SNPLSNPL 11

GAUSS ELIMINATION GAUSS ELIMINATION

& BACK & BACK SUBSTITUTIONSUBSTITUTION

Dayun YuDayun YuSeungmuk JiSeungmuk Ji

SNPLSNPL 22

Gauss EliminationGauss EliminationGauss EliminationGauss Elimination nn matrix : A , n-vector : x and b,

Ax = b

The solution vector x can be obtained without difficulty in case A is upper-triangular with all diagonal.

Ax = b

nnnnnn

n

n

b

b

b

x

x

x

aaa

aaa

aaa

2

1

2

1

21

22221

11211

nnnn

n

n

b

b

b

x

x

x

a

aa

aaa

2

1

2

1

222

11211

00

0b x A

SNPLSNPL 33

Back SubstitutionBack SubstitutionBack SubstitutionBack Substitutionann 0, We must have

We now know xn, the second last equation

Similarly,

In general

nn

nn ab

x

1,1

,11

1

1,11,111,1 0,

nn

nnnn

n

nnnnnnnnn

a

xabx

abxaxa

0, 2,2

2,2

,211,22

2

nn

nn

nnnnnnn

n aa

xaxabx

0,1

kk

kk

n

kjjkjk

k aa

xabx

SNPLSNPL 44

Define matrix Define matrix WWDefine matrix Define matrix WW

Define matrix W in the system Ax = b

nnnn

n

n

baa

baa

baa

1

2221

1111

W

SNPLSNPL 55

PivotingPivotingPivotingPivoting

Partial-pivoting is interchanging row.

☞ If first column(a11=0) of first row is zero, we have to interchange other row .

Full-pivoting is interchanging row and columns.

☞ If is satisfy, we must find .

So we execute interchanging row ( row i row

p).

ki max

nki pi aa i p

SNPLSNPL 66

ExampleExampleExampleExampleEx)

132

3344

532

321

2321

1321

xxx

pxxx

pxxx

15500

7120

5132

6260

7120

5132

1132

3344

5132

21pp

W

SNPLSNPL 77

Algorithm 1Algorithm 1Algorithm 1Algorithm 1(Upper-triangular)Initialize the n-vector p to have pi = i, i = 1,,n. For k = 1, ,n-1∙∙∙ , do; For i = k+1, ,n∙∙∙ , do; set m and , m equal to .

For j = k+1, , n+1, do;∙∙∙ set .

kpiw

kp

kp

k

i

w

w

jpjpjp kii

mwww

SNPLSNPL 88

Algorithm 2Algorithm 2Algorithm 2Algorithm 2(Back substitution) For k = n, n-1,∙∙∙,1 , do;

calculate .kk

n

kj jkjk

k a

xabx

1

SNPLSNPL 99

Programming with CProgramming with CProgramming with CProgramming with C#include <stdio.h>#include <math.h>#include <conio.h>main(){

int i, j, k, t;int n, N;float m, s;float w[3][4]={{2,3,-1, 5},{4,4,-3, 3},{-2,3,-1, 1}};float p[3]={0.0};clrscr();

n = 3;N = 4;

printf("\n\n***** Matrix W ****\n\n"); for(i = 0; i < n; i++) { for(j = 0; j < N; j++) printf("%10.3f", w[i]

[j]);printf("\n");

p[i] = i; }

for(k = 0; k < n-1; k ++) { for(j = k; j < N; j++) { if(j == n) {

printf(" W is not invertible ! "); break; } else if(w[p[j]][k] == 0) continue; else { for(t = k; t < N; t++) {

s = w[p[j]][t];w[p[j]][t] = w[p[k]][t];w[p[k]][t] = s;

} } break;

}for(i = k+1; i < n; i++) { q = w[p[i]][k] / w[p[k]][k]; for(j = k; j < N; j++) w[p[i]][j] = w[p[i]][j] - q*w[p[k]][j]; } }

SNPLSNPL 1010

Programming with CProgramming with CProgramming with CProgramming with Cprintf("\n\n"); for(i = 0; i < n; i++) { for(j =0; j < N; j++) printf("%10.3f", w[p[i]][j]);

printf("\n"); }

}getch();return(0);

}

SNPLSNPL 1111

ResultResultResultResult

***** Matrix W ****

2.000 3.000 -1.000 5.000 4.000 4.000 -3.000 3.000 -2.000 3.000 -1.000 1.000

2.000 3.000 -1.000 5.000 0.000 -2.000 -1.000 -7.000 0.000 0.000 -5.000 -15.000

SNPLSNPL 1212

General program with CGeneral program with CGeneral program with CGeneral program with C#include <stdio.h>#include <conio.h>#include <math.h>#include <stdlib.h>#define MAXROW 20

typedef struct{ int Row,Column; double Element[MAXROW][MAXROW+1];} Matrix;

Matrix M;void gainp(int a, int b);void gapiv(int row, int i);void gacal(int row, int col);

int main(){ int a,b; clrscr(); printf(“Row is :”); scanf(“%2d”,&a); printf(“Column is :”); scanf(“%2d”,&b);

gainp(a,b); gacal(a,b); getch(); return 0;}

void gainp(int a, int b){ int i,j; double tmp; printf(“\n”); printf("Enter the matrix (%2d,%2d) \n",a,b); for(i=0;i<a;i++) { for(j=0;j<b;j++){ printf("Enter for element (%2d,%2d) :", i+1,j+1); scanf(" %lf",&tmp); M.Element[i][j]=tmp; } printf("\n"); }}

SNPLSNPL 1313

General program with CGeneral program with CGeneral program with CGeneral program with Cvoid gapiv(int row, int i){ int j,ii; double tmp; for(ii=i+1;ii<row;ii++){ if(M.Element[ii][i]!=0.0){ for(j=i;j<row+1;j++){

tmp=M.Element[i][j]; M.Element[i][j]=M.Element[ii][j]; M.Element[ii][j]=tmp;

} break; } }}void gacal(int row, int col){ int i,j,k, N; double div, sum, X[MAXROW]; for(i=0;i<row;i++){ if(M.Element[i][i]==0.0) gapiv(row,i);

for(j=i+1;j<row;j++){ div=M.Element[j][i]/M.Element[i][i]; for(k=0;k<col;k++)

M.Element[j][k]-=M.Element[i][k]*div; } }

N=row; X[N-1]=M.Element[N-1][N]/M.Element[N-1][N-1]; for(i=N-2;i>=0;i--){ sum=0.0; for(j=i+1;j<N;j++) sum += M.Element[i][j]*X[j]; X[i]=(M.Element[i][N]-sum)/M.Element[i][i]; }

printf("Solution is :\n"); for(i=0;i<N;i++) printf(" X%-d = %f\n",i+1,X[i]);}