C-PROGRAM TO SOLVE L INEAR EQUATIONS

Prepared by:

Sendash Pangambam

Dept. of Computer Science

Banaras Hindu University

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MATHEMATICAL CONCEPTS:

Given equations:

1. a11x+a12y+a13z=c1

2. a21x+a22y+a23z=c2

3. a31x+a32y+a33z=c3

Three equations in three variables x, y and z.

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MATHEMATICAL CONCEPTS:

In matrix form:

a11 a12 a13 x c1

a21 a22 a23 y = c2

a31 a32 a33 z c3

=>AX = C

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MATHEMATICAL CONCEPTS:

AX = C

= (A-1A)X = A-1C

=> IX = A-1C

=> X = A-1C

Thus, one can find the values of elements of X i.e. x, y

and z by comparing the two matrices X and A-1C.

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BASIC STEPS:

Find the determinant of A.

• If det.A≠0, system of eqns. has a unique solution.

• If det.A=0, no solution. A is singular.

Find Inverse of A.

Solve by using matrix method.

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NOW LETS GO TO

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VARIABLES USED IN MY PROGRAM

Co-efficients of X,Y and Z in the three eqns. are stored in an array a[3][3] and constant

terms in the array c[3].

A[3][3] is first used to store the co-factors, and later is used to store the Adjoint of the

matrix a[3][3]

det is used to store the value of determinant of a[3][3].

X[3] is used to store the value of the three variables X,Y and Z.

In the swap() function, arguments are *a and *b, temp is a temporary variable used

while swapping.

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PROGR AM WIT H E X PLANAT IONS(1 )

/*SOLVE A SYSTEM OF LINEAR EQUATION WITH 3 VARIABLES BY MATRIX METHOD*/

#include<stdio.h>

#include<conio.h> Header files are included here.

#include<stdlib.h>

/*FUNCTION TO SWAP*/

void swap(float *a,float *b)

{

float temp; Function to swap two numbers

temp=*a; To be used while calculating Adjoint/Transpose

*a=*b;

*b=temp;

}

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PROGRAM WITH EXPLANATIONS(2)

void main() main() function starts here.

{

int i,j; Variable declaration.

float a[3][3],c[3],A[3][3],INV[3][3],det=0.0,X[3];

clrscr();

/*ENTER THE CO-EFFICIENTS AND CONSTANT TERMS*/

for(i=0;i<3;i++) Equation

{ taken as

printf("\nEnter the X,Y,Z co-efficients and the constant term of equation no. input

%d: ",i+1); here.

scanf("%f%f%f%f",&a[i][0],&a[i][1],&a[i][2],&c[i]);

}

clrscr();

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PROGRAM WITH EXPLANATIONS(3)

/*PRINTING THE EQUATIONS*/

printf("\nThe equations are: \n"); Printing

for(i=0;i<3;i++) the eqns.

printf("%.2f X + %.2f Y + %.2f Z = %.2f \n",a[i][0],a[i][1],a[i][2],c[i]); for surety.

/*FINDING CO-FACTORS*/

for(i=0;i<3;i++) Finding cofactors

{ to use while calculating

for(j=0;j<3;j++) determinant, adjoint,

A[i][j]=(a[(i+1)%3][(j+1)%3]*a[(i+2)%3][(j+2)%3]) - inverse, etc.

(a[(i+1)%3][(j+2)%3]*a[(i+2)%3][(j+1)%3]);

}

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PROGRAM WITH EXPLANATIONS(4)

/*FINDING DETERMINANT OF THE MATRIX*/

for(i=0;i<3;i++) Finding determinant using 1stcolumn

det+=a[i][0]*A[i][0];

printf("\n\n\n");

if(det==0)

{

printf("SINGULAR MATRIX !\nCANNOT FIND THE Check whether the

SOLUTION\n\n\tPress any key to exit."); matrix is singular or not.

getch();

exit(1);

}

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PROGRAM WITH EXPLANATIONS(5)

/*FINDING ADJOINT MATRIX*/

for(i=0;i<2;i++) Calculation of adjoint.

{ Adjoint=Transpose of the matrix formed

for(j=i+1;j<3;j++) by co-factors of corresponding element.

swap(&A[i][j], &A[j][i]);

}

/*FINDING INVERSE*/

for(i=0;i<3;i++) Calculation of inverse.

{

for(j=0;j<3;j++) Inverse of A=Adjoint of A/det. A

INV[i][j]=A[i][j]/det;

}

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PROGRAM WITH EXPLANATIONS(6)

/*SOLVING MATRIX EQUATION*/

for(i=0;i<3;i++)

{

X[i]=0.0; Finding the value of X = A-1C

for(j=0;j<3;j++) And find the values of elements of X

{ by comparing corresponding elements

X[i]+=INV[i][j]*c[j]; of X and A-1C

}

}

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PROGRAM WITH EXPLANATIONS(7)

/*PRINTING THE SOLUTIONS*/

printf("Solution of the equation:

\n\tX=%.2f\n\tY=%.2f\n\tZ=%.2f",X[0],X[1],X[2]); Printing the solutions

getch();

} End of main()

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RUN THE PROGRAM

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THUS THE PROBLEM IS SOLVED

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