diff --git a/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf b/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf
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diff --git a/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex b/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex
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@@ -0,0 +1,63 @@
+\documentclass[UTF8, 12pt]{ctexart}
+% UTF8编码，ctexart现实中文
+\usepackage{color}
+% 使用颜色
+\usepackage{geometry}
+\setcounter{tocdepth}{4}
+\setcounter{secnumdepth}{4}
+% 设置四级目录与标题
+\geometry{papersize={21cm,29.7cm}}
+% 默认大小为A4
+\geometry{left=3.18cm,right=3.18cm,top=2.54cm,bottom=2.54cm}
+% 默认页边距为1英尺与1.25英尺
+\usepackage{indentfirst}
+\setlength{\parindent}{2.45em}
+% 首行缩进2个中文字符
+\usepackage{setspace}
+\renewcommand{\baselinestretch}{1.5}
+% 1.5倍行距
+\usepackage{amssymb}
+% 因为所以
+\usepackage{amsmath}
+% 数学公式
+\usepackage[colorlinks,linkcolor=black,urlcolor=blue]{hyperref}
+% 超链接
+\author{Didnelpsun}
+\title{多元函数积分学}
+\date{}
+\begin{document}
+\maketitle
+\pagestyle{empty}
+\thispagestyle{empty}
+\tableofcontents
+\thispagestyle{empty}
+\newpage
+\pagestyle{plain}
+\setcounter{page}{1}
+\section{二重积分}
+
+\subsection{交换积分次序}
+
+\subsubsection{直角坐标系}
+
+\textbf{例题：}交换积分次序$\int_0^1\textrm{d}x\int_0^{x^2}f(x,y)\,\textrm{d}y+\int_1^3\textrm{d}x\int_0^{\frac{1}{2}(3-x)}f(x,y)\,\textrm{d}y$。
+
+解：已知积分区域分为两个部分。将$X$型变为$Y$型。画出图形可以知道$y\in(0,1)$，$x$的上下限由$y=x^2$和$y=\dfrac{1}{2}(3-x)$转化为$\sqrt{y}$和$3-2y$。
+
+所以转换为$\int_0^1\textrm{d}y\int_{\sqrt{y}}^{3-2y}f(x,y)\,\textrm{d}x$。
+
+\subsubsection{极坐标系}
+
+\subsection{极直互化}
+
+\textbf{例题：}将$I=\int_0^{\frac{\sqrt{2}}{2}R}e^{-y^2}\textrm{d}y\int_0^ye^{-x^2}\,\textrm{d}x+\int_{\frac{\sqrt{2}}{2}R}^Re^{-y^2}\,\textrm{d}y\int_0^{\sqrt{R^2-y^2}}e^{-x^2}\,\textrm{d}x$转换为极坐标系并计算结果。
+
+解：首先根据积分上下限得到积分区域$D=\left\{0\leqslant y\leqslant\dfrac{\sqrt{2}}{2}R,0\leqslant x\leqslant y\right\}\cup\left\{\dfrac{\sqrt{2}}{2}R\leqslant y\leqslant R,0\leqslant x\leqslant\sqrt{R^2-y^2}\right\}$，$D$为一个八分之一圆的扇形。
+
+根据$x=r\cos\theta$，$y=r\sin\theta$替换得到$D=\left\{(x,y)\bigg|0\leqslant r\leqslant R,\dfrac{\pi}{4}\leqslant\theta\leqslant\dfrac{\pi}{2}\right\}$。
+
+又$e^{-y^2}\cdot e^{-x^2}=e^{-(x^2+y^2)}=e^{-r^2}$。
+
+$\therefore I=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\textrm{d}\theta\int_0^Re^{-r^2}r\,\textrm{d}r$。
+
+\end{document}
diff --git a/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf b/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf
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diff --git a/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex b/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex
index d71b49b..b1b5ec4 100644
--- a/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex
+++ b/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex
@@ -182,7 +182,7 @@ $\therefore2I=\displaystyle{\iint\limits_D\dfrac{a\sqrt{f(x)}+b\sqrt{f(y)}}{\sqr
 
 $\therefore2I=\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\displaystyle{\iint\limits_D(x^2+y^2)\textrm{d}x\textrm{d}y}$，$\therefore I=\dfrac{1}{2}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\displaystyle{\iint\limits_D(x^2+y^2)\textrm{d}\sigma}$。
 
-根据公式三转换为极坐标系：$I=\dfrac{1}{2}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\int_0^{2\pi}\textrm{d}\theta\int)^Rr^2r\,\textrm{d}r$。
+根据公式三转换为极坐标系：$I=\dfrac{1}{2}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\int_0^{2\pi}\textrm{d}\theta\int_0^Rr^2r\,\textrm{d}r$。
 
 即$I=\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\dfrac{\pi R^4}{4}$。
 
diff --git a/linear-algebra/exercise/5-similar/similar.pdf b/linear-algebra/exercise/5-similar/similar.pdf
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diff --git a/linear-algebra/exercise/5-similar/similar.tex b/linear-algebra/exercise/5-similar/similar.tex
index db8b881..3b87436 100644
--- a/linear-algebra/exercise/5-similar/similar.tex
+++ b/linear-algebra/exercise/5-similar/similar.tex
@@ -265,4 +265,24 @@ $\therefore AP=PB$，$P^{-1}AP=B$，$A\sim B$。
 
