third order matrix

how do I program a simple algorithm in Delphi/pascal for solving a system of three equations with 3 unknowns?

Here is an example:

1st Equation: 3x+4y+5z=19
2nd Eq.       7x+5y-6z=23
3d  Eq.       9x-3y+2z=17

I need to program a matrix so as to obtain the
unknowns x, y and z. Please use Delphi/pascal.
Who is Participating?
VSFConnect With a Mentor Commented:
Glad to help!

Do you have some kind of an example that you could post for us to look at? This could help us to figure out how to taylor something that might work for your situation. :>)

The Crazy One

I don't have the time to write it for you, but maybe you can do somthing with the information below.

You can solve it with the Gauss-Jordan Elimination Algorithm.

The matrix of the example is

  3  4  5 19
  7  5 -6 23
  9 -3  2 17

1.     Read in a matrix of size mxn.(m rows and n columns)
2.     Begin working down the main diagonal starting at row one, column one.
3.     Check for a zero in the working position.
     o     If working position is zero, interchange the working row with the row directly beneath it. If you are on row m (last row) then the system is unsolvable.
     o     Repeat step 3.
4.     Divide the working row by the number in the working position. This will force the diagonal element to have a value equal to one.
5.     Replace the rows beneath the working row by the following:
     1.     Store the value in the working column and next row.
     2.     Multiply working position by the stored number.
     3.     Subtract the product(step 2) from the from the next row entry.
     4.     Move column one to the right.
     5.     Multiply entry in working row new column by stored number.
     6.     Repeat steps 2 - 5 until last column have been operated on.
     7.     Keeping the working row constant, repeat steps 1-6 coupling the next row in line with the working row until all rows have been operated on. If this is done correctly, all the entries below the working position will be zero.
6.     Move to the next position in the diagonal.
7.     Repeat steps 3-7 until the last element in the diagonal has been operated on.
8.     Your matrix is now in row echelon form.
9.     Now begin the back substitution.
10.     For the sake of not sounding redundant, back substitution is doing steps 3-7 going back up the diagonal.

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DidierD described exactly what the Gauss process is.

I have a complete system with the Gauss elimination process implemented in delphi 5 to solve linear equations systems as big as 50x50 (can be greater if you want to)

Freeware with sources
Portuguese and English support
Readme file attached!

Info about the program:

It can be downloaded at:

Mirror Site (Planet Delphi):

Contact me on my e-mail if help is needed!
Here is a complete Pascal Source Code:

Program Gauss_Elimination;

Uses Crt,Printer;

{------------------VarIABLE DECLARATION and  DEFinITIONS--------------------}

  MAXROW = 50; (* Maximum # of rows in a matrix    *)
  MAXCOL = 50; (* Maximum # of columns in a matrix *)

  Mat_Array = Array[1..MAXROW,1..MAXCOL] of Real; (* 2-D Matrix of Reals *)
  Col_Array = Array[1..MAXCOL] of Real; (* 1-D Matrix of Real numbers    *)
  Int_Array = Array[1..MAXCOL] of Integer; (* 1-D Matrix of Integers     *)

  N_EQNS      : Integer;   (* User Input : Number of equations in system  *)
  COEFF_MAT   : Mat_Array; (* User Input : Coefficient Matrix of system   *)
  COL_MAT     : Col_Array; (* User Input : Column matrix of Constants     *)
  X_MAT       : Col_Array; (* OutPut : Solution matrix For unknowns       *)
  orDER_VECT  : Int_Array; (* Defined to pivot rows where necessary       *)
  SCALE_VECT  : Col_Array; (* Defined to divide by largest element in     *)
                           (* row For normalizing effect                  *)
  I,J,K       : Integer;   (* Loop control and Array subscripts           *)
  Ans         : Char;      (* Yes/No response to check inputted matrix    *)


{>>>>>>>>>>>>>>>>>>>>>>>>>   ProcedureS    <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<}

Procedure Home;  (* clears screen and positions cursor at (1,1)            *)
end; (* Procedure Home *)


Procedure Instruct;  (* provides user instructions if wanted               *)

  Ans : Char;  (* Yes/No answer by user For instructions or not            *)

