# 线性代数教学改革研究Research on Reform of Teaching Linear Algebra

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This paper first refers to the current state of teaching linear algebra, based on the author’s expe-riences of teaching linear algebra in recent years. It then gives some advices about textbook and teaching, with a deeper understanding of linear algebra. Indeed, this paper aims to build the im-age of how easy linear algebra is, so that it can play a more important role in college education.

1. 引言

2. 线性代数教学的现状

2.1. 对线性代数发展史了解不深

2.2. 基础理论与实践操作不连贯

2.3. 对线性代数课程的认识不清

3. 线性代数教学改革建议

3.1. 在教学中处理好与线性代数发展史的关系

$\begin{array}{rrr}\hfill 1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 5& \hfill 6\end{array}$ ，称之为矩阵。为了表示它们是一个整体，一般用小括号或者中括号括起来，即 $\left(\begin{array}{rrr}\hfill 1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 5& \hfill 6\end{array}\right)$

$\left(\begin{array}{cccc}{a}_{11}& {a}_{12}& \dots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{m1}& {a}_{m2}& \dots & {a}_{mn}\end{array}\right)\left(\begin{array}{cccc}{b}_{11}& {b}_{12}& \dots & {b}_{1s}\\ {b}_{21}& {b}_{22}& \cdots & {b}_{2s}\\ ⋮& ⋮& \ddots & ⋮\\ {b}_{n1}& {b}_{n2}& \dots & {b}_{ns}\end{array}\right)=\left(\begin{array}{cccc}{c}_{11}& {c}_{12}& \dots & {c}_{1s}\\ {c}_{21}& {c}_{22}& \cdots & {c}_{2s}\\ ⋮& ⋮& \ddots & ⋮\\ {c}_{m1}& {c}_{m2}& \dots & {c}_{ms}\end{array}\right),$

${c}_{ij}=\underset{k=1}{\overset{n}{\sum }}{a}_{ik}{b}_{kj}={a}_{i1}{b}_{1j}+\cdots +{a}_{in}{b}_{nj}=\left(\begin{array}{cccc}{a}_{i1}& {a}_{i2}& \cdots & {a}_{in}\end{array}\right)\left(\begin{array}{c}{b}_{1j}\\ {b}_{2j}\\ ⋮\\ {b}_{nj}\end{array}\right).$

$\left\{\begin{array}{c}{x}_{1}+2{x}_{2}=3\\ 4{x}_{1}+5{x}_{2}=6\end{array}\stackrel{矩阵形式}{\to }\left(\begin{array}{c}{x}_{1}+2{x}_{2}\\ 4{x}_{1}+5{x}_{2}\end{array}\right)=\left(\begin{array}{c}3\\ 6\end{array}\right)\stackrel{系数与未知元分离}{\to }\left(\begin{array}{cc}1& 2\\ 4& 5\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right)=\left(\begin{array}{c}3\\ 6\end{array}\right).$

3.2. 基础理论要为实践操作服务

3.3. 建立线性代数通俗易懂的观念

$\left(\begin{array}{c}1\\ 2\end{array}\right)+\left(\begin{array}{c}2\\ 3\end{array}\right)=\left(\begin{array}{c}3\\ 5\end{array}\right)无非是\text{\hspace{0.17em}}\text{\hspace{0.17em}}1+2=3,\text{\hspace{0.17em}}2+3=5.$

4. 结论

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