 \subsection{正交相似}
 
+\textbf{例题：}已知$A$是三阶实对称矩阵，若正交矩阵$Q$使得$Q^{-1}AQ=\left[\begin{array}{ccc}
+    3 & 0 & 0 \\
+    0 & 3 & 0 \\
+    0 & 0 & 6
+\end{array}\right]$，如果$\alpha_1=(1,0,-1)^T$和$\alpha_2=(0,1,1)^T$是矩阵$A$属于特征值$\lambda=3$的特征向量，求$Q$。
+
+解：首先由正交矩阵就可以知道各特征值正交。令$\alpha_3=(x_1,x_2,x_3)^T$。对应的$\lambda_3=6$。
+
+$\alpha_3^T\alpha_1=x_1-x_3=0$，$\alpha_3^T\alpha_2=x_2+x_3=0$，求$\lambda_3$的特征值，则不如令$x_3=1$，则解得$\alpha_3=(1,-1,1)^T$。
+
+这样$Q=\left[\begin{array}{ccc}
+    1 & 0 & 1 \\
+    0 & 1 & -1 \\
+    -1 & 1 & 1
+\end{array}\right]$，还需要将$Q$正交单位化。可知$\alpha_3$根据正交规律求出来，一定是正交的，而$\alpha_1^T\alpha_2=-1\neq0$所以需要正交。
+
+令$\beta_1=\alpha_1=(1,0,-1)^T$，$\beta_2=\alpha_2-\dfrac{<\alpha_2,\beta_1>}{<\beta_1,\beta_1>}\beta_1=(0,1,1)^T+\dfrac{1}{2}(1,0,-1)^T=(\dfrac{1}{2},1,\dfrac{1}{2})^T$。
+
+最后对整个$Q$进行单位化：$\gamma_1=\dfrac{1}{\sqrt{2}}(1,0,-1)^T$，$\gamma_2=\dfrac{1}{\sqrt{6}}(1,2,1)^T$，$\gamma_3=\dfrac{1}{\sqrt{3}}(1,-1,1)^T$。
+
 \end{document}
diff --git a/linear-algebra/exercise/6-quadratic-form/quadratic-form.pdf b/linear-algebra/exercise/6-quadratic-form/quadratic-form.pdf
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index 0000000..60398d3
--- /dev/null
+++ b/linear-algebra/exercise/6-quadratic-form/quadratic-form.tex
@@ -0,0 +1,38 @@
+\documentclass[UTF8, 12pt]{ctexart}
+% UTF8编码，ctexart现实中文
+\usepackage{color}
+% 使用颜色
+\usepackage{geometry}
+\setcounter{tocdepth}{4}
+\setcounter{secnumdepth}{4}
+% 设置四级目录与标题
+\geometry{papersize={21cm,29.7cm}}
+% 默认大小为A4
+\geometry{left=3.18cm,right=3.18cm,top=2.54cm,bottom=2.54cm}
+% 默认页边距为1英尺与1.25英尺
+\usepackage{indentfirst}
+\setlength{\parindent}{2.45em}
+% 首行缩进2个中文字符
+\usepackage{setspace}
+\renewcommand{\baselinestretch}{1.5}
+% 1.5倍行距
+\usepackage{amssymb}
+% 因为所以
+\usepackage{amsmath}
+% 数学公式
+\usepackage[colorlinks,linkcolor=black,urlcolor=blue]{hyperref}
+% 超链接
+\author{Didnelpsun}
+\title{二次型}
+\date{}
+\begin{document}
+\maketitle
+\pagestyle{empty}
+\thispagestyle{empty}
+\tableofcontents
+\thispagestyle{empty}
+\newpage
+\pagestyle{plain}
+\setcounter{page}{1}
+\section{正定二次型}
+\end{document}
diff --git a/linear-algebra/knowledge/5-similar/similar.pdf b/linear-algebra/knowledge/5-similar/similar.pdf
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diff --git a/linear-algebra/knowledge/5-similar/similar.tex b/linear-algebra/knowledge/5-similar/similar.tex
index 17c80fa..73b81da 100644
--- a/linear-algebra/knowledge/5-similar/similar.tex
+++ b/linear-algebra/knowledge/5-similar/similar.tex
@@ -286,6 +286,18 @@ $\therefore\lambda_1=\lambda_2=1$，$\lambda_3=10$。
 
 所以内积是一个数值。
 
+单位化是保持向量方向不变，将其长度化为1。
+
+正交化是指将线性无关向量系转化为正交系的过程。
+
+\subsubsection{施密特正交化}
+
+将一个线性无关向量组变为一个标准正交向量组。
+
+对于线性无关向量组$\alpha_1,\alpha_2,\cdots,\alpha_n$，令$\beta_1=\alpha_1$，$\beta_2=\alpha_2-\dfrac{<\alpha_2,\beta_1>}{<\beta_1,\beta_1>}\beta_1$，$\beta_3=\alpha_3-\dfrac{<\alpha_3,\beta_1>}{<\beta_1,\beta_1}\beta_1-\dfrac{<\alpha_3,\beta_2>}{<\beta_2,\beta_2>}\beta_2$，$\cdots$，$\beta_n=\alpha_n-\dfrac{<\alpha_n,\beta_1>}{<\beta_1,\beta_1>}\beta_1-\dfrac{<\alpha_n,\beta_2>}{<\beta_2,\beta_2>}\beta_2-\cdots-\dfrac{<\alpha_n,\beta_{n-1}}{<\beta_{n-1},\beta_{n-1}>}\beta_{n-1}$。其中$<n,n>$代表$n,n$的内积。
+
+最后单位化：$\gamma_i=\dfrac{\beta_i}{\Vert\beta_i\Vert}$。
+
 \subsubsection{定义}
 
 \textcolor{violet}{\textbf{定义：}}$A^T=A$则$A$就是对称矩阵，若$A$的元素都是实数，则$A$是实对称矩阵。