LU分解求九维矩阵逆 C语言实现

LU分解是一种常用于求矩阵逆的方法，下面是一个使用C语言实现九维矩阵逆的示例代码：

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

#define N 9

void luDecomposition(double A[N][N], int P[N]) {
    int i, j, k;
    double max, tmp;
    for (i = 0; i < N; i++) {
        P[i] = i; // 初始化置换矩阵
        max = 0.0;
        for (j = 0; j < N; j++) {
            if (abs(A[i][j]) > max) {
                max = abs(A[i][j]); // 找出第i行绝对值最大的元素
            }
        }
        if (max == 0.0) { // 矩阵奇异，无法进行LU分解
            printf("Singular matrix, cannot do LU decomposition.\n");
            return;
        }
        for (j = i; j < N; j++) {
            A[i][j] /= max; // 将第i行除以绝对值最大的元素，以避免在后面的计算中出现大量小数
        }
        for (j = i + 1; j < N; j++) {
            max = A[j][i]; // 找出第i列下方绝对值最大的元素
            k = i;
            for (int l = i + 1; l < N; l++) {
                if (abs(A[l][i]) > max) {
                    max = abs(A[l][i]);
                    k = l;
                }
            }
            if (max == 0.0) { // 矩阵奇异，无法进行LU分解
                printf("Singular matrix, cannot do LU decomposition.\n");
                return;
            }
            if (k != j) { // 交换第j行和第k行
                for (int l = 0; l < N; l++) {
                    tmp = A[k][l];
                    A[k][l] = A[j][l];
                    A[j][l] = tmp;
                }
                tmp = P[k];
                P[k] = P[j];
                P[j] = tmp;
            }
            A[j][i] /= A[i][i];
            for (int l = i + 1; l < N; l++) {
                A[j][l] -= A[j][i] * A[i][l];
            }
        }
    }
}

void luSolve(double A[N][N], int P[N], double b[N], double x[N]) {
    int i, j;
    double sum;
    for (i = 0; i < N; i++) {
        sum = b[P[i]];
        for (j = 0; j < i; j++) {
            sum -= A[i][j] * x[j];
        }
        x[i] = sum;
    }
    for (i = N - 1; i >= 0; i--) {
        sum = x[i];
        for (j = i + 1; j < N; j++) {
            sum -= A[i][j] * x[j];
        }
        x[i] = sum / A[i][i];
    }
}

void matrixInverse(double A[N][N], double A_inv[N][N]) {
    int P[N];
    double L[N][N], U[N][N], b[N], x[N];
    int i, j, k;
    luDecomposition(A, P);
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            if (i == j) {
                L[i][j] = 1.0;
                U[i][j] = A[i][j];
            } else if (i > j) {
                L[i][j] = A[i][j];
                U[i][j] = 0.0;
            } else {
                L[i][j] = 0.0;
                U[i][j] = A[i][j];
            }
        }
    }
    for (k = 0; k < N; k++) {
        for (i = 0; i < N; i++) {
            b[i] = 0.0;
        }
        b[k] = 1.0;
        luSolve(L, P, b, x);
        for (i = 0; i < N; i++) {
            A_inv[i][k] = x[i];
        }
        luSolve(U, P, x, b);
        for (i = 0; i < N; i++) {
            A_inv[i][k] = x[i];
        }
    }
}

void printMatrix(double A[N][N], char *name) {
    int i, j;
    printf("%s = [\n", name);
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            printf("%8.4f ", A[i][j]);
        }
        printf("\n");
    }
    printf("]\n");
}

int main() {
    double A[N][N] = {{0.8147, 0.0975, 0.1576, 0.1419, 0.6557, 0.7577, 0.7431, 0.3922, 0.6555},
                      {0.9058, 0.2785, 0.9706, 0.4218, 0.0357, 0.7431, 0.3922, 0.7060, 0.1712},
                      {0.1270, 0.5469, 0.9572, 0.9157, 0.8491, 0.3922, 0.7060, 0.0318, 0.7060},
                      {0.9134, 0.9575, 0.4854, 0.7922, 0.9340, 0.0318, 0.0318, 0.2769, 0.0318},
                      {0.6324, 0.9649, 0.8003, 0.9595, 0.6787, 0.2769, 0.2769, 0.0462, 0.2769},
                      {0.0975, 0.1576, 0.1419, 0.6557, 0.7577, 0.7431, 0.3922, 0.6555, 0.3922},
                      {0.2785, 0.9706, 0.4218, 0.0357, 0.7431, 0.3922, 0.7060, 0.1712, 0.7060},
                      {0.5469, 0.9572, 0.9157, 0.8491, 0.3922, 0.7060, 0.0318, 0.7060, 0.0318},
                      {0.9575, 0.4854, 0.7922, 0.9340, 0.0318, 0.0318, 0.2769, 0.0318, 0.2769}};
    double A_inv[N][N];
    printMatrix(A, "A");
    matrixInverse(A, A_inv);
    printMatrix(A_inv, "A_inv");
    return 0;
}

其中，`luDecomposition`函数用于进行LU分解，`luSolve`函数用于求解线性方程组，`matrixInverse`函数用于求矩阵逆，`printMatrix`函数用于打印矩阵。在`main`函数中，我们定义了一个九维矩阵`A`，并求出它的逆矩阵`A_inv`，然后分别打印出`A`和`A_inv`。

需要注意的是，LU分解的实现中需要使用置换矩阵，这里我们用一个长度为N的数组`P`来表示。此外，在进行LU分解和求解线性方程组时，需要避免除以0的情况，因此要特判矩阵奇异的情况。