   Home; (* calls Home Procedure *)
   GotoXY(22,8); Writeln('STEVE`S GAUSSIAN ELIMinATION Program');
   GotoXY(36,10); Writeln('2-12-92');
   GotoXY(31,18); Write('Instructions(Y/N):');
   GotoXY(31,49); readln(Ans);
   if Ans in ['Y','y'] then
     Home; (* calls Home Procedure *)
     Writeln('  Welcome to Steve`s Gaussian elimination Program.  With this');
     Writeln('Program you will be able to enter the augmented matrix of    ');
     Writeln('your system of liNear equations and have returned to you the ');
     Writeln('solutions For each unknown.  The Computer will ask you to    ');
     Writeln('input the number of equations in your system and will then   ');
     Writeln('have you input your coefficient matrix and then your column  ');
     Writeln('matrix.  Please remember For n unknowns, you will need to    ');
     Writeln('have n equations.  ThereFore you should be entering a square ');
     Writeln('(nxn) coefficient matrix.  Have FUN!!!!                      ');
     Writeln('(hit <enter> to continue...)');  (* Delay *)


Procedure Initialize_Array( Var Coeff_Mat : Mat_Array ;
                            Var Col_Mat,X_Mat, Scale_Vect : Col_Array;
                            Var order_Vect : Int_Array);

(*** This Procedure initializes all matrices to be used in Program       ***)
(*** ON ENTRY : Matrices have undefined values in them                   ***)
(*** ON Exit  : All Matrices are zero matrices                           ***)

  MAXROW = 50; { maximum # of rows in matrix    }
  MAXCOL = 50; { maximum # of columns in matrix }

  I : Integer; { I & J are both loop control and Array subscripts }
  J : Integer;

  For I :=  1 to MaxRow do   { row indices }
    Col_Mat[I]    := 0;
    X_Mat[I]      := 0;
    order_Vect[I] := 0;
    Scale_Vect[I] := 0;
    For J := 1 to MaxCol do   { column indices }
      Coeff_Mat[I,J] := 0;
end; (* Procedure initialize_Array *)


Procedure Input(Var N : Integer;
                Var Coeff_Mat1 : Mat_Array;
                Var Col_Mat1 : Col_Array);

(*** This Procedure lets the user input the number of equations and the  ***)
(*** augmented matrix of their system of equations                       ***)
(*** ON ENTRY : N => number of equations : UNDEFinED
                Coeff_Mat1 => coefficient matrix : UNDEFinED
                Col_Mat1 => column matrix :UNDEFinED
     ON Exit  : N => # of equations input by user
                Coeff_Mat1 => defined coefficient matrix
                Col_Mat1 => defined column matrix input by user          ***)

  I,J : Integer;  (* loop control and Array indices *)

  Home; (* calls Procedure Home *)
  Write('Enter the number of equations in your system: ');
  Writeln('Now you will enter your coefficient and column matrix:');
  For I := 1 to N do     { row indice }
    Writeln('ROW #',I);
    For J := 1 to N do   {column indice }
      readln(Coeff_Mat1[I,J]);    {input of coefficient matrix}
    readln(Col_Mat1[I]);          {input of Constant matrix}
end;  (* Procedure Input *)


Procedure Check_Input( Coeff_Mat1 : Mat_Array;
                          N : Integer; Var Ans : Char);

(*** This Procedure displays the user's input matrix and asks if it is  ***)
(*** correct.                                                           ***)
(*** ON ENTRY : Coeff_Mat1 => inputted matrix
                N => inputted number of equations
                Ans => UNDEFinED                                        ***)
(*** ON Exit  : Coeff_Mat1 => n/a
                N => n/a
                Ans => Y,y or N,n                                       ***)

  I,J   : Integer;  (* loop control and Array indices *)

  Home; (* calls Home Procedure *)
  Writeln; Writeln('Your inputted augmented matrix is:');Writeln;Writeln;

  For I := 1 to N do   { row indice }
    For J := 1 to N do { column indice }
  Writeln; Write('Is this your desired matrix?(Y/N):'); (* Gets Answer *)
end;  (* Procedure Check_Input *)


