A$ 为 $\displaystyle 3$ 阶正定矩阵, $\displaystyle v\_1,\cdots,v\_k$ 为 $\displaystyle \mathbb\{R\}^3$ 中的 $\displaystyle k$ 个列向量, 满足
\begin\{aligned\} \forall\ i,j\in\left\\{1,\cdots,k\right\\}: i\neq j, v\_i^\mathrm\{T\} Av\_j\leq-\frac\{1\}\{2023\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle v\_i^\mathrm\{T\}$ 表示 $\displaystyle v\_i$ 的转置. 证明: $\displaystyle k\leq 4$. (中山大学2023年高等代数考研试题) [欧氏空间 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 我们给出一般的结论. 在 $\displaystyle n$ 维欧几里得空间中, 设 $\displaystyle \alpha\_1,\cdots,\alpha\_m$ 中的任意两个不同向量之间的夹角均为钝角. 证明: $\displaystyle m\leq n+1$. 用反证法. 若 $\displaystyle m\geq n+2$, 则
\begin\{aligned\} (\alpha\_i,\alpha\_j) < 0, \forall\ 1\leq i\neq j\leq m. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们可以证明其中任意 $\displaystyle n+1$ 个向量线性无关. 比如我们证明 $\displaystyle \alpha\_1,\cdots,\alpha\_\{n+1\}$ 线性无关. 用反证法. 若存在不全为 $\displaystyle 0$ 的 $\displaystyle k\_i\in\mathbb\{R\}$, 使得
\begin\{aligned\} \sum\_\{i=1\}^\{n+1\}k\_i\alpha\_i=0\Rightarrow \sum\_\{i=1\}^\{n+1\}k\_i(\alpha\_i,\alpha\_\{n+2\})=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
注意到 $\displaystyle (\alpha\_i,\alpha\_\{n+2\}) < 0$, 我们知 $\displaystyle k\_1,\cdots,k\_\{n+1\}$ 有正有负,
\begin\{aligned\} &\sum\_\{k\_i < 0\}k\_i\alpha\_i+\sum\_\{k\_i > 0\}k\_i\alpha\_i=0\\\\ \Rightarrow&\alpha=\sum\_\{k\_i > 0\}k\_i\alpha\_i=-\sum\_\{k\_i < 0\}k\_i\alpha\_i\\\\ \Rightarrow&(\alpha,\alpha)=\left(\sum\_\{k\_i > 0\}k\_i\alpha\_i, -\sum\_\{k\_j < 0\}k\_j\alpha\_j\right) =-\sum\_\{k\_i > 0\}\sum\_\{k\_j < 0\}k\_ik\_j(\alpha\_i,\alpha\_j) < 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这是一个矛盾. 故 $\displaystyle \alpha\_1,\cdots,\alpha\_\{n+1\}$ 线性无关, $\displaystyle \dim V\geq n+1$, 这与题设 $\displaystyle \dim V=n$ 矛盾. 故有结论. (2)、 回到题目. 考虑 $\displaystyle \mathbb\{R\}^3$ 在内积 $\displaystyle (v,w)=v^\mathrm\{T\} Aw$ 下构成欧氏空间, 而由第 1 步知 $\displaystyle k\leq 3+1=4$.跟锦数学微信公众号. [在线资料](
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1590、 12、 (15 分) 在 $\displaystyle \mathbb\{R\}^4$ 上规定内积
\begin\{aligned\} \left((a\_1,a\_2,a\_3,a\_4)^\mathrm\{T\}, (b\_1,b\_2,b\_3,b\_4)^\mathrm\{T\}\right)=\sum\_\{i=1\}^4 a\_ib\_i. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
已知 $\displaystyle \gamma\_1=(1,0,0,0)^\mathrm\{T\}, \gamma\_2=\left(0,\frac\{1\}\{2\},\frac\{1\}\{2\},\frac\{\sqrt\{2\}\}\{2\}\right)^\mathrm\{T\}$. 求 $\displaystyle \gamma\_3,\gamma\_4$ 使得 $\displaystyle \gamma\_1,\gamma\_2,\gamma\_3,\gamma\_4$ 为 $\displaystyle \mathbb\{R\}^4$ 的一个标准正交基. (重庆师范大学2023年高等代数考研试题) [欧氏空间 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 易知
\begin\{aligned\} &\alpha\_1=\gamma\_1=(1,0,0,0)^\mathrm\{T\}, \alpha\_2=\gamma\_2=\left(0,\frac\{1\}\{2\},\frac\{1\}\{2\},\frac\{\sqrt\{2\}\}\{2\}\right)^\mathrm\{T\},\\\\ &\alpha\_3=(0,0,1,0)^\mathrm\{T\}, \alpha\_3=(0,0,0,1)^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是 $\displaystyle \mathbb\{R\}^4$ 的一个基, 对他们施行 Gram-Schmidt 标准正交化过程即得 $\displaystyle \mathbb\{R\}^4$ 的一组标准正交基
