# Quod Erat Demonstrandum

## 2014/11/28

### 某 m2 題

Filed under: NSS — johnmayhk @ 4:34 下午
Tags: ,

Part (c) 的建議答案，考慮

## 2014/11/14

### M2 堂偶拾

Filed under: NSS,Pure Mathematics — johnmayhk @ 8:19 下午
Tags: , ,

1.

$\cos^{-1}\frac{4}{5}+2\tan^{-1}\frac{1}{2}=90^o$

## 2014/11/03

### 存在非平凡解的齊次線性方程組

Filed under: NSS,Teaching — johnmayhk @ 4:08 下午
Tags:

$\left \{ \begin{array}{ll} 2x+(2+k)y+2z=0\\(4+k)x+2y+5z=0\\7x+3y+(6+k)z=0\end{array}\right.$

$k$ 值。

$\left|\begin{array}{ccc}2 & 2+k & 2\\4+k & 2 & 5\\7 & 3 & 6+k\end{array}\right|=0$

$k=1,-1,-12$

$\left(\begin{array}{cccc}2 & 2+k & 2 & 0\\4+k & 2 & 5 & 0\\7 & 3 & 6+k & 0\end{array}\right)$

~$\left(\begin{array}{cccc}1 & \frac{k}{2}+1 & 1 & 0\\7 & 3 & 6+k & 0\\4+k & 2+k & 5 & 0\end{array}\right)$

~$\left(\begin{array}{cccc}1 & \frac{k}{2}+1 & 1 & 0\\ 0 & -\frac{7k}{2}-4 & k-1 & 0\\ 0 & -\frac{k^2}{2}-3k-2 & 1-k & 0\end{array}\right)$

~$\left(\begin{array}{cccc}1 & \frac{k}{2}+1 & 1 & 0\\ 0 & -\frac{7k}{2}-4 & k-1 & 0\\ 0 & -\frac{k^2}{2}-\frac{13k}{2}-6 & 0 & 0\end{array}\right)$

$-\frac{k^2}{2}-\frac{13k}{2}-6=0$

$k=-1,-12$

::: 停一停，想一想 ::: (more…)