Welcome to a lesson
on homogeneous systems of linear equations.
The goals of the video are to define
a homogeneous system of linear equations
and also to solve a homogeneous system of linear equations.
A system of linear equations is homogeneous
if the constant term equals zero,
meaning the system must fit this form here.
And again the main thing to remember
is the constant term must be zero.
A homogeneous system of equations
will always have either one trivial solution,
when all the variables are equal to zero,
or it will have an infinite number of solutions.
If there are m equations with n unknowns
and n is greater than m, meaning the number of variables
is greater than the number of equations,
then there are infinite solutions.
And there will also be n minus m free variables.
The number of variables minus the number of equations
will tell us how many free variables we have,
which are the variables that can take on any value
and we'll assign a parameter for them.
Let's take a look at our examples.
We want to solve the system of equations
using an augmented matrix,
and notice how because the constants are equal to zero,
this is a homogeneous system of linear equations.
Let's go and set up the augmented matrix.
So we have two, negative three, zero.
And then we have six, negative eight, zero.
We want to write this in row echelon form,
but let's focus on getting a zero here.
Notice if we multiply a positive two by negative three,
that'd give us a negative six.
Then we could add that to row two.
So we're going to place row two
with negative three times row one plus row two.
So the first row stays the same.
Then we'll have negative three times two.
That's negative six plus six, that's zero.
Negative three times negative three is positive nine,
plus negative eight is positive one.
And this is still zero.
Going back up to our system just for a moment,
notice how we have two variables and two equations,
which means for this system, we have zero free variables.
I think this is enough work to solve our system.
Let's go ahead and re-write this.
Now we would have two x sub one
minus three times x sub two equals zero.
And also x sub two must equal zero.
Well notice if x sub two equals zero,
the first equation becomes two times x sub one equals zero,
which means x sub one must equal zero.
So just as defined in the previous slide,
a homogeneous system either has the trivial solution
when the variables are equal to zero,
or has an infinite number of solutions.
Here we have the trivial solution
where x sub one equals zero and x sub two equals zero.
Let's take a look at a second example.
Notice in this homogeneous system we have
three variables and two equations,
which means we'll have one free variable
and an infinite number of solutions.
So not only will we have the trivial solution
when all the variables are equal to zero.
We'll have an infinite number of solutions
which we'll express parametrically.
Let's first manipulate this system though
using an augmented matrix.
So we'll have one, two, negative three, zero
and two, one, six, zero.
Let's start by obtaining a zero here.
So we're going to replace this row
with negative two times row one plus row two.
So the first row stays the same.
Second row is going to be negative two times one plus two.
That's zero.
Negative two times positive two plus one,
that's negative three.
Negative two times negative three plus six,
that's six plus six or 12.
We have zero.
Now to obtain a one here,
let's multiply row two by negative 1/3.
First row stays the same.
Second row would be zero, one,
this would be negative four, and zero.
Now let's go ahead and rewrite this
as a system of equations.
Notice how the first equation didn't change.
So we have x sub one plus two times x sub two
minus three times x sub three equals zero.
Second equation is now x sub two
minus four x sub three equals zero.
Again because we have three variables and two equations,
we have one free variable,
which we'll assign a parameter to.
So let's let t equal some real number.
And then looking at our equations,
looks like it'll be easiest if we let x sub three equal t.
So because we're letting x sub three be t,
let's go ahead and solve this first equation for x sub one
and the second equation for x sub two.
So we would have x sub one equals
three x sub three minus two x sub two.
And then x sub two is going to be equal
to four times x sub three.
Now let's go ahead and put all these pieces together
to represent all of our solutions parametrically.
Again we're going to start by letting x sub three equal t,
which means x sub two would equal four t.
So x sub one is going to be equal to three times x sub three,
which is going to be three t minus two times x sub two.
x sub two is four t.
So x sub one is equal to three t
minus eight t is negative five t.
This would be the parametric representation
of the solution to the system of equations
that has an infinite number of solutions.
Okay, hope you found this helpful.