\begin\{aligned\} &\gamma\_1=(1,0,0,0)^\mathrm\{T\}, \gamma\_2=\left(0,\frac\{1\}\{2\},\frac\{1\}\{2\},\frac\{\sqrt\{2\}\}\{2\}\right)^\mathrm\{T\},\\\\ &\gamma\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}0,-\frac\{1\}\{2\sqrt\{3\}\},\frac\{\sqrt\{3\}\}\{2\}, -\frac\{1\}\{\sqrt\{6\}\}\end\{array\}\right)^\mathrm\{T\}, \gamma\_4=\left(0,-\frac\{2\}\{\sqrt\{6\}\},0,\frac\{1\}\{\sqrt\{3\}\}\right)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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1591、 7、 分为几问. (1)、 计算外积. (2)、 求曲面切平面. (3)、 求柱面或旋转面方程. (4)、 与直纹面有关的关系. (5)、 判断二次曲面的类型. (北京师范大学2023年高等代数与解析几何考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /跟锦数学微信公众号. [在线资料](
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1592、 8、 给定两条一面直线, 求通过公垂线中点且与公垂线垂直的平面方程. (北京师范大学2023年高等代数与解析几何考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /跟锦数学微信公众号. [在线资料](
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1593、 9、 给定二次型曲面, 将其转化为标准方程, 并判断类型. (北京师范大学2023年高等代数与解析几何考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /跟锦数学微信公众号. [在线资料](
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1594、 8、 (20 分) 在直角坐标系下, 已知一点 $\displaystyle M\_0(1,2,0)$ 和一条直线
\begin\{aligned\} L: \left\\{\begin\{array\}\{llllllllllll\}x-y-z+2=0,\\\\ 2x-3y+3=0.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle M\_0$ 到 $\displaystyle L$ 的距离, 并写出过 $\displaystyle M\_0$ 且与 $\displaystyle L$ 垂直相交的直线方程. (东北师范大学2023年高等代数与解析几何考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle L$ 的第 $\displaystyle 2$ 个方程知 $\displaystyle \frac\{x\}\{3\}=\frac\{y-1\}\{2\}=t$. 代入第 $\displaystyle 1$ 个方程得 $\displaystyle z=1+t$. 故 $\displaystyle L$ 的点向式方程为
\begin\{aligned\} \frac\{x\}\{3\}=\frac\{y-1\}\{2\}=\frac\{z-1\}\{1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而 $\displaystyle L$ 过点 $\displaystyle P(0,1,1)$, 方向向量为 $\displaystyle \vec\{t\}=\left\\{3,2,1\right\\}$. $\displaystyle M\_0$ 到 $\displaystyle L$ 的距离
\begin\{aligned\} d=\left|\overrightarrow\{PM\}\times\frac\{\vec\{t\}\}\{|\vec\{t\}|\}\right| =\sqrt\{\frac\{13\}\{7\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
过 $\displaystyle M\_0$ 且与 $\displaystyle L$ 垂直相交的直线 $\displaystyle \ell$ 的方向向量为
\begin\{aligned\} \frac\{(\overrightarrow\{PM\}\times \vec\{t\})\times \vec\{t\}\}\{2\}=\left\\{-1,-3,9\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle \ell$ 的方程为 $\displaystyle \frac\{x-1\}\{-1\}=\frac\{y-2\}\{-2\}=\frac\{z\}\{9\}$.跟锦数学微信公众号. [在线资料](
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---
1595、 9、 (10 分) 设函数 $\displaystyle f: \mathbb\{R\}^3\to\mathbb\{R\}$,
\begin\{aligned\} f(M)=Ax+By+Cz+D,\quad \forall\ M(x,y,z)\in\mathbb\{R\}^3, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle A,B,C,D$ 是不全为零的实数. 证明: 如果三点 $\displaystyle M\_0,M\_1,M\_2$ 共线, 且
\begin\{aligned\} \overrightarrow\{M\_1M\_0\}=\lambda\overrightarrow\{M\_0M\_2\}, \lambda\in\mathbb\{R\}, \lambda\neq -1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
那么 $\displaystyle f(M\_0)=\frac\{f(M\_1)+\lambda f(M\_2)\}\{1+\lambda\}$. (东北师范大学2023年高等代数与解析几何考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle M\_i (x\_i,y\_i,z\_i), i=0,1,2$, 则由题中等式知
\begin\{aligned\} x\_0-x\_1=\lambda(x\_2-x\_0)\Rightarrow x\_0=\frac\{x\_1+\lambda x\_2\}\{1+\lambda\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理, $\displaystyle y\_0=\frac\{y\_1+\lambda y\_2\}\{1+\lambda\}, z\_0=\frac\{z\_1+\lambda z\_2\}\{1+\lambda\}$. 代入 $\displaystyle f$ 的表达式即知 $\displaystyle f(M\_0)=\frac\{f(M\_1)+\lambda f(M\_2)\}\{1+\lambda\}$.跟锦数学微信公众号. [在线资料](
