diff --git a/advanced-math/exercise/4-indefinite-integral-and-definite-integral/indefinite-integral-and-definite-integral.pdf b/advanced-math/exercise/4-indefinite-integral-and-definite-integral/indefinite-integral-and-definite-integral.pdf
index 107f2bf..415727a 100644
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diff --git a/advanced-math/exercise/4-indefinite-integral-and-definite-integral/indefinite-integral-and-definite-integral.tex b/advanced-math/exercise/4-indefinite-integral-and-definite-integral/indefinite-integral-and-definite-integral.tex
index 4b7e68a..9214141 100644
--- a/advanced-math/exercise/4-indefinite-integral-and-definite-integral/indefinite-integral-and-definite-integral.tex
+++ b/advanced-math/exercise/4-indefinite-integral-and-definite-integral/indefinite-integral-and-definite-integral.tex
@@ -473,6 +473,14 @@ $=-\dfrac{4x+3}{2(x^2+x+1)}-\dfrac{6}{\sqrt{3}}\arctan\dfrac{2x+1}{\sqrt{3}}+C$
 
 $\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\dfrac{1}{n+i}=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\dfrac{1}{n+\dfrac{i}{n}}\dfrac{1}{n}=\displaystyle{\int_0^1\dfrac{1}{1+x}\,\textrm{d}x}=[\ln(1+x)]_0^1=\ln2$。
 
+\textbf{例题：}求$\lim\limits_{n\to\infty}\left(\dfrac{n+1}{n^2+1}+\dfrac{n+2}{n^2+4}+\cdots+\dfrac{n+n}{n^2+n^2}\right)$。
+
+即求$\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\dfrac{n+i}{n^2+i^2}=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\dfrac{n^2+ni}{n^2+i^2}\cdot\dfrac{1}{n}=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\dfrac{1+\dfrac{i}{n}}{1+\left(\dfrac{i}{n}\right)^2}\cdot\dfrac{1}{n}$
+
+$=\displaystyle{\int_0^1\dfrac{1+x}{1+x^2}\textrm{d}x}=\displaystyle{\int_0^1\dfrac{1}{1+x^2}\textrm{d}x+\int_0^1\dfrac{x}{1+x^2}\textrm{d}x}$
+
+$=\left[\arctan x+\dfrac{1}{2}\ln(1+x^2)\right]_0^1$。
+
 \subsection{变限积分}
 
 \subsection{牛莱公式}
diff --git a/linear-algebra/knowledge/2-matrix/matrix.pdf b/linear-algebra/knowledge/2-matrix/matrix.pdf
index 5dfaf9c..c1f043a 100644
Binary files a/linear-algebra/knowledge/2-matrix/matrix.pdf and b/linear-algebra/knowledge/2-matrix/matrix.pdf differ
diff --git a/linear-algebra/knowledge/2-matrix/matrix.tex b/linear-algebra/knowledge/2-matrix/matrix.tex
index 3df06bd..fb4f4fe 100644
--- a/linear-algebra/knowledge/2-matrix/matrix.tex
+++ b/linear-algebra/knowledge/2-matrix/matrix.tex
@@ -747,4 +747,78 @@ $$(a_1,a_2,\cdots,a_n)\left(\begin{array}{c}
 
