(Partial) Solutions to Homework 2
Problem 4.1
Write down examples of augmented matrices corresponding
to each of the five types of solution sets for systems of equations with three
unknowns. (no solution, one solution, a line of solutions, a plane of solutions,
all of
R
3
.)
Answer:
The easiest way to write down these systems is using reduced row
echelon form matrices (or even just row echelon form). Why is that? Remember,
the number of free variables and pivot variables has something to do with the
solution set. Let’s look at an example to figure out what they have to do with
each other.
Example: Let
M
=
1
0
1
0
1
0
0
0
0
, let
V
=
5
2
0
and let’s look for solutions
to the system
MX
=
V
. Our matrix
M
is in reduced row echelon form, so it’s
easy to solve. We have one free variable, corresponding to column 2. Let’s say
our variables are
x
1
, x
2
,
and
x
3
, so that
x
1
and
x
3
are pivot variables, and
x
2
is a free variable. Then to write the solution set, we give
x
2
an arbitrary value,
say
x
2
=
λ
, and then we can write
x
1
and
x
3
in terms of
λ
(although in this
case,
x
3
does not need to be written in terms of
λ
, we know in fact that
x
3
= 2).
So any solution has
x
3
= 2,
x
2
=
λ
, and
x
1
= 5

λ
. In vector form, this can be
written as:
{
5

λ
λ
2
}
=
{
5
0
2
+
λ

1
1
0
}
What does this set look like, i.e., what does it represent? There is one degree
of freedom, that is, we can make
λ
whatever we want it to be. Think about it
– if we have a vector
v
, and we’re allowed to have any multiple of
v
, then that
looks like a line.
So this solution set looks like a line.
You can do examples
with different numbers of free variables to get a better idea of how they relate
to geometric interpretation, but it turns out that the number of free variables
is the dimension of the solution set. So in three variables, we have the following
correspondence.
If there is at least one solution
, then
•
3 pivot variables, no free variables
↔
one unique solution
•
2 pivot variables, 1 free variable
↔
line of solutions = one dimension of
solutions
•
1 pivot variable, 2 free variables
↔
plane of solutions = two dimensions
of solutions
•
0 pivot variables, 3 free variables
↔
three dimensions of solutions
But wait, what does 3 free variables mean? If we only have a 3x3 matrix,
how do we have three free variables? Remember what a free variable is – it’s