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1596、 10、 (20 分) 证明: 双曲抛物面同族的任意两条直母线必是异面直线, 且同族的全体直母线平行于同一个平面. (东北师范大学2023年高等代数与解析几何考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 双曲抛物面 $\displaystyle z=\frac\{x^2\}\{a^2\}-\frac\{y^2\}\{b^2\} =\left(\frac\{x\}\{a\}+\frac\{y\}\{b\}\right)\left(\frac\{x\}\{a\}-\frac\{y\}\{b\}\right), a,b > 0$ 的两组直母线为 (1)、 $\displaystyle (I)\_\lambda$:
\begin\{aligned\} \frac\{x\}\{a\}+\frac\{y\}\{b\}=\lambda, z=\lambda\left(\frac\{x\}\{a\}-\frac\{y\}\{b\}\right) \Leftrightarrow\frac\{x\}\{a\}=\frac\{y-b\lambda\}\{-b\}=\frac\{z+\lambda^2\}\{2\lambda\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
过点 $\displaystyle P\_\lambda(0,b\lambda,-\lambda^2)$, 方向向量为 $\displaystyle \vec\{v\}\_\lambda=\left\\{a,-b,2\lambda\right\\}$; (2)、 $\displaystyle (II)\_\mu$:
\begin\{aligned\} \frac\{x\}\{a\}-\frac\{y\}\{b\}=\mu, z=\mu\left(\frac\{x\}\{a\}+\frac\{y\}\{b\}\right) \Leftrightarrow\frac\{x\}\{a\}=\frac\{y+b\mu\}\{b\}=\frac\{z+\mu^2\}\{2\mu\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
过点 $\displaystyle P\_\mu(0,-b\mu,-\mu^2)$, 方向向量为 $\displaystyle \vec\{v\}\_\mu=\left\\{a,b,2\mu\right\\}$. 由
\begin\{aligned\} \left(\overrightarrow\{P\_\lambda P\_\{\lambda'\}\},\vec\{v\}\_\lambda,\vec\{v\}\_\{\lambda'\}\right)=-2ab(\lambda-\lambda')^2\neq 0,\\\\ \left(\overrightarrow\{Q\_\mu Q\_\{\mu'\}\},\vec\{w\}\_\mu,\vec\{w\}\_\{\mu'\}\right)=2ab(\mu-\mu')^2\neq 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知双曲抛物面同族的任意两条直母线必是异面直线. 又由
\begin\{aligned\} \vec\{v\}\_\lambda\cdot \left\\{b,a,0\right\\}=0, \vec\{w\}\_\mu\cdot \left\\{-b,a,0\right\\}=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知同族的全体直母线平行于同一个平面.跟锦数学微信公众号. [在线资料](
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---
1597、 6、 (15 分) 证明:
\begin\{aligned\} \left(\vec\{\alpha\}\times \vec\{\beta\},\vec\{\beta\}\times \vec\{\gamma\}, \vec\{\gamma\}\times \vec\{\alpha\}\right)=\left|\begin\{array\}\{cccccccccc\}\vec\{\alpha\}\cdot\vec\{\alpha\}&\vec\{\alpha\}\cdot \vec\{\beta\}&\vec\{\alpha\}\cdot \vec\{\gamma\}\\\\ \vec\{\beta\}\cdot\vec\{\alpha\}&\vec\{\beta\}\cdot \vec\{\beta\}&\vec\{\beta\}\cdot \vec\{\gamma\}\\\\ \vec\{\gamma\}\cdot\vec\{\alpha\}&\vec\{\gamma\}\cdot \vec\{\beta\}&\vec\{\gamma\}\cdot \vec\{\gamma\}\end\{array\}\right|, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \left(\vec\{\alpha\}\times \vec\{\beta\},\vec\{\beta\}\times \vec\{\gamma\}, \vec\{\gamma\}\times \vec\{\alpha\}\right)$ 表示向量 $\displaystyle \vec\{\alpha\}\times \vec\{\beta\},\vec\{\beta\}\times \vec\{\gamma\}$ 和 $\displaystyle \vec\{\gamma\}\times \vec\{\alpha\}$ 的混合积. (吉林大学2023年高等代数与解析几何考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{左端\}=&\left\[\left(\vec\{\alpha\}\times \vec\{\beta\}\right)\times \left(\vec\{\beta\}\times \vec\{\gamma\}\right)\right\] \cdot \left(\vec\{\gamma\}\times \vec\{\alpha\}\right)\\\\ =&\left\\{ \left\[(\vec\{\beta\}\times \vec\{\gamma\})\cdot\vec\{\alpha\}\right\]\vec\{\beta\} -\left\[\vec\{\beta\}\cdot \left(\vec\{\beta\}\times \vec\{\gamma\}\right)\right\]\vec\{\alpha\} \right\\}\cdot(\vec\{\gamma\}\times \vec\{\alpha\})\\\\ =&\left\[(\vec\{\beta\}\times \vec\{\gamma\})\cdot\vec\{\alpha\}\right\] \cdot \left\[(\vec\{\gamma\}\times \vec\{\alpha\})\cdot \vec\{\beta\}\right\]-0\\\\ =&\left|(\vec\{\alpha\}\times \vec\{\beta\})\cdot \vec\{\gamma\}\right|^2 =|(\alpha,\beta,\gamma)|^2=\left|\begin\{array\}\{cccccccccc\}\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha^\mathrm\{T\}\\\\ \beta^\mathrm\{T\}\\\\ \gamma^\mathrm\{T\}\end\{array\}\right)(\alpha,\beta,\gamma)\end\{array\}\right|\\\\ =&\left|\begin\{array\}\{cccccccccc\}\alpha^\mathrm\{T\}\alpha&\alpha^\mathrm\{T\}\beta&\alpha^\mathrm\{T\} \gamma\\\\ \beta^\mathrm\{T\}\alpha&\beta^\mathrm\{T\}\beta&\beta^\mathrm\{T\} \gamma\\\\ \gamma^\mathrm\{T\}\alpha&\gamma^\mathrm\{T\}\beta&\gamma^\mathrm\{T\} \gamma\end\{array\}\right| =\left|\begin\{array\}\{cccccccccc\}\vec\{\alpha\}\cdot\vec\{\alpha\}&\vec\{\alpha\}\cdot \vec\{\beta\}&\vec\{\alpha\}\cdot \vec\{\gamma\}\\\\ \vec\{\beta\}\cdot\vec\{\alpha\}&\vec\{\beta\}\cdot \vec\{\beta\}&\vec\{\beta\}\cdot \vec\{\gamma\}\\\\ \vec\{\gamma\}\cdot\vec\{\alpha\}&\vec\{\gamma\}\cdot \vec\{\beta\}&\vec\{\gamma\}\cdot \vec\{\gamma\}\end\{array\}\right|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1598、 7、 (15 分) 求单叶双曲面 $\displaystyle \frac\{x^2\}\{9\}+\frac\{y^2\}\{16\}-z^2=1$ 上过点 $\displaystyle (3,4,1)$ 的直母线方程. (吉林大学2023年高等代数与解析几何考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} &\frac\{x^2\}\{9\}-z^2=1-\frac\{y^2\}\{16\} \Leftrightarrow \left(\frac\{x\}\{3\}+z\right)\left(\frac\{x\}\{3\}-z\right)=\left(1+\frac\{y\}\{4\}\right)\left(1-\frac\{y\}\{4\}\right)\\\\ \Leftrightarrow&\left\\{\begin\{array\}\{llllllllllll\}\lambda\left(\frac\{x\}\{3\}+z\right)=\mu\left(1+\frac\{y\}\{4\}\right),\\\\ \mu\left(\frac\{x\}\{3\}-z\right)=\lambda\left(1-\frac\{y\}\{4\}\right);\end\{array\}\right.\mbox\{或\} \left\\{\begin\{array\}\{llllllllllll\}\lambda\left(\frac\{x\}\{3\}+z\right)=\mu\left(1-\frac\{y\}\{4\}\right),\\\\ \mu\left(\frac\{x\}\{3\}-z\right)=\lambda\left(1+\frac\{y\}\{4\}\right).\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle P\_0(x\_0,y\_0,z\_0)=(3,4,1)$, 两族直母线方程的参数分别满足 $\displaystyle \lambda=\mu, \lambda=0$, 而它们分别为
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}\frac\{x\}\{3\}-\frac\{y\}\{4\}+z-1=0,\\\\ \frac\{x\}\{3\}+\frac\{y\}\{4\}-z-1=0;\end\{array\}\right.\quad \left\\{\begin\{array\}\{llllllllllll\}y=4,\\\\ x-3z=0.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1599、 8、 (20 分) 证明曲面
\begin\{aligned\} x^2+2y^2+3z^2-2xz+4yz-2z+4=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
与任意一个平面的交集不可能是双曲线. (吉林大学2023年高等代数与解析几何考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 二次型 $\displaystyle x^2+2y^2+3z^2-2xz+4yz$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&2&2\\\\ -1&2&3\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 3+\sqrt\{3\},3-\sqrt\{3\},0$. 故这是一张椭圆抛物面, 与任意平面的交线是椭圆或抛物线, 一定不是双曲线.跟锦数学微信公众号. [在线资料](
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---
1600、 (2)、 已知直线
\begin\{aligned\} x-a=\frac\{y-5\}\{-3\}=\frac\{z-8\}\{5\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
与
\begin\{aligned\} \frac\{x+3\}\{-3\}=\frac\{y+4\}\{12\}=\frac\{z+7\}\{-14\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