 对于任何矩阵都能通过初等列变换变为行阶梯形矩阵和行最简形矩阵，再通过列变换可以变为\textbf{标准形}：左上角是一个单位矩阵，其他元全部是0。
 
-\end{document} 
\ No newline at end of file
+\subsection{初等变换性质}
+
+\textcolor{aqua}{\textbf{定理：}}设$AB$都是$m\times n$矩阵：\begin{enumerate}
+    \item $A\overset{r}{\thicksim}B$的充要条件是存在$m$阶可逆矩阵$P$，使得$PA=B$。
+    \item $A\overset{c}{\thicksim}B$的充要条件是存在$n$阶可逆矩阵$Q$，使得$AQ=B$。
+    \item $A\thicksim B$的充要条件是存在$m$阶可逆矩阵$P$和$n$阶可逆矩阵$Q$，使得$PAQ=B$。
+\end{enumerate}
+
+初等矩阵\textcolor{violet}{\textbf{定义：}}由单位矩阵$E$经过一次初等变换得到的矩阵。
+
+初等矩阵包括：\begin{enumerate}
+    \item 第$ij$行对换：$E_m(ij)A$，第$ij$列变换：$AE_n(ij)$。
+    \item 数$k$乘第$i$行：$E_m(i(k))A$，数$k$乘第$i$列：$AE_n(i(k))$。
+    \item 数$k$乘第$j$行加到$i$行：$E_m(ij(k))A$，数$k$乘第$j$列加到$i$列：$AE_n(ij(k))$。
+\end{enumerate}
+
+\textcolor{aqua}{\textbf{定理：}}设$A$是一个$m\times n$矩阵，对$A$进行一次初等行变换，相当于在$A$左乘对应$m$阶初等矩阵；对$A$进行一次列变换，相当于在$A$右乘对应$n$阶初等矩阵。
+
+\textcolor{aqua}{\textbf{定理：}}方阵$A$可逆的充分必要条件是存在有限个初等矩阵$P_i$使得$A=\prod\limits_{i=1}^nP_i$。
+
+\textcolor{aqua}{\textbf{定理：}}方阵$A$可逆的充要条件是$A\overset{r}{\thicksim}E$。（即$A$方阵所代表的线性方程组能通过初等计算得到最后的解）
+
+\subsection{初等行变换求逆}
+
+已知$A^{-1}=\dfrac{A^*}{\vert A\vert}$，但是伴随矩阵计算非常麻烦，并且若矩阵在三阶以上计算就很难办到，所以还有一种方法，就是若该矩阵$A$是可逆矩阵，就将$AX=B$的增广矩阵$(A,B)$化为最简形矩阵，从而得到方程解。
+
+$(A\vdots B)\overset{r}{\thicksim}(E\vdots A^{-1})$，$\left(\begin{array}{c}
+    A \\
+    \cdots \\
+    B
+\end{array}\right)\overset{c}{\thicksim}\left(\begin{array}{c}
+    E \\
+    \cdots \\
+    A^{-1}
+\end{array}\right)$。
+
+\textbf{例题：}求解矩阵方程$AX=B$，$A=\left(\begin{array}{ccc}
+    2 & 1 & -3 \\
+    1 & 2 & -2 \\
+    -1 & 3 & 2
+\end{array}\right)$，$B=\left(\begin{array}{cc}
+    1 & -1 \\
+    2 & 0 \\
+    -2 & 5
+\end{array}\right)$。
+
+因为$AX=B$，所以左乘$A^{-1}$：$A^{-1}AX=EX=A^{-1}B$，增广矩阵行变换：
+
+$$(A,B)=\left(\begin{array}{ccccc}
+    2 & 1 & -3 & 1 & -1 \\
+    1 & 2 & -2 & 2 & 0 \\
+    -1 & 3 & 2 & -2 & 5
+\end{array}\right)\thicksim\left(\begin{array}{ccccc}
+    1 & 2 & -2 & 2 & 0 \\
+    0 & -3 & 1 & -3 & -1 \\
+    0 & 5 & 0 & 0 & 5
+\end{array}\right)$$
+
+$\thicksim\left(\begin{array}{ccccc}
+    1 & 2 & -2 & 2 & 0 \\
+    0 & 1 & 0 & 0 & 1 \\
+    0 & 0 & 1 & -3 & 2
+\end{array}\right)\thicksim\left(\begin{array}{ccccc}
+    1 & 0 & 0 & -4 & 2 \\
+    0 & 1 & 0 & 0 & 1 \\
+    0 & 0 & 1 & -3 & 2
+\end{array}\right)$，从而$X=\left(\begin{array}{cc}
+    -4 & 2 \\
+    0 & 1 \\
+    -3 & 2
+\end{array}\right)$。
+
+\section{矩阵秩}
+
+\end{document}
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