Linear Algebra

Systems of Linear Equations

The program determines the solution vector of a system of linear equations (SLE) with n equations and n variables.

First enter then number of equations and then the coefficients of the system. The system must be transformed to :

a1,1 ·x1 + ... + a1,n ·xn = b1

  :                                  :

an,1 ·x1 + ... + an,n ·xn = bn

As an intermediate result the ref (Row Echelon Form) and the rref (Reduced Row Echelon Form) can be reported.

Example with unique solution:

      
 1·x1 + 1·x2 + 1·x3  =   3
 4·x1 + 2·x2 + 1·x3  =   1
16·x1 + 4·x2 + 1·x3  =   9

L = { ( 2; -8; 9; ) }

Example with one-dim. solution:

 2·x1 + 3·x2 + 4·x3  =   0
 1·x1 - 1·x2 - 1·x3  =   1
 3·x1 + 2·x2 + 3·x3  =   1

L = { ( 0,6-0,2t; -0,4-1,2t; t ) | t ∈ R }}

Example with two-dim. solution:

 0·x1 + 0·x2 + 2·x3 - 1·x4  =  1
 1·x1 + 1·x2 + 1·x3 + 1·x4  =  4
 2·x1 + 2·x2 - 4·x3 + 5·x4  =  5
 1·x1 + 1·x2 - 7·x3 + 5·x4  =  0

L = { ( 3,5-s-1,5t; s; 0,5+0,5t; t ) | s,t ∈ R }

Demonstration for the first example:

If you are looking for a parabola through  P(1|3), Q(2|1) and R(4|9), you have to solve the following system of equations.

Approach:   f(x) = a·x2 + b·x + c

P(1|3) ∈ Cf :         1·x1 + 1·x2 + 1·x3  =   3

Q(2|1) ∈ Cf :         4·x1 + 2·x2 + 1·x3  =   1

R(4|9) ∈ Cf :       16·x1 + 4·x2 + 1·x3  =   9

The solution vector is:     (2, -8, 9)

Thus the parabola is discribed by   y = 2x2 - 8x + 9.

Pop-up Menu:

Right click to open a local menu, which offers you the following functions to manage the matrix.

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