相交, 则参数 $\displaystyle a=\underline\{\ \ \ \ \ \ \ \ \ \ \}$, 交点到平面 $\displaystyle 2x-5y+19z=0$ 的距离为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国科学技术大学2023年高等代数考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 第一条直线过点 $\displaystyle P(a,5,8)$, 方向向量为 $\displaystyle \vec\{l\}\_1=\left\\{1,-3,5\right\\}$; 第二条直线过点 $\displaystyle Q (-3,-4,-7)$, 方向向量为 $\displaystyle \vec\{l\}\_2=\left\\{-3,-4,-7\right\\}$. 由题设,
\begin\{aligned\} 0=\left(\overrightarrow\{PQ\},\vec\{l\}\_1,\vec\{l\_2\}\right) =\left|\begin\{array\}\{cccccccccc\}-3-a&-9&-15\\\\ 1&-3&5\\\\ -3&12&-14\end\{array\}\right|=18(a+1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle a=-1$. 代入第一个方程利用 $\displaystyle x$ 表示出 $\displaystyle y,z$ 代入第二个方程, 解得交点为 $\displaystyle (-18,56,-77)$. 它到题中平面的距离利用公式即为 $\displaystyle 593\sqrt\{\frac\{3\}\{130\}\}$.跟锦数学微信公众号. [在线资料](
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---
1601、 (3)、 平面上线性变换
\begin\{aligned\} \mathscr\{A\} \left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}3&2\\\\ 2&3\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将单位圆 $\displaystyle C$ 变成椭圆 $\displaystyle E$, 椭圆 $\displaystyle E$ 长半轴的长度为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国科学技术大学2023年高等代数考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mathscr\{A\}(e\_1,e\_2)=(e\_1,e\_2)A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}3&2\\\\ 2&3\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle A$ 的特征值为 $\displaystyle 5,1$ 知 $\displaystyle \mathscr\{A\}$ 把单位圆 $\displaystyle C$ 变成的了椭圆 $\displaystyle E$, 长半轴长度为 $\displaystyle 5$, 短半轴长度为 $\displaystyle 1$.跟锦数学微信公众号. [在线资料](
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---
1602、 2、 解答题 (共 90 分, 需写出详细的解答过程). (1)、 (15 分) 设给定直角坐标系下二次曲面的方程为
\begin\{aligned\} xy-2yz+2y-3z-1=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
试用正交变换和平移将其化为标准方程, 并判断这是什么类型的曲面? (中国科学技术大学2023年高等代数考研试题) [解析几何 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 二次项的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&\frac\{1\}\{2\}&0\\\\ \frac\{1\}\{2\}&0&-1\\\\ 0&-1&0\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle \frac\{\sqrt\{5\}\}\{2\},-\frac\{\sqrt\{5\}\}\{2\},0$. 由
\begin\{aligned\} \frac\{\sqrt\{5\}\}\{2\}E-A&\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&\frac\{1\}\{2\}\\\\ 0&1&\frac\{\sqrt\{5\}\}\{2\}\\\\ 0&0&0 \end\{array\}\right), -\frac\{\sqrt\{5\}\}\{2\}E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&\frac\{1\}\{2\}\\\\ 0&1&-\frac\{\sqrt\{5\}\}\{2\}\\\\ 0&0&0 \end\{array\}\right),\\\\ 0E-A\to&\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&-2\\\\ 0&1&0\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A$ 的属于特征值 $\displaystyle \frac\{\sqrt\{5\}\}\{2\},-\frac\{\sqrt\{5\}\}\{2\},0$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\-\sqrt\{5\}\\\\2 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\\sqrt\{5\}\\\\2 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\0\\\\1 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令
\begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\} -\frac\{1\}\{\sqrt\{10\}\}&-\frac\{1\}\{\sqrt\{10\}\}&\frac\{2\}\{\sqrt\{5\}\}\\\\ -\frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{2\}\}&0\\\\ \frac\{2\}\{\sqrt\{10\}\}&\frac\{2\}\{\sqrt\{10\}\}&\frac\{1\}\{\sqrt\{5\}\} \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle P$ 正交, 且
\begin\{aligned\} P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}\left(\frac\{\sqrt\{5\}\}\{2\},-\frac\{\sqrt\{5\}\}\{2\},0\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故经过正交变换 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\\\\z\end\{array\}\right)=P\left(\begin\{array\}\{cccccccccccccccccccc\}u\\\\v\\\\w\end\{array\}\right)$ 后, 二次项化为
\begin\{aligned\} \frac\{\sqrt\{5\}\}\{2\}u^2-\frac\{\sqrt\{5\}\}\{2\}v^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此时, 一次项化为
\begin\{aligned\} 2y-3z=&(0,2,-3)\left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\\\\z\end\{array\}\right) =(0,2,-3)P\left(\begin\{array\}\{cccccccccccccccccccc\}u\\\\v\\\\w\end\{array\}\right)\\\\ =&\left(-\frac\{6\}\{\sqrt\{10\}\}-\sqrt\{2\}\right)u +\left(-\frac\{6\}\{\sqrt\{10\}\}+\sqrt\{2\}\right)v -\frac\{3w\}\{\sqrt\{5\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再作平移变换
\begin\{aligned\} u-\left(\frac\{2\}\{\sqrt\{10\}\}+\frac\{3\sqrt\{2\}\}\{5\}\right)=U, v-\left(\frac\{2\}\{\sqrt\{10\}\}-\frac\{3\sqrt\{2\}\}\{5\}\right)=V, w=W, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则二次曲线化为
\begin\{aligned\} \frac\{\sqrt\{5\}\}\{2\}U^2-\frac\{\sqrt\{5\}\}\{2\}V^2-\frac\{3\}\{\sqrt\{5\}\}W=\frac\{17\}\{5\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这是一张双曲抛物面.跟锦数学微信公众号. [在线资料](
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---
1603、 6、 给定 $\displaystyle x\_0 > 0$ 以及 $\displaystyle [0,+\infty)$ 上连续函数 $\displaystyle f(x)$. 证明: 至多具有一个定义于 $\displaystyle [0,+\infty)$ 上的连续函数 $\displaystyle y(x)$ 满足对于任意的 $\displaystyle x > 0$, 有
\begin\{aligned\} \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}=-y^3+f(x), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle y(0)=y\_0$. (复旦大学2023年分析(第6,7,8,9,10题没做)考研试题) [常微分方程 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /跟锦数学微信公众号. [在线资料](
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---
1604、 7、 分析 $\displaystyle \left\\{(x,y); x^2+y^2 < 1\right\\}$ 上的实系统
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \frac\{\mathrm\{ d\} x\}\{\mathrm\{ d\} t\}=f(x,y),\\\\ \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} t\}=g(x,y),\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} f(x,y)=&\left\\{\begin\{array\}\{llllllllllll\}-x+\frac\{2y\}\{\ln(x^2+y^2)\},&(x,y)\neq (0,0),\\\\ 0,&(x,y)=(0,0);\end\{array\}\right.\\\\ g(x,y)=&\left\\{\begin\{array\}\{llllllllllll\}-y+\frac\{2x\}\{\ln(x^2+y^2)\},&(x,y)\neq (0,0),\\\\ 0,&(x,y)=(0,0)\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的所有奇点, 并确定其类型, 并画出取点附近的大致相图, 并与之对应的一次近似系统做比较. (复旦大学2023年分析(第6,7,8,9,10题没做)考研试题) [常微分方程 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /跟锦数学微信公众号. [在线资料](
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---
1605、 7、 第 3 题 20 分, 其余各题 10 分, 共 60 分. (1)、 求解微分方程 $\displaystyle y=(y'-1)\mathrm\{e\}^\{y'\}$. (山东大学2023年高等代数与常微分方程考研试题) [常微分方程 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle y'=p$, 则 $\displaystyle y=(p-1)\mathrm\{e\}^p$, 关于 $\displaystyle x$ 求导有
