## Homogeneous and Non Homogeneous Systems of Linear Equations

A homogeneous system of linear equations is any system of equations where the coefficients are constant and each equation is equal to zero.
Example:
$2x_1-2x_2=0$

$4x_1+5x_2=0$

Any system of homogeneous equation has a zero solution if the number of equations is greater than the number of variables, and cannot be reduced to a number of equations less than the number of variables. Such equations are said to be independent. The equations above have only the solution
$x_2=x_2=0$
, because the equations are independent. We can write a homogeneous system of equations in the form
$A \mathbf{x}= \mathbf{0}$
. If the equations are independent then
1. If the number of equations is greater than or equal to the number of variables then the only solution is where each variable is equal to zero. This is called the trivial solution.
2. If the number of equations is less than the number of variables then there are an infinite number of solutions. The equation
$x_1+x_2=0$
has the infinite solution set
$x_1=s, x_2=-s$
where
$s$
is arbitrary.
A non homogeneous system of linear equations is any system of equations where the coefficients are constant and at least one equation is not equal to zero.
Example:
$2x_1-2x_2=3$

$4x_1+5x_2=0$

We can write a non homogeneous system of equations in the form
$A \mathbf{x}= \mathbf{b}$
. If the equations are independent then
1. If the number of equations is greater than the number of variables then there is no solution.
Example:
$x_1+x_2=5$

$x_1-x_2=7$

$x_1=0$

The equations are inconsistent. Adding half the first to half the second gives
$x_1=6$
which is contradicted by the third equation.
2. If the number of equations is equal to the number of variables, and the equations are consistent, then there is a unique solution. If the equations are inconsistent then there is no solution.
Example:
$x_1+x_2=5$

$x_1-x_2=7$

The solution is
$x_1=6, \: x_2 =-1$
.
2. If the number of equations is less than the number of variables then there are an infinite number of solutions. The equation
$x_1+x_2=3$
has the infinite solution set
$x_1=s, x_2=3-s$
where
$s$
is arbitrary.