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:
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.
