Systems of linear equations

Gauss method

ExampleLet the following system of linear equations is given:

1 2

1 2

2 3 7

4 5 13

x x

x x

To solve this system we’ll remove variable x1 from second equation: (multiply first equation with –4/2 and add to second)

1 2

1 2

2 3 7

0 1 1

x x

x x

21 2 1 1

7 32 3 7 2

2

xx x x x

Find x2 from second equation:

2 21 1.x x

Now find x1 from first equation (using x2):

General Algorithm:(direct step)

• Exclude x1 from all equations starting with 2nd

• Exclude x2 from all equations starting with 3th

............

• Exclude xi from all equations starting with (i+1)th

............

• Exclude xn-1 from equation with index n

General Algorithm:(inverse step)

• Find xn from equation with index n

• Find xn-1 from equation with index n-1........

• Find xi from equation with index i ........

• Find x1 from equation with index 1

Let the following system is given:

1,1 1 1,2 2 1, 1

2,1 1 2,2 2 2, 2

,1 1 ,2 2 2,

,1 1 ,2 2 ,

...

n n

n n

i i n n i

n n n n n n

a x a x a x b

a x a x a x b

a x a x a x b

a x a x a x b

Our task is to transform it to triangular form:

* * * *1,1 1 1,2 2 1, 1

* * *2,2 2 2, 2

* * *, ,

* *,

...

...

...

...

...

n n

n n

i i i i n n i

n n n n

a x a x a x b

a x a x b

a x a x b

a x b

To transform initial system to triangular form we’ll use following algorithm:

For equations from 1 to n-1

For equations from i+1 to n

Multiply all coefficients of equation (i) with -aj,i/ai,i then add equation (i) to equation (j).

Notes:

1. If ai,i=0 then change equation (i) with equation (k), k > i, in which ak,i 0

2. Equation (i) of system will not be changed after multiplication

After transformations we have system:

* * * *1,1 1 1,2 2 1, 1

* * *2,2 2 2, 2

* * *, ,

* *,

...

...

...

...

...

n n

n n

i i i i n n i

n n n n

a x a x a x b

a x a x b

a x a x b

a x b

Our task is to find solutions xn, xn-1, ... x1

*

*,

* *1 1,

1 *1, 1

* *

1

*

...

nn

n n

n n n nn

n n

n

i ij jj i

iii

bx

a

b a xx

a

b a x

xa

To solve this problem we’ll use following algorithm:

• From equation with index (n) find xn

• In equation (n-1) place value of xn and find xn-1

...

• In equation (i) place values of xn, xn-1, ... , xi+1 and find xi

...

• In equation (1) place values of xn, xn-1, ... , x2 and find x1

Data structures:

1. Two-dimensional array A[n,n+1] – for system coefficients and free terms of equations. Line i of array will contain coefficients of equation i of first n places and free term of this equation on n+1 place

2. One-dimensional array X[n] – for system solutions. Element X[i] of array will store solution component xi.

Sample program:program sys_equ; const max = 50; type mat = array[1..max+1, 1..max] of real; Sol = array[1..max] of real; var a: mat; x: Sol; n, i, j: integer;

procedure exclude(var AA:mat;i,j:integer); var k: integer; r: real; begin r:=-AA[j,i]/AA[i,i]; for k:=i to n+1 do AA[j,k]:=AA[j,k]+AA[i,k]*r; end;

function find(AA:mat;i:integer):integer; var k:integer; begin find:=i; for k:=i+1 to n do if AA[k,i]<>0 then find:=k; end;

procedure change(var AA:mat;i,j:integer);var r:real; k:integer;begin for k:=1 to n do begin r:=a[i,k]; a[i,k]:=a[j,k]; a[j,k]:=r; end;end;

procedure transform(var AA:mat); var i,j,l:integer; begin for i:=1 to n-1 do begin if AA[i,i]= 0 then begin l:=find(AA,i); if l<>i then change(AA,i,l) else begin write('Not unique solution'); readln; halt; end; end; for j:=i+1 to n do exclude(AA,i,j); end; end;

procedure readdata(var AA:mat; var n:integer);var i,j : integer; f : text;begin assign(f, 'gauss.in'); reset(f); readln(f,n); for i:=1 to n do for j:=1 to n+1 do read(f, AA[i,j]); close(f);end;

procedure outputsol(X:sol);var i: integer;begin for i:=1 to n do writeln('x[',i,']=',X[i]:0:3); readln;end;

procedure calculate(AA:mat; var X:sol); var s : real; i,j : integer; begin for i:=n downto 1 do begin s:=0; for j:=i+1 to n do s:=s+AA[i,j]*x[j]; x[i]:=(AA[i,n+1]-s)/AA[i,i]; end; end;

begin readdata(A,n); transform(A); calculate(A,X); outputsol(X);end.