My linear algebra library (the value of the inverse matrix, determinant)

#include "stdafx.h"

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

// N可以随意指定
#define N 4

void main()
{
    double a[N][N];
    double L[N][N], U[N][N];
    double r[N][N], u[N][N];
    double out[N][N], out1[N][N], E[N][N];
    memset(a, 0, sizeof(a));    // 初始化
    memset(L, 0, sizeof(L));
    memset(U, 0, sizeof(U));
    memset(r, 0, sizeof(r));
    memset(u, 0, sizeof(u));
    int n = N;
    int k,i,j;
    double s,t;
    srand(time(0));

    // 输入矩阵
    printf("input A=\n");
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            //scanf("%f", &a[i][j]);
            do{
                a[i][j] = rand() % 100;
            }while(a[i][j] == 0);
        }
    }
    
    the printf ( " input matrix: \ n- " );
     for (I = 0 ; I <n-; I ++ ) {
         for (J = 0 ; J <n-; J ++ ) {
            printf(" %lf", a[i][j]);
        }
        printf("\n");
    }

    // first row of the matrix U unchanged
     // for (J = 0; J <n-; J ++) {
     //     A [0] [J] = A [0] [J];             // Calculation of the matrix U line
     // } 
    for (I = . 1 ; I <n-; I ++ ) {
        A [I] [ 0 ] = A [I] [ 0 ] / A [ 0 ] [ 0 ];     // first column of matrix U computing 
    }
    
    for(k=1;k<n;k++){
        for(j=k;j<n;j++){
            s = 0;
            for(i=0;i<k;i++){
                s += a[k][i] * a[i][j];    //累加
            }
            A [K] [J] = A [K] [J] - S;     // other matrix elements calculated U 
        }
         for (I = K + . 1 ; I <n-; I ++ ) {
            t = 0;
            for(j=0;j<k;j++){
                T + A = [I] [J] * A [J] [K];     // accumulating 
            }
            A [I] [K] = (A [I] [K] - T) / A [K] [K];     // Other matrix element calculation L 
        }
    }

    for (I = 0 ; I <n-; I ++ ) {
         for (J = 0 ; J <n-; J ++ ) {
             IF (I> J) {
                 // if i> j, description line is greater than columns, calculated by the following triangular matrix portion, obtained value of L, U is 0 
                L [I] [J] = a [I] [J];
                U[i][j] = 0;
            }else if(i == j){
                U[i][j] = a[i][j];
                L[i][j] = 1;
            } The else {
                 // Otherwise i <j rows than the column description, the upper triangular part of the matrix calculation, the value of U derived, L is 0 
                U [I] [J] = A [I] [J];
                L[i][j] = 0;
            }
        }
    }

    double temp = 1.0;
    for(i=0;i<n;i++){
        temp *= U[i][i];
    }
    if(temp == 0){
        the printf ( " \ n-matrix does not exist " );
         return ;
    }


    // inverse matrix of L and U

    // find the inverse matrix U 
    for (I = 0 ; I <n-; I ++ ) {
        U [I] [I] = . 1 / the U-[I] [I];     // value of the inverse of the diagonal elements of the direct access 
        for (K = I- . 1 ; K> = 0 ; K-- ) {
            s = 0;
            for(j=k+1;j<=i;j++){
                s += U[k][j] * u[j][i];
            }
            U [K] [I] = -s / the U-[K] [K];     // iteration, once obtained by the reverse of each column value 
        }
    }

    // find the inverse matrix L 
    for (I = 0 ; I <n-; I ++ ) {
        R & lt [I] [I] = . 1 ;     // value of the diagonal elements directly taking the reciprocal of, here. 1 
        for (K = I + . 1 ; K <n-; K ++ ) {
             for (J = I; J <= - K- . 1 ; J ++ ) {
                R & lt [K] [I] - = L [K] [J] R & lt * [J] [I];     // iteration, once obtained sequentially according to each of the column value 
            }
        }
    }


    // Print L, the U- 
    the printf ( " \ nL matrix: " );
     for (I = 0 ; I <n-; I ++ ) {
        printf("\n");
        for(j=0;j<n;j++){
            printf(" %lf", L[i][j]);
        }
    }
    printf("\nU矩阵:");
    for(i=0;i<n;i++){
        printf("\n");
        for(j=0;j<n;j++){
            printf(" %lf", U[i][j]);
        }
    }
    printf("\n");

