## 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 :

_{1,1} ·x_{1} + ... + a_{1,n} ·x_{n} = b_{1}

: :

a_{n,1} ·x_{1} + ... + a_{n,n} ·x_{n} = b_{n}

As an intermediate result the **ref** (Row Echelon Form) and the
**rref** (Reduced Row Echelon Form)

### 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: ^{2} + b·x + c

P(1|3) ∈ C_{f} : _{1} + 1·x_{2} + 1·x_{3} = 3

Q(2|1) ∈ C_{f} : _{1} + 2·x_{2} + 1·x_{3} = 1

R(4|9) ∈ C_{f} : _{1} + 4·x_{2} + 1·x_{3} = 9

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

Thus the parabola is discribed by y = 2x^{2} - 8x + 9.

### Pop-up Menu:

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

*Cut Matrix , Copy Matrix*and*Paste Matrix*With this you may copy the matrix to the clipboard and paste it into "Matrix multiplication".

*Export Matrix*and*Import Matrix*Exports or imports the matrix in CSV format (Comma separated values), which is used to exchange data with Excel.