Procedure order(Var Scale_Vect1 : Col_Array;
                Var order_Vect1 : Int_Array;
                Var Coeff_Mat1  : Mat_Array;
                    N           : Integer);

(*** This Procedure finds the order and scaling value For each row of the
     inputted coefficient matrix.                                        ***)
(*** ON ENTRY : Scale_Vect1 => UNDEFinED
                order_Vect1 => UNDEFinED
                Coeff_Mat1  => as inputted
                N           => # of equations
     ON Exit  : Scale_Vect1 => contains highest value For each row of the
                               coefficient matrix
                order_Vect1 => is assigned the row number of each row from
                               the coefficient matrix in order
                Coeff_Mat   => n/a
                N           => n/a                                      ***)

  I,J : Integer;  {loop control and Array indices}

For I := 1 to N do
    order_Vect1[I] := I;  (* ordervect gets the row number of each row *)
    Scale_Vect1[I] := Abs(Coeff_Mat1[I,1]); (* gets the first number of each row *)
    For J := 2 to N do { goes through the columns }
      begin  (* Compares values in each row of the coefficient matrix and
                stores this value in scale_vect[i] *)
        if Abs(Coeff_Mat1[I,J]) > Scale_Vect1[I] then
           Scale_Vect1[I] := Abs(Coeff_Mat1[I,J]);
end;  (* Procedure order *)


Procedure Pivot(Var Scale_Vect1 : Col_Array;
                    Coeff_Mat1  : Mat_Array;
                Var order_Vect1 : Int_Array;
                    K,N         : Integer);

(*** This Procedure finds the largest number in each column after it has been
     scaled and Compares it With the number in the corresponding diagonal
     position. For example, in column one, a(1,1) is divided by the scaling
     factor of row one. then each value in the matrix that is in column one
     is divided by its own row's scaling vector and Compared With the
     position above it. So a(1,1)/scalevect[1] is Compared to a[2,1]/scalevect[2]
     and which ever is greater has its row number stored as pivot. Once the
     highest value For a column is found, rows will be switched so that the
     leading position has the highest possible value after being scaled. ***)

(*** ON ENTRY : Scale_Vect1 => the normalizing value of each row
                Coeff_Mat1  => the inputted coefficient matrix
                order_Vect1 => the row number of each row in original order
                K           => passed in from the eliminate Procedure
                N           => number of equations
     ON Exit  : Scale_Vect  => same
                Coeff_Mat1  => same
                order_Vect  => contains the row number With highest scaled
                k           => n/a
                N           => n/a                                      ***)

  I           : Integer; {loop control and Array indice }
  Pivot, Idum : Integer; {holds temporary values For pivoting }
  Big,Dummy   : Real; {used to Compare values of each column }
  Pivot := K;
  Big := Abs(Coeff_Mat1[order_Vect1[K],K]/Scale_Vect1[order_Vect1[K]]);
  For I := K+1 to N do
    Dummy := Abs(Coeff_Mat1[order_Vect1[I],K]/Scale_Vect1[order_Vect1[I]]);
    if Dummy > Big then
      Big := Dummy;
      Pivot := I;
  Idum := order_Vect1[Pivot];              { switching routine }
  order_Vect1[Pivot] := order_Vect1[K];
  order_Vect1[K] := Idum;
end; { Procedure pivot }


Procedure Eliminate(Var Col_Mat1, Scale_Vect1 : Col_Array;
                    Var Coeff_Mat1 : Mat_Array;
                    Var order_Vect1 : Int_Array;
                    N : Integer);

  I,J,K       : Integer;
  Factor      : Real;

 For K := 1 to N-1 do
   Pivot (Scale_Vect1,Coeff_Mat1,order_Vect1,K,N);
   For I := K+1 to N do
     Factor := Coeff_Mat1[order_Vect1[I],K]/Coeff_Mat1[order_Vect1[K],K];
     For J := K+1 to N do
       Coeff_Mat1[order_Vect1[I],J] := Coeff_Mat1[order_Vect1[I],J] -
   Col_Mat1[order_Vect1[I]] := Col_Mat1[order_Vect1[I]] - Factor*Col_Mat1[order_Vect1[K]];


