摘要：
使用namespacestd；#定义N 10//将方阵的最大阶定义为10//函数floatMatDet（flow*p；//定义矩阵flow[N][N]；row；buffer=（flow*）calloc（num；i<；row；“row；
- 矩阵求逆c++实现
高斯消元法可以用来找出一个可逆矩阵的逆矩阵。设A 为一个N * N的矩阵，其逆矩阵可被两个分块矩阵表示出来。将一个N * N单位矩阵 放在A 的右手边，形成一个N * 2N的分块矩阵B = [A,I] 。经过高斯消元法的计算程序后，矩阵B 的左手边会变成一个单位矩阵I ，而逆矩阵A ^(-1) 会出现在B 的右手边。假如高斯消元法不能将A 化为三角形的格式，那就代表A 是一个不可逆的矩阵。应用上，高斯消元法极少被用来求出逆矩阵。高斯消元法通常只为线性方程组求解。
实验结果：123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263//********************************//*** 求任何一个实矩阵的逆***//********************************#include"stdafx.h"#include <math.h>#include <malloc.h>#include <iostream>#include <iomanip>using namespace std;#define N10//定义方阵的最大阶数为10//函数的声明部分floatMatDet(float*p,intn);//求矩阵的行列式floatCreat_M(float*p,intm,intn,intk);//求矩阵元素A(m, n)的代数余之式voidprint(float*p,intn);//输出矩阵n*nbool Gauss(floatA[][N],floatB[][N],intn);//采用部分主元的高斯消去法求方阵A的逆矩阵Bintmain(){float*buffer, *p;//定义数组首地址指针变量introw, num;//定义矩阵的行数和矩阵元素个数inti, j;floatdeterm;//定义矩阵的行列式floata[N][N], b[N][N];intn;cout <<"采用逆矩阵的定义法求矩阵的逆矩阵! ";cout <<"请输入矩阵的行数: ";cin >> row;num =2* row * row;buffer = (float*)calloc(num, sizeof(float));//分配内存单元p = buffer;if(NULL != p){for(i =0; i < row; i++){cout <<"Please input the number of "<< i+1<<" row: ";for(j =0; j < row; j++){cin >> *p++;}}}else{cout <<"Can't distribute memory ";}cout <<"The original matrix : ";print(buffer, row);//打印该矩阵determ = MatDet(buffer, row);//求整个矩阵的行列式p = buffer + row * row;if(determ !=0){cout <<"The determinant of the matrix is "<< determ << endl;for(i =0; i < row; i++)//求逆矩阵{for(j =0; j < row; j++){*(p+j*row+i) = Creat_M(buffer, i, j, row)/determ;}}cout <<"The inverse matrix is: "<< endl;print(p, row);//打印该矩阵}else{cout <<"The determinant is 0, and there is no inverse matrix! ";}free(buffer);//释放内存空间cout <<"采用部分主元的高斯消去法求方阵的逆矩阵! ";cout <<"请输入方阵的阶数: ";cin >> n;cout <<"请输入"<< n <<"阶方阵: ";//输入一个n阶方阵for(i =0; i < n; i++){for(j =0; j < n; j++){cin >> a[i][j];}}//运用高斯消去法求该矩阵的逆矩阵并输出if(Gauss(a, b, n)){cout <<"该方阵的逆矩阵为: ";for(i =0; i < n; i++){cout << setw(4);for(j =0; j < n; j++){cout << b[i][j] << setw(10);}cout << endl;}}getchar();return0;}//-----------------------------------------------//功能: 求矩阵(n*n)的行列式//入口参数: 矩阵的首地址，矩阵的行数//返回值: 矩阵的行列式值//----------------------------------------------floatMatDet(float*p,intn){intr, c, m;intlop =0;floatresult =0;floatmid =1;if(n !