c语言用伴随矩阵法求矩阵的逆

伴随矩阵法是一种求解矩阵逆的方法，它适用于方阵。假设矩阵A是一个n阶方阵，它的伴随矩阵为Adj(A)，则矩阵A的逆矩阵A^-1可以用以下公式求解：

A^-1 = (1/det(A)) * Adj(A)

其中，det(A)为矩阵A的行列式。

以下是用伴随矩阵法求解矩阵逆的C语言代码：

#include <stdio.h>
#include <stdlib.h>

// 求解矩阵行列式
float determinant(float **a, int n) {
    int i,j,j1,j2;
    float det = 0;
    float **m = NULL;

    if (n < 1) { /* Error */

    } else if (n == 1) { /* Shouldn't get used */
        det = a[0][0];
    } else if (n == 2) {
        det = a[0][0] * a[1][1] - a[1][0] * a[0][1];
    } else {
        det = 0;
        for (j1=0;j1<n;j1++) {
            m = (float **) malloc((n-1)*sizeof(float *));
            for (i=0;i<n-1;i++)
                m[i] = (float *) malloc((n-1)*sizeof(float));
            for (i=1;i<n;i++) {
                j2 = 0;
                for (j=0;j<n;j++) {
                    if (j == j1)
                        continue;
                    m[i-1][j2] = a[i][j];
                    j2++;
                }
            }
            det += pow(-1.0,j1+2.0) * a[0][j1] * determinant(m,n-1);
            for (i=0;i<n-1;i++)
                free(m[i]);
            free(m);
        }
    }
    return(det);
}

// 求解伴随矩阵
void adjoint(float **a, int n, float **adj) {
    int i,j,ii,jj,i1,j1;
    float det;
    float **b;

    b = (float **) malloc((n-1)*sizeof(float *));
    for (i=0;i<n-1;i++)
        b[i] = (float *) malloc((n-1)*sizeof(float));

    for (j=0;j<n;j++) {
        for (i=0;i<n;i++) {
            // Form the adjoint a_ij
            i1 = 0;
            for (ii=0;ii<n;ii++) {
                if (ii == i)
                    continue;
                j1 = 0;
                for (jj=0;jj<n;jj++) {
                    if (jj == j)
                        continue;
                    b[i1][j1] = a[ii][jj];
                    j1++;
                }
                i1++;
            }
            // Calculate the determinant
            det = determinant(b,n-1);
            // Fill in the elements of the adjoint
            adj[j][i] = pow(-1.0,i+j+2.0) * det;
        }
    }
    for (i=0;i<n-1;i++)
        free(b[i]);
    free(b);
}

// 求解矩阵逆
void inverse(float **a, int n, float **inv) {
    int i,j;
    float det;
    float **adj;

    adj = (float **) malloc(n*sizeof(float *));
    for (i=0;i<n;i++)
        adj[i] = (float *) malloc(n*sizeof(float));

    // Calculate the adjoint matrix
    adjoint(a,n,adj);

    // Calculate the determinant
    det = 0;
    for (i=0;i<n;i++)
        det += a[0][i] * adj[0][i];

    // Calculate the inverse matrix
    for (i=0;i<n;i++) {
        for (j=0;j<n;j++)
            inv[i][j] = adj[i][j] / det;
    }

    for (i=0;i<n;i++)
        free(adj[i]);
    free(adj);
}

int main() {
    int i,j,n;
    float **a, **inv;

    printf("Enter the value of n: ");
    scanf("%d",&n);

    // Allocate memory for matrices
    a = (float **) malloc(n*sizeof(float *));
    for (i=0;i<n;i++)
        a[i] = (float *) malloc(n*sizeof(float));
    inv = (float **) malloc(n*sizeof(float *));
    for (i=0;i<n;i++)
        inv[i] = (float *) malloc(n*sizeof(float));

    // Read in the matrix
    printf("Enter the elements of the matrix:\n");
    for (i=0;i<n;i++) {
        for (j=0;j<n;j++) {
            scanf("%f",&a[i][j]);
        }
    }

    // Calculate the inverse matrix
    inverse(a,n,inv);

    // Display the inverse matrix
    printf("The inverse matrix is:\n");
    for (i=0;i<n;i++) {
        for (j=0;j<n;j++) {
            printf("%f ",inv[i][j]);
        }
        printf("\n");
    }

    // Free memory
    for (i=0;i<n;i++)
        free(a[i]);
    free(a);
    for (i=0;i<n;i++)
        free(inv[i]);
    free(inv);

    return 0;
}