正方矩阵求逆 (C++)

核心思想

将矩阵A 以单位矩阵作为陪矩阵 [A | I]。通过同时变换矩阵A 和单位矩阵I ,使得A 变为单位矩阵,I 经过得到相应的变换得到的矩阵就是矩阵A 的逆矩阵A-1

 

例子

矩阵 A 为:image  ,

添上等大的单位矩阵作为伴随矩阵:image

最终计算得到矩阵A 的逆矩阵:image

 

流程

image

代码

 1: Matrix8g Inverse( Matrix8g &mat){

 2:     Matrix8g mat_inv;

 3:     Matrix8g mat_cpy;

 4:     int dim;

 5:     int i, nzero_row,j;

 6:     uint8_t temp, jtimes;

 7:

 8:     assert(mat.rr == mat.cc);

 9:     dim = mat.rr;

 10:     /* mat_cpy = mat;*/

 11:     Copy(mat_cpy , mat);

 12:

 13:     mat_inv.Make_identity(mat.rr, mat.cc);

 14:

 15:     /* from column 0 to column dim-1;

 16:  * dim = this.rr = this.cc */

 17:     for(i = 0; i < dim; i++){

 18:         nzero_row = i;

 19:         /* matrix(i, i) == 1 add the nzero_row th row 

 20:  * which elems[nzero_row*dim +i] != 0 to the ith row*/

 21:         if(mat_cpy.Get(i,i) == 0){

 22:             do{

 23:                 ++nzero_row;

 24:                 if(nzero_row >= dim)

 25:                     ERROR("Non-full rank matrix!");

 26:                 temp = mat_cpy.Get(nzero_row,i);

 27:             }while((temp == 0)&&(nzero_row < dim));

 28:             mat_cpy.Swap_rows(i,nzero_row);

 29:             mat_inv.Swap_rows(i,nzero_row);

 30:         }

 31:         /* matrix(i, i) != 0 now */

 32:         for(j = 0; j < dim; j++){

 33:             /* if matrix(j,i) == 0; then */

 34:             if(mat_cpy.Get(j,i) == 0)

 35:                 continue;

 36:             if(j != i){

 37:                 jtimes = (uint8_t)galois_single_divide(mat_cpy.Get(j,i), mat_cpy.Get(i,i), 8);

 38:                 mat_cpy.Row_plus_irow(j, i, jtimes);

 39:                 mat_inv.Row_plus_irow(j, i, jtimes);

 40:             }

 41:             else{

 42:                 jtimes = (uint8_t)galois_inverse(mat_cpy.Get(i, i), 8);

 43:                 mat_cpy.Row_to_irow(i , jtimes);

 44:                 mat_inv.Row_to_irow(i , jtimes);

 45:             }

 46:

 47:         }

 48:     }

 49:     return mat_inv;

 50: }

同样的另一篇关于Guass – Jardon 消元计算逆矩阵:http://www.codelast.com/?p=1288

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