=1){lop = (n ==2) ?1: n;//控制求和循环次数,若为2阶，则循环1次，否则为n次for(m =0; m < lop; m++){mid =1;//顺序求和, 主对角线元素相乘之和for(r =0, c = m; r < n; r++, c++){mid = mid * (*(p+r*n+c%n));}result += mid;}for(m =0; m < lop; m++){mid =1;//逆序相减, 减去次对角线元素乘积for(r =0, c = n-1-m+n; r < n; r++, c--){mid = mid * (*(p+r*n+c%n));}result -= mid;}}elseresult = *p;returnresult;}//----------------------------------------------------------------------------//功能: 求k*k矩阵中元素A(m, n)的代数余之式//入口参数: k*k矩阵的首地址，矩阵元素A的下标m,n,矩阵行数k//返回值: k*k矩阵中元素A(m, n)的代数余之式//----------------------------------------------------------------------------floatCreat_M(float*p,intm,intn,intk){intlen;inti, j;floatmid_result =0;intsign =1;float*p_creat, *p_mid;len = (k-1)*(k-1);//k阶矩阵的代数余之式为k-1阶矩阵p_creat = (float*)calloc(len, sizeof(float));//分配内存单元p_mid = p_creat;for(i =0; i < k; i++){for(j =0; j < k; j++){if(i != m && j != n)//将除第i行和第j列外的所有元素存储到以p_mid为首地址的内存单元{*p_mid++ = *(p+i*k+j);}}}sign = (m+n)%2==0?1: -1;//代数余之式前面的正、负号mid_result = (float)sign*MatDet(p_creat, k-1);free(p_creat);returnmid_result;}//-----------------------------------------------------//功能: 打印n*n矩阵//入口参数: n*n矩阵的首地址,矩阵的行数n//返回值: 无返回值//-----------------------------------------------------voidprint(float*p,intn){inti, j;for(i =0; i < n; i++){cout << setw(4);for(j =0; j < n; j++){cout << setiosflags(ios::right) << *p++ << setw(10);}cout << endl;}}//------------------------------------------------------------------//功能: 采用部分主元的高斯消去法求方阵A的逆矩阵B//入口参数: 输入方阵，输出方阵，方阵阶数//返回值: true or false//-------------------------------------------------------------------bool Gauss(floatA[][N],floatB[][N],intn){inti, j, k;floatmax, temp;floatt[N][N];//临时矩阵//将A矩阵存放在临时矩阵t[n][n]中for(i =0; i < n; i++){for(j =0; j < n; j++){t[i][j] = A[i][j];}}//初始化B矩阵为单位阵for(i =0; i < n; i++){for(j =0; j < n; j++){B[i][j] = (i == j) ? (float)1:0;}}for(i =0; i < n; i++){//寻找主元max = t[i][i];k = i;for(j = i+1; j < n; j++){if(fabs(t[j][i]) > fabs(max)){max = t[j][i];k = j;}}//如果主元所在行不是第i行，进行行交换if(k != i){for(j =0; j < n; j++){temp = t[i][j];t[i][j] = t[k][j];t[k][j] = temp;//B伴随交换temp = B[i][j];B[i][j] = B[k][j];B[k][j] = temp;}}//判断主元是否为0, 若是, 则矩阵A不是满秩矩阵,不存在逆矩阵if(t[i][i] ==0){cout <<"There is no inverse matrix!";returnfalse;}//消去A的第i列除去i行以外的各行元素temp = t[i][i];for(j =0; j < n; j++){t[i][j] = t[i][j] / temp;//主对角线上的元素变为1B[i][j] = B[i][j] / temp;//伴随计算}for(j =0; j < n; j++)//第0行->第n行{if(j != i)//不是第i行{temp = t[j][i];for(k =0; k < n; k++)//第j行元素 - i行元素*j列i行元素{t[j][k] = t[j][k] - t[i][k]*temp;B[j][k] = B[j][k] - B[i][k]*temp;}}}}getchar();returntrue;}</iomanip></iostream></malloc.h></math.h>