如何反转n * n矩阵 [英] How do I invert an n*n matrix

问题描述

我想得到n * n方法的反演，所以使用增广方法，所以我做了单位矩阵，然后做了列减少，然后发现反向

i want to get the inversion of n*n method so using augmented method so i made the identity matrix then made column reduction then found the inverse

for(i=0;i<n;i++)
 {
    t=a[i][i];
    for(j=0;j<2*n;j++)
    {a[i][j]=a[i][j]/t;}     // divide all the row column by a[i][i] element

    for(int m=0;m<n;m++)
    {
       if(i!=m)
       {
          t=a[m][i];
          for(k=0;k<2*n;k++)
              a[m][k]-=t*a[i][k];
        }
    }
 }

需要此部分的解释



我尝试了什么：



i尝试分析它并找到：

1.通过此子主的对角元素分割所有行元素

2.减少一行行，这部分无法理解

need explanation for this part

What I have tried:

i tried to analyze it and found:
1.dividing all row element by the main diagonal element of this roe
2.make a row reduction and this part can not understand it

推荐答案

我有一种方法可以解决矩阵的逆矩阵，逐列求解：

高精度原生高斯消除 [ ^ ]



我也认为你会从阅读矩阵中受益，不只是反位。矩阵在很多方面都非常有用，但是掌握和理解它们起初可能有点棘手。

I have one way of solving the inverse of a matrix, solving it column by column:
High precision native Gaussian Elimination[^]

I also think you would benefit from reading up on matrixes in general, not just the inverse bit. Matrices are very useful in many areas, but mastering and understanding them can be a little tricky at first.

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