; learn how to test for linear independence.
always has a trivial solution
. has nontrivial solutions if there is at least one
free variable. If 
is the coefficient matrix after a
reduction.
of the system to reduced echelon form, the non-trivial solutions
to are linear combinations of vectors obtained from those columns
of associated with the free variables. The required vectors are
not quite the same as the associated columns of ; 1's or 0's
need to be inserted in the places for the free variables. An easy
way to determine the parametric form is to write out the
equations for 
, move all of the free variable terms
to the right-hand side, insert an equation of the form 
for each
free variable, and then rewrite your solution in vector form with 
on
the left and a vector term for each free variable (with the free variable as
a multiplier) on the right. The parametric vector form for solutions to
is found by adding a (particular) solution 
to the parametric solution (with free variables as parameters) for .
An easy way to find a particular solution to is to
set any free variables (in the reduced system) equal to 0, and then determine
the basic variable values from the right-hand sides.
Linear independence is sometimes difficult to understand initially, and you
might find it easier to think first about dependence. A set of vectors is a
dependent set if one of the vectors 
can be written as a linear
combination of ( depends on) the other vectors in the set.
The simplest check for dependence is to see if any vector in the
set is a multiple of one of the other vectors; if so, the set is dependent.
If not, a more careful analysis is needed to decide whether the set is
a dependent set or an independent set.
If the vectors are put (as columns) into a matrix A, then the set
is dependent if and only iff has some nontrivial solutions.
A general purpose check for dependence is to put the vectors into a matrix
and reduce the matrix. If every column has a pivot, then the vectors are
independent; otherwise they are dependent. Another simple check for
dependence is to count components: if the vectors are n-vectors and
there are more than n of them, they must be dependent. Unfortunately, the
converse is not true: a set of m n-vectors with 
could be
dependent or independent. The columns of the 
identity matrix
form a nice set of independent vectors.