Procedure Substitute(Var Col_Mat1, X_Mat1 : Col_Array;
                         Coeff_Mat1 : Mat_Array;
                     Var order_Vect1 : Int_Array;
                     N : Integer);

(*** This Procedure will backsubstitute to find the solutions to your
     system of liNear equations.
     ON ENTRY : Col_Mat => your modified Constant column matrix
                X_Mat1  => UNDEFinED
                Coeff_Mat1 => modified into upper triangular matrix
                order_Vect => contains the order of your rows
                N          => number of equations
     ON Exit  : Col_Mat => n/a
                X_MAt1  => your solutions !!!!!!!!!!!!!
                Coeff_Mat1 => n/a
                order_Vect1 => who cares
                N           => n/a                                      ***)

  I, J  : Integer; (* loop and indice of Array control *)
  Sum   : Real;    (* used to sum each row's elements *)

  X_Mat1[N] := Col_Mat1[order_Vect1[N]]/Coeff_Mat1[order_Vect1[N],N];
  (***** This gives you the value of x[n] *********)

  For I := N-1 downto 1 do
    Sum := 0.0;
    For J := I+1 to N do
      Sum := Sum + Coeff_Mat1[order_Vect1[I],J]*X_Mat1[J];
    X_Mat1[I] := (Col_Mat1[order_Vect1[I]] - Sum)/Coeff_Mat1[order_Vect1[I],I];
end;   (** Procedure substitute **)


Procedure Output(X_Mat1: Col_Array; N : Integer);

(*** This Procedure outputs the solutions to the inputted system of     ***)
(*** equations                                                          ***)
(*** ON ENTRY : X_Mat1 => the solutions to the system of equations
                N => the number of equations
     ON Exit  : X_Mat1 => n/a
                N => n/a                                                ***)

  I    : Integer; (* loop control and Array indice *)

  Writeln;Writeln;Writeln; (* skips lines *)
  Writeln('The solutions to your sytem of equations are:');
  For I := 1 to N do
  Writeln('X(',I,') := ',X_Mat1[I]);
end;   (* Procedure /output *)

(*                                                                         *)
(*                                                                         *)
(*                                                                         *)


    Instruct;  (* calls Procedure Instruct *)
    Initialize_Array(Coeff_Mat, Col_Mat, X_Mat, Scale_Vect, order_Vect);
             (* calls Procedure Initialize_Array *)
      Input(N_EQNS, Coeff_Mat, Col_Mat); (* calls Procedure Input *)
      Check_Input(Coeff_Mat,N_EQNS,Ans); (* calls Procedure check_Input *)
    Until Ans in ['Y','y']; (* loops Until user inputs correct matrix *)

    order(Scale_Vect,order_Vect,Coeff_Mat,N_EQNS); (* calls Procedure order *)
    Eliminate(Col_Mat,Scale_Vect,Coeff_Mat,order_Vect,N_EQNS);   (*etc..*)
    Substitute(Col_Mat,X_Mat,Coeff_Mat,order_Vect,N_EQNS);       (*etc..*)
    Output(X_Mat,N_EQNS);                                        (*etc..*)

    Write('Do you wish to solve another system of equations?(Y/N):');
  Until Ans in ['N','n'];

end. (*************** end of Program GAUSS_ELIMinATION *******************)

casalAuthor Commented:
I wish to thank VSF too, but I do not finde the <accept> bar.
Are you having problems with Experts-Exchange <accept> bar?
If so, please report here or to the community support.

Hi there...
I'm still waiting for the points!
Are you having problems with EE engine?!
Please report if so.

tic tac tic tac .....Rinnnnnngggggggg time is up :-)

He accepted all his other questions, so i think he will accept your answer eventually.
Keep trying :-)

Good luck,

casalAuthor Commented:
Yes yes  I do accept.
Let me know if you did'nt get your 200 points.
casalAuthor Commented:
I haven't been able to find an accept "BAR", only an accept "radio button". Am I doing something wrong??
Please tell me haow to do the right thing!
casalAuthor Commented:
My thanks, and my ACCEPTANCE go to both VSF and DidierD.
Does pressing the "accept" radio button not suffice?

Needing anything just call for help!

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