System of Linear Equations Solver
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Matrices eigenvalues and rotation matrix.
Solutions of the linear system of equations
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System of linear equations Print ellipse
A system of  m  linear equations in  n  unknowns has a solution if and only if the rank  r  of the augmented matrix equals that of the coefficient matrix.
If the two matrices have the same rank r and   r = n,   the solution is unique.
If the two matrices have the same rank r and   r < n,   then at least one set of  r  of the unknowns can be solved in terms of the remaining   (n   r)   unknowns.
The general form of a linear system of equations can be presented by the formula:     A xi = Bi
Coefficient matrix is the matrix which contains the variables coefficiens (A matrix in the formula).
Constant matrix is the array of the free column values (Bi vector array).
Augmented matrix is the combined matrix which contains the Coefficient matrix and Constant matrix side by side (ABi matrix).
Cramer's rule
Another way to solve linear system of equation is by Cramer's rule which involves only determinants. Consider the system of equations:
Linear system of equations
The solution is given by the equation:   Solution of linear system of equations
Di    is the determinant esteblished by changing the ith column with the free values column bi
D    is the determinant of the Coefficient matrix.
Examples of solving linear system of equations
Example: Solve the system of linear equations of three variables x, y and z:
x   y + 2z =   2
3x   2y + 4z =   5
2y   3z = 2
Consider the system of linear equations containing three variables x, y and z. this equations cab be written in matrix notation as:

One way to solve the equations is by row transformation on the augmented matrix.

From the third line it is obvious that     z = 0
From second line:    y   2z = 1    →    y = 1
From first row:        x   y + 2z =   2    →    x =   1
Solving the above equations by Cramer' s rule.
Solve the above equations by Cramer' s rule.
Example: Solve the system of the linear equations:
x + 2y   3z = 4
x + 3y + z = 11
2x + 5y   4z = 13
2x + 6y + 2z = 22
After performing rows operations we get the upper triangular matrix:

It is seen that r = n (rank = number of variables) therfore the equations have a unique solution:     z = 1     y = 3     x = 1     or in vector representation (1, 3, 1)
Example: Solve the system of the linear equations:
x + 2y   2z + 3w = 2
2x + 4y   3z + 4w = 5
5x + 10y   8z + 11w = 12

The result is that     r < n (r = 2 n = 4)
r   Rank of the matrix   total non zero rows
n   Number of variables
Because    r   n = 4   2 = 2    two variables are dependent on the other two variables and we have to choose certain values for two variables
for example:     w = a     and     y = b     (a, b are any number) then:

z = 1 + 2a
x = 4   2b + a
The solution space vector is     (4 + a   2b,   b,   1 + 2a,   a)
for example if we choose:     a = 1    b =   1    then the solution is:
(7,   1, 3, 1).
Example: Solve the system of the linear equations:
2x + y   2z + 3w = 1
3x + 2y   z + 2w = 4
3x + 3y + 3z   3w = 5

From the last row it is observed that: 0 =   8 this is clearly impossible,and this set of equations don't have a solution.