Solving the Linear Equation by Using the Gauss

Solving the Linear Equation by Using the Gauss Elimination & the Back-Substitution Soongsin Joo • System of Linear Equations • Method of Gauss Elimination & Back-Substitution 1

Gauss Elimination A: n n matrix , x and b: n-vector Gauss Elimination 2 : Upper-Triangular Matrix

Back-Substitution 3

Algorithm (Gauss Elimination) Initialize the n-vector p to have pi = i (i=0, , n-1). For j = 0, ∙∙∙, n-2, do; Find (the smallest) k≥j such that. If (w[p[k]][j]==0) A is not invertiable and stop. else pk pj (Exchange the contents) For k = j, ∙∙∙, n-1, do; set and r equal to. For i = j, ∙∙∙, n, do; set. If (w[p[k]][j]==0), A is not invertiable and stop. 4

Algorithm [Input the Upper-triangular Matrix W] Define matrix W in the system Ax = b. Work-Matrix: W=(A b), W: n (n+1) (Back substitution) For k = n-2, n-3, ∙∙∙, 0 , do; calculate . 5

Sloving the Linear Equation (1/3) /* Solving the Linear Equation by Using the Gauss Elimination & the Back-Substitution */ /* Include Header Files */ #include <stdio. h> #include <math. h> #include <conio. h> /* Define Global variable */ int i, j, k, l; int n; int p[3]; /* Pivot Vector */ /* Working Matrix */ float w[3][4] = {{2, 3, -1, 5}, {4, 4, -3, 3}, {-2, 3, -1, 1}}; float A[3][3], b[3], x[3]; char FOUTN[20]; FILE *FOUT; /* System of the Linear Equations */ /*Gauss function is calculated for the upper-triangular working matrix*/ void Gauss(); /*Bsub function is calculated for the solution vecter x*/ void Bsub(); /* Main Routine */ void main(){ n=3; /* No. of Rows in W */ /*Input the Name of the Output File and Open It */ printf("n. INPUT THE NAME OF THE OUTPUT FILEn"); scanf("%s", FOUTN); if((FOUT=fopen(FOUTN, "w"))==NULL){ printf("FILE OPEN ERROR. . . n"); getch(); exit(-1); } /* Print Initial Matrix W */ printf("n Initial Matrix n"); fprintf(FOUT, "n Initial Matrix n"); for(i=0; i<n; i++){ for(j=0; j<n+1; j++){ printf("%10. 2 f", w[i][j]); fprintf(FOUT, "%10. 2 f", w[i][j]); } printf("n"); fprintf(FOUT, "n"); } 6

Sloving the Linear Equation (2/3) Gauss(); /*Working matrix W is divided into A[][] and b[]*/ for(i=0; i<n; i++){ for(j=0; j<n; j++){ A[i][j]=w[p[i]][j]; } } for(k=0; k<n; k++){ b[k]=w[p[k]][n]; } Bsub(); /* Print the Output */ printf("n Result of Gauss Elimination n"); for(i=0; i<n; i++){ for(j=0; j<n+1; j++){ printf("%10. 2 f", w[p[i]][j]); } printf("n"); } printf("n - Solution -n"); for(j=0; j<n; j++){ printf(" x%d =%5. 2 fn", j+1, x[j]); } printf("nn******* END *******n"); /* Fileout the Result */ fprintf(FOUT, "n Result of Gauss Elimination n") for(i=0; i<n; i++){ for(j=0; j<n+1; j++){ fprintf(FOUT, "%10. 2 f", w[p[i]][j]); } fprintf(FOUT, "n"); } fprintf(FOUT, "n - Solution -n"); for(j=0; j<n; j++){ fprintf(FOUT, " x%d =%5. 2 fn", j+1, x[j]); } fprintf(FOUT, "nn******* END *******n"); fclose(FOUT); getch(); } void Gauss(){ float r, s; /* Initialize the Pivot vector */ for(i=0; i<n; i++){ p[i]=i; } 7

Sloving the Linear Equation (3/3) for(j=0; j<n-1; j++){ for(k=j; k<n+1; k++){ if(k==n){ printf(" W is not invertible!!"); getch(); exit(-1); } else if(w[p[k]][j]==0)continue; else{ for(l=j+1; l<n+1; l++){ s=w[p[k]][l]; /* Exchange the contents of p[k] and p[j] */ w[p[k]][l]=w[p[j]][l]; w[p[j]][l]=s; } break; } } /* for i=j+1, . . , n-1 do; set r=w[p[i]][j]/w[p[j]][j] */ for(i=k+1; i<n; i++){ r=w[p[i]][j]/w[p[j]][j]; /* For k=j, . . . . , n do; set w[p[i]][k] = w[p[i]][k]-r*w[p[j]][k] */ /* starting value of j is j+1. the source elements are not counted zero. Thus, starting value of k is j as it calculated to give all the elements. */ for(k=j; k<n+1; k++){ w[p[i]][k] = w[p[i]][k] - r*w[p[j]][k]; } } void Bsub(){ int t; x[n-1]=b[n-1]/A[n-1]; for(k=n-2; k>=0; k--){ t=0; for(j=k+1; j<n; j++){ t+=A[k][j]*x[j]; x[k]=(b[k]-t)/A[k][k]; } } } 8