    // Print L, U inverse matrix of 
    the printf ( " \ nL inverse of a matrix: " );
     for (I = 0 ; I <n-; I ++ ) {
        printf("\n");
        for(j=0;j<n;j++){
            printf(" %lf", r[i][j]);
        }
    }
    the printf ( " \ nU inverse of a matrix: " );
     for (I = 0 ; I <n-; I ++ ) {
        printf("\n");
        for(j=0;j<n;j++){
            printf(" %lf", u[i][j]);
        }
    }
    printf("\n");

    // validate multiplying U and L, the original matrix to obtain 
    the printf ( " \ nL matrix and the matrix product U \ n- " );
     for (I = 0 ; I <n-; I ++ ) {
         for (J = 0 ; J <n-; ++ J ) {
             OUT [I] [J] = 0 ;
        }
    }
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            for(k=0;k<n;k++){
                out[i][j] += L[i][k]*U[k][j];
            }
        }
    }
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            printf(" %lf", out[i][j]);
        }
        printf("\n");
    }

    // The r and u multiplied by the inverse matrix of 
    the printf ( " \ n inverse matrix of the original matrix \ n " );
     for (I = 0 ; I <n; I ++ ) {
         for (J = 0 ; J <n; ++ J ) {
            out1[i][j] = 0;
        }
    }
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            for(k=0;k<n;k++){
                out1[i][j] += u[i][k]*r[k][j];
            }
        }
    }
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            printf(" %lf", out1[i][j]);
        }
        printf("\n");
    }

    // validate 
    the printf ( " \ n-former matrix and the inverse matrix product \ n- " );
     for (I = 0 ; I <n-; I ++ ) {
         for (J = 0 ; J <n-; J ++ ) {
            E[i][j] = 0;
        }
    }
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            for(k=0;k<n;k++){
                E[i][j] += out[i][k]*out1[k][j];
            }
        }
    }
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            printf(" %lf", E[i][j]);
        }
        printf("\n");
    }

    double det = 1;
    for(i=0;i<n;i++){
        the * = U [i] [i];
    }
    the printf ( " \ nA% of the determinant F = \ n- " , DET);
}

 

#include "stdafx.h"
#include <stdlib.h>#include <stdio.h>#include <time.h>#include <string.h>
#define N 4
void main(){double a[N][N];double L[N][N], U[N][N];double r[N][N], u[N][N];double out[N][N], out1[N][N], E[N][N];memset(a, 0, sizeof(a));// 初始化memset(L, 0, sizeof(L));memset(U, 0, sizeof(U));memset(r, 0, sizeof(r));memset(u, 0, sizeof(u));int n = N;int k,i,j;double s,t;srand(time(0));
// 输入矩阵printf("input A=\n");for(i=0;i<n;i++){for(j=0;j<n;j++){//scanf("%f", &a[i][j]);do{a[i][j] = rand() % 100;}while(a[i][j] == 0);}}printf("输入矩阵:\n");for(i=0;i<n;i++){for(j=0;j<n;j++){printf(" %lf", a[i][j]);}printf("\n");}
The first row of the constant matrix U // // for (j = 0; j <n; j ++) {// a [0] [j] = a [0] [j]; // Calculation of the matrix U line //} for (i = 1; i <n; i ++) {a [i] [0] = a [i] [0] / a [0] [0]; // calculate a first matrix U column} for (k = 1; k <n; k ++) {for (j = k; j <n; j ++) {s = 0; for (i = 0; i <k; i ++) {s + = a [ k] [i] * a [ i] [j]; // accumulating} a [k] [j] = a [k] [j] - s; // calculate other elements of the matrix U} for (i = k +1; i <n; i ++ ) {t = 0; for (j = 0; j <k; j ++) {t + = a [i] [j] * a [j] [k]; // accumulating} a [i] [k] = (a [i] [k] - t) / a [k] [k]; other elements // L is calculated matrices}}
for (I = 0; I <n-; I ++) {for (j = 0; j <n; j ++) {if (i> j) {// if i> j, is greater than the described rows columns, calculated by the following matrix of triangular section, the value of L obtained, U is 0L [i] [j] = a [i] [j]; U [i] [j] = 0;} else if (i == j) {U [i] [j] = a [i] [j] ; L [i] [j] = 1;} else {// otherwise i <j rows than the column description, the triangular part of the matrix calculation, the value of U derived, L is 0U [i] [j] = a [ I] [J]; L [I] [J] = 0;}}}
Double TEMP = 1.0; for (I = 0; I <n-; I ++) {TEMP * = the U-[I] [I];} IF (temp == 0) {printf ( "\ n matrix does not exist"); return;}

