线性方程组 [英] Linear Equations

问题描述

any1可以帮助我用C/C ++进行程序设计以解决5级线性方程吗?的方程组是5，而5个变量是dere ... ...给我一些想法...... div ="h2_lin">解决方案

有很多网站只是在谷歌上搜索这个高斯消去C ++" .热门歌曲之一将我引导到了这里: http://www.programmersheaven.com/download/29309/Download.aspx [ ^ ].

我想向您推荐由两位作者写的一本非常不错的书:《伊尔娅·尼古拉·耶维奇·布朗斯坦和康斯坦丁·阿道夫·威奇·塞门贾杰》，但是不得不发现我找不到它的英文译本.这是我在德国学习计算机科学时所用的书，其中有一章很好地论述了求解n度线性方程式的算法.也许您可以在Google上搜索更多内容并找到翻译.

我发布的链接是C ++代码的一种，专门用于称为"Gauss-Seidel"算法的"Gaussian Elimination"方法.

希望对您有所帮助！

最好的问候，
曼弗雷德(Manfred)

此处，曼弗雷德(Manfred)是100％，这是指向多个受尊敬的"实现的有用链接:
面向对象的数字页面 [ ^ ]

可能有一天会派上用场，这当然是一个很好的学习资源.

问候
Espen Harlinn

您可以使用高斯消除方法.以下是主要步骤
1.创建系数矩阵.
2.使用行或列明智的操作将其简化为上三角形式
3.现在，您可以从最后到第一个开始等效系数.

:)

Can any1 help me wid the designing of a program in C/C++ to solve a linear equation of degree 5 i.e. the no. of equations is 5 and 5 variables are dere... Suggest me some ideas man...

解决方案

There are lots of sites just google this "gaussian elimination C++". One of the hits directed me there: http://www.programmersheaven.com/download/29309/Download.aspx[^].

I would have like to recommend you a very good book by two authors "Ilja Nikolajewitsch Bronstein and Konstantin Adolfowitsch Semendjajew", but had to discover I couldn''t find an english translation of it. It''s a book that I used during my studies of computer science in Germany and has a great chapter on algorithms for solving linear equations of n degrees. Maybe you can google some more and find a translation.

The link I posted is to C++ code for a specialization of the "Gaussian Elimination" method called "Gauss-Seidel" algorithm.

Hope this helps some!

Best Regards,
Manfred

While Manfred is 100% right here is a useful link to several "respected" implementations:
The Object-Oriented Numerics Page[^]

May come in handy one day, and it''s certainly a great learning source.

Regards
Espen Harlinn

You can use gauss elimination method. following are the major steps
1. create a Coefficient matrix.
2.reduce to upper triangular form, using row, or column wise operations
3. now u can start equating Coefficient from last to first.

:)

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