\begin\{aligned\} p=p\mathrm\{e\}^p\cdot\frac\{\mathrm\{ d\} p\}\{\mathrm\{ d\} x\}\Leftrightarrow \mathrm\{e\}^p\mathrm\{ d\} p=\mathrm\{ d\} x\Leftrightarrow x=\mathrm\{e\}^p+C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故原方程的通解为 $\displaystyle x=\mathrm\{e\}^p+C, y=(p-1)\mathrm\{e\}^p, C$ 为任意常数.跟锦数学微信公众号. [在线资料](
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---
1606、 (2)、 求解微分方程 $\displaystyle y''-y=x\mathrm\{e\}^x\cos x$. (山东大学2023年高等代数与常微分方程考研试题) [常微分方程 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 先求 $\displaystyle y''-y=0$ 的通解. 它的特征方程为 $\displaystyle \lambda^2-1=0$, 而通解为 $\displaystyle y=C\_1\mathrm\{e\}^x+C\_2\mathrm\{e\}^\{-x\}$. 由常数变易法知原方程的通解可设为
\begin\{aligned\} y=C\_1(x)\mathrm\{e\}^x+C\_2(x)\mathrm\{e\}^\{-x\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} &C\_1'(x)\mathrm\{e\}^x+C\_2'(x)\mathrm\{e\}^\{-x\}=0, C\_1'(x)\mathrm\{e\}^x-C\_2'(x)\mathrm\{e\}^\{-x\}=x\mathrm\{e\}^x \cos x\\\\ \Rightarrow&C\_1'(x)=\frac\{1\}\{2\}x\cos x, C\_2'(x)=-\frac\{1\}\{2\}x\mathrm\{e\}^\{2x\}\cos x\\\\ \Rightarrow&C\_1(x)=\frac\{1\}\{2\}(\cos x+x\sin x)+c\_1,\\\\ &C\_2(x)=-\frac\{1\}\{50\}\mathrm\{e\}^\{2x\}\left\[(10x-3)\cos x+(5x-4)\sin x\right\]+c\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故原方程的通解为
\begin\{aligned\} y=&c\_1\mathrm\{e\}^x+c\_2\mathrm\{e\}^\{-x\} +\frac\{1\}\{25\}\mathrm\{e\}^x\left\[(14-5x)\cos x+2(1+5x)\sin x\right\], \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle c\_1,c\_2$ 为任意常数.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1607、 (3)、 求 $\displaystyle x'=Ax$ 的基本解组, 其中 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&4&2\\\\ 0&-3&4\\\\ 0&4&3\end\{array\}\right)$. (山东大学2023年高等代数与常微分方程考研试题) [常微分方程 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 易知 $\displaystyle A$ 的特征值为 $\displaystyle 5,1,-5$. 由
\begin\{aligned\} &5E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&1&-\frac\{1\}\{2\}\\\\ 0&0&0 \end\{array\}\right), E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0\\\\ 0&0&1\\\\ 0&0&0 \end\{array\}\right),\\\\ &-5E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&1&2\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A$ 的属于特征值 $\displaystyle 5,1,-5$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\1\\\\2 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\0\\\\0 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-2\\\\1 \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle x'=Ax$ 的一个基本解组为
\begin\{aligned\} \mathrm\{e\}^\{5t\}\eta\_1=\mathrm\{e\}^\{5t\}\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\1\\\\2\end\{array\}\right), \mathrm\{e\}^t\eta\_2=\mathrm\{e\}^t\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\0\\\\0\end\{array\}\right), \mathrm\{e\}^\{-5t\}\eta\_3=\mathrm\{e\}^\{-5t\}\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-2\\\\1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1608、 (4)、 解方程 $\displaystyle y=\frac\{3\}\{2\}(y')^2-2y'x+x^2$. (山东大学2023年高等代数与常微分方程考研试题) [常微分方程 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle y'=p$, 则 $\displaystyle y'=\frac\{3\}\{2\}p^2-2px+x^2$. 关于 $\displaystyle x$ 求导有