// inverse matrices L and U
// find the inverse matrix U for (i = 0; i < n; i ++) {u [i] [i] = 1 / U [i] [i]; // value of the diagonal elements of the inverse of the direct access for ( k = i-1; k> = 0; k -) {s = 0; for (j = k + 1; j <= i; j ++) {s + = U [k] [j] * u [j ] [i];} u [ k] [i] = -s / U [k] [k]; // iteration, once obtained by the reverse of each column value}}
// find the inverse matrix L for (i = 0; i <n; i ++) {r [i] [i] = 1; // value of the diagonal elements directly taking the reciprocal of, here 1for (k = i + 1; k <n; k ++) {for ( j = i; j <= k -1; j ++) {r [k] [i] - = L [k] [j] * r [j] [i]; iteration // in columns sequentially once obtained per a value}}}

// Print L, uprintf ( "\ nL matrix:"); for (i = 0; i <n; i ++) {printf ( "\ n"); for (j = 0; j <n ; j ++) {printf ( " % lf", L [i] [j]);}} printf ( "\ nU matrix:"); for (i = 0; i <n; i ++) {printf ( "\ n "); for (J = 0; J <n-; J ++) {the printf ("% LF ", U [I] [J]);}} the printf (" \ n-");
// Print L, the inverse of U matrix printf ( "\ nL inverse of a matrix:"); for (i = 0; i <n; i ++) {printf ( "\ n"); for (j = 0; j <n; j ++) {printf ( "% lf", r [i ] [j]);}} printf ( "\ nU inverse matrix of the matrix:"); for (i = 0; i <n;i++){printf("\n");for(j=0;j<n;j++){printf(" %lf", u[i][j]);}}printf("\n");
// validate multiplying U and L, to obtain the original matrix printf ( "\ nL matrix and the matrix product U \ n"); for (i = 0; i <n; i ++) {for (j = 0; j <n; j ++) {out [i] [j] = 0;}} for (i = 0; i <n; i ++) {for (j = 0; j <n; j ++) {for (k = 0; k <n ; k ++) {out [i ] [j] + = L [i] [k] * U [k] [j];}}} for (i = 0; i <n; i ++) {for (j = 0 ; J <n-; J ++) {the printf ( "% LF", OUT [I] [J]);} printf ( "\ n-");}
// multiplying r and u, to obtain an inverse matrix printf ( "\ an inverse matrix of n original matrix \ n "); for (i = 0; i <n; i ++) {for (j = 0; j <n; j ++) {out1 [i] [j] = 0;}} for (i = 0; i <n ; i ++) {for (j = 0; j <n; j ++) {for (k = 0; k <n; k ++) {out1 [i] [j] + = u [i ] [k] * r [k ] [j];}}} for (i = 0; i <n; i ++) {for (j = 0; j <n; j ++) {printf ( "% lf", out1 [I] [J]);} printf ( "\ n");}
// verify printf ( "\ n of the original matrix and the inverse matrix product \ n"); for (i = 0; i <n; i ++) {for (j = 0; j <n; j ++) {E [i] [j] = 0;}} for (i = 0; i <n; i ++) {for (j = 0; j <n; j ++ ) {for (k = 0; k <n; k ++) {E [i] [j] + = out [i] [k] * out1 [k] [j];}}} for (i = 0; i <n; i ++) {for (j = 0; j <n;j++){printf(" %lf", E[i][j]);}printf("\n");}
double det = 1; for (i = 0; i <n; i ++) {det * = U [i] [i];} printf ( "\ nA determinant value =% f \ n", det);}