---
tags: Math, AMSIT, Homework
title: AMSIT, Homework
author: Vo Hoang Anh - SPyofgame
license: Private Use
---

$$
\begin{array}{c}
\Huge \texttt{Applied Mathematics and Statistics}\\
\Huge \texttt{for Information Technology}\\
\Huge \textbf{Quiz Week 2}\\
\end{array}
$$

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$$
\begin{array}{c}
\textbf{Name: } \texttt{Võ Hoàng Anh}&&&&&&&&&&&&&
\textbf{Course: } \texttt{MTH00051-22CLC04}\\
\textbf{Student-ID: } \texttt{22127022}&&&&&&&&&&&&&
\textbf{Semester: } \texttt{2024-III}\\
\end{array}
$$

If the PDF file has $\texttt{errors}$, use this $\texttt{hackmd link}$: https://hackmd.io/@spyofgame/quiz-2

The solutions are $\texttt{verified}$ with my own $\texttt{C++ solver}$: https://ideone.com/GuQZHl

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## Summary

#### Bài 2.g

:::danger
$$
\begin{array}{lll}
\mathbf{G} = \left[
\begin{array}{r}
1 & -1 & 2\\
1 & 0 & -1\\
-1 & 1 & 2\\
0 & 1 & 1\\
\end{array}
\right]\\
\end{array}
$$
:::

:::success
Dễ tính được $(u_1, u_2, u_3)$ có giá trị như sau

$\begin{cases}\begin{array}{llll}
&u_1 &=& ( 1, 1,-1, 0)\\
&u_2 &=& (-1, 0, 1, 1)\\
&u_3 &=& ( 2, -1, 2, 2)\\
\end{array}\end{cases}$

Hệ trực giao $\{v_1, v_2, v_3\}$ được tính như sau

$\begin{cases}\begin{array}{llll}
&v_1
&= u_1
&= (1, 1, -1, 0)
\\
&v_2
&= u_2 - \large\frac{\langle u_2,v_1\rangle}{||v_1||^2}v_1
&= \left(-\frac{1}{3}, \frac{2}{3}, \frac{1}{3}, 1\right)
\\
&v_3
&= u_3 - \large\frac{\langle u_3,v_1\rangle}{||v_1||^2}v_1 - \large\frac{\langle u_3,v_2\rangle}{||v_2||^2}v_2
&= \left(\frac{12}{5}, -\frac{4}{5}, \frac{6}{5}, \frac{4}{5}\right)
\\
\end{array}\end{cases}$

Hệ trực chuẩn $\{q_1, q_2, q_3\}$ được tính theo $q_k = \frac{1}{||v_k||}v_k$ hay

$\begin{cases}\begin{array}{llll}
&q_1
&= \frac{1}{||v_1||}v_1
&= \large\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, 0\right)
\\
&q_2
&= \frac{1}{||v_2||}v_2
&= \large\left(-\frac{1}{\sqrt{15}}, \frac{2}{\sqrt{15}}, \frac{1}{\sqrt{15}}, \frac{\sqrt{15}}{5}\right)
\\
&q_3
&= \frac{1}{||v_3||}v_3
&= \large\left(\frac{\sqrt{15}}{5}, -\frac{1}{\sqrt{15}}, \frac{2}{\sqrt{15}}, \frac{1}{\sqrt{15}}\right)
\\
\end{array}\end{cases}$

$\Rightarrow \mathbf{Q} = \frac{1}{\sqrt{15}} \cdot \left[
\begin{array}{ccc}
\sqrt{5} & 1 & 3\\
\sqrt{5} & 2 & -1\\
-\sqrt{5} & 1 & 2\\
0 & 3 & 1\\
\end{array}
\right]$

Mà $\begin{cases}
\langle q_1, u_1 \rangle = \sqrt{3}\\
\langle q_1, u_2 \rangle = -\frac{2\sqrt{3}}{3}\\
\langle q_1, u_3 \rangle = -\frac{1}{\sqrt{3}}\\
\langle q_2, u_2 \rangle = \frac{\sqrt{15}}{3}\\
\langle q_2, u_3 \rangle = \frac{\sqrt{15}}{15}\\
\langle q_3, u_3 \rangle = \frac{4\sqrt{15}}{5}\\
\end{cases} \Rightarrow \mathbf{R} = \left[
\begin{array}{rrr}
\sqrt{3} & -\frac{2\sqrt{3}}{3} & -\frac{1}{\sqrt{3}}\\
0 & \frac{\sqrt{15}}{3} & \frac{\sqrt{15}}{15}\\
0 & 0 & \frac{4\sqrt{15}}{5}\\
\end{array}
\right]$
:::

:::info
$$\mathbf{Q} = \frac{1}{\sqrt{15}} \cdot \left[
\begin{array}{ccc}
\sqrt{5} & 1 & 3\\
\sqrt{5} & 2 & -1\\
-\sqrt{5} & 1 & 2\\
0 & 3 & 1\\
\end{array}
\right]$$

$$\mathbf{R} = \left[
\begin{array}{rrr}
\sqrt{3} & -\frac{2\sqrt{3}}{3} & -\frac{1}{\sqrt{3}}\\
0 & \frac{\sqrt{15}}{3} & \frac{\sqrt{15}}{15}\\
0 & 0 & \frac{4\sqrt{15}}{5}\\
\end{array}
\right]$$
:::

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