esProc Solves Multivariable Linear Equations

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For a given system of linear equations in n unknowns, there must be n equations to form the system. e.g.:




Well, how do you solve this system of linear equations with n variables with esProc? First, the system of equations should be written in a standard form, and parameters are filled in a text file. Parameters in the same line are separated with Tab. As listed below: 


How to solve systems of linear equations in n unknowns by programming? The method of elimination by addition and subtraction is widely used. See the following system of equations:






If a11 is not equal to zero, we multiply each term of the first equation simultaneously by a21/a11, subtract the first equation from the second one, then the coefficient of x1 in the second equation becomes zero. If a11 equals zero, we need to find an equation in which the coefficient of x1 is not zero, and make it the first. In this way, we can eliminate all coefficients of x1 in the equations after the first one. See below:






Now let’s look at the case of b22. In the same method, we continue to eliminate coefficients of x2 in all equations after the second one… We do this continuously until the coefficient of xn-1 in the nth equation is eliminated. At this point, the system of equations will be like the following:






And we can solve it step by step from the nth equation:






If, during the calculation, zero appears in parameters like a11,b22,c33,the system of equations has no solution or infinitely many solutions.

The following shows how to solve a system of linear equations in n unknowns with esProc:

  A B C D
1 =file(“EquInput.txt”) =A1.import() =B1.len() >result=[0]*C1
2 =B1.((C1+1).(B1.~.field(#)))    
3 for C1-1 =0    
4   for A3,C1 if A2(B4)(A3)!=0 >B3=B4
5       break
6   if B3==0 || A3==C1 next  
7   =A2(B3) >A2(B3)=A2(A3) >A2(A3)=B7
8   for A3+1,C1 if A2(B8)(A3)==0 next
9     for C1+1,A3 >A2(B8)(C9)-=B7(C9)*A2(B8)(A3)/B7(A3)
10 for C1,1 =C1.sum(result(#)*A2(A10)(#))  
11   if A2(A10)(A10)!=0 >result(A10)=round((A2(A10)(C1+1)-B10)/A2(A10)(A10),4)
12   else >result=”Error” break
13 =result      

In B1, data is read from the text file; Parameters of the system of equations are transformed into sequence groups in A2:


C1 seeks the number of equations in the system of equations, then the result gets looping execution in A3, and coefficients of xn will be gradually eliminated with the method of elimination by addition and subtraction. During looping execution in B4, we look for the first equation in which the coefficient of xk is not zero from the kth equation; if we cannot find one like this, it means that the term cannot get eliminated. Among the code in line 7, move the equation in which the coefficient of xk is not zero to line k. In line 8 and 9, we eliminate coefficients of xk in equations starting from line k+1.

Having done the elimination of every step, we can start to work out solution gradually from the last equation in A10, during which a sequence called result will be used to store solutions. If there is no solution or infinite many solutions, return Error. Considering that there could be errors with double-precision number during computation, the solutions will be kept to four decimal places.

Result can be viewed in A13 after all computations:


If new system of equations is required to be solved:




x1+x2+x3 =1

We only need to modify the EquInput.txt file:


With the same program, we compute and read result in A13.


In addition, list of parameters is unnecessary to be imported from a file. Instead, they can be written directly in esProc. The computing progress is similar to that we’ve explained.

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