\begin\{aligned\} &p=3p\frac\{\mathrm\{ d\} p\}\{\mathrm\{ d\} x\}-2\frac\{\mathrm\{ d\} p\}\{\mathrm\{ d\} x\}x-2p+2x \Leftrightarrow (3p-2x)\frac\{\mathrm\{ d\} p\}\{\mathrm\{ d\} x\}=3p-2x\\\\ \Leftrightarrow&3p-2x=0\mbox\{或\} \frac\{\mathrm\{ d\} p\}\{\mathrm\{ d\} x\}=1 \Leftrightarrow p=\frac\{2x\}\{3\}\mbox\{或\} x+C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故原方程的通解为
\begin\{aligned\} y=&\frac\{3\}\{2\}(x+C)^2-2(x+C)x+x^2=\frac\{1\}\{2\}(x^2+2Cx+3C^2), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且有特解 $\displaystyle y=\frac\{x^2\}\{3\}$ ($p=\frac\{2x\}\{3\}$ 对应的特解).跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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1609、 (5)、 求区域 $\displaystyle G$, 使得当 $\displaystyle (x\_0,y\_0)\in G$ 时, 初值问题 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}y'=2\sqrt\{1+y\},\\\\ y(x\_0)=y\_0\end\{array\}\right.$ 的解存在且唯一. (山东大学2023年高等代数与常微分方程考研试题) [常微分方程 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} &y'=2\sqrt\{1+y\}\Rightarrow \frac\{\mathrm\{ d\} y\}\{2\sqrt\{1+y\}\}=\mathrm\{ d\} x \Rightarrow \mathrm\{ d\} \sqrt\{1+y\}=\mathrm\{ d\} x\\\\ \Rightarrow& \sqrt\{1+y\}=x+C \stackrel\{y(x\_0)=y\_0\}\{\Rightarrow\} \sqrt\{1+y\_0\}=x\_0+C\\\\ \Rightarrow& \sqrt\{1+y\}=x+\sqrt\{1+y\_0\}-x\_0 \Rightarrow y=\left(x+\sqrt\{1+y\_0\}-x\right)^2-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle G=\left\\{(x,y); x\in\mathbb\{R\}, y > -1\right\\}$. 注意当 $\displaystyle y\_0=-1$ 时解不唯一:
\begin\{aligned\} y(x)\equiv -1, y(x)=\left\\{\begin\{array\}\{llllllllllll\}\left(x+\sqrt\{1-y\_0\}-x\right)^2-1,&x < x\_0-\sqrt\{1+y\_0\},\\\\ -1,&x\geq x\_0-\sqrt\{1+y\_0\}.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1610、 (2)、 (15 分) 解线性常微分方程组
\begin\{aligned\} \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} \frac\{\mathrm\{ d\} x\}\{\mathrm\{ d\} t\}=&7x&+&9y,\\\\ \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} t\}=&-6x&-&8y, \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle x=x(t)$ 及 $\displaystyle y=y(t)$. (中国科学技术大学2023年高等代数考研试题) [常微分方程 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}7&9\\\\ -6&-8\end\{array\}\right)$, 则 $\displaystyle A$ 的特征值为 $\displaystyle 1,-2$. 由
\begin\{aligned\} E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&\frac\{3\}\{2\}\\\\ 0&0\end\{array\}\right),\quad -E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1\\\\ 0&0\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A$ 的属于特征值 $\displaystyle 1,-2$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-3\\\\2\end\{array\}\right),\quad \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle P=(\xi\_1,\xi\_2)$, 则
\begin\{aligned\} &P^\{-1\}AP=\mathrm\{diag\}(1,-2), P^\{-1\}\mathrm\{e\}^\{At\}P=\mathrm\{diag\}(\mathrm\{e\}^t, \mathrm\{e\}^\{-2t\}),\\\\ &\left(\begin\{array\}\{cccccccccccccccccccc\}x(t)\\\\y(t)\end\{array\}\right)=\mathrm\{e\}^\{At\}\left(\begin\{array\}\{cccccccccccccccccccc\}a\\\\b\end\{array\}\right) =\left(\begin\{array\}\{cccccccccccccccccccc\}3(a+b)\mathrm\{e\}^t-(2a+3b)\mathrm\{e\}^\{-2t\}\\\\ -2(a+b)\mathrm\{e\}^t+(2a+3b)\mathrm\{e\}^\{-2t\}\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle a,b$ 为任意常数.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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发表于 2023-3-5 13:27:15
