Welcome to a lesson
in homogeneous systems of equations.
A system of equations is called homogeneous
if each equation is equal to zero.
A homogeneous system has the form shown below
and this is not as complicated as it looks.
a sub 1, 1 through a sub mn are the scalars or coefficients
and x sub 1 through x sub n are the variables.
And there are two examples of homogeneous systems below.
Notice, in both systems the equations are equal to zero,
which is why they're considered homogeneous systems
of equations.
A homogeneous system of equations always has the solution
where all the variables are equal to zero.
So for the first system on the left,
because we have two unknowns,
zero comma zero is the trivial solution
to the homogeneous system.
Notice, when x is zero and y is zero,
both equations are true or are satisfied.
For the system on the right,
because we have three variables,
the order triple, zero comma zero comma zero,
is the trivial solution.
Once again, notice how when x, y, and z are all zero,
all three equations are satisfied or are true.
A solution in which not all the variables are equal to zero
is called a nontrivial solution.
Let's take a look at an example
of determining the nontrivial solutions
to a homogeneous system of equations.
Notice, here we have two equations with three unknowns,
and therefore the trivial solution
is zero comma zero comma zero,
but we're looking for the nontrivial solutions.
To find the nontrivial solutions, if they exist,
we set up an augmented matrix.
Because we have two equations with three unknowns,
we have a 2 by 4 augmented matrix.
Equation 1 gives us the first row of 2, 1, -1, zero.
The second equation gives us
the second row of 1, 2, -2, zero.
Next we write the augmented matrix
in reduced row echelon form shown here on the right.
Once we have the augmented matrix in this form,
the nonzero leading entries in each row
are the pivot positions.
Notice, in this case,
we have a pivot position here,
as well as here.
These are nonzero leading entries in the rows,
which then in case of the pivot columns
are column 1 and column 2,
and therefore the variables x and y are the basic variables
or leading variables.
And the remaining variable in this case, z,
is a free variable.
From row 1 we know that 1x equals zero x equals zero.
From row 2 we know 1y minus 1z equals zero
or y minus z equals zero.
And because z is a free variable, z can be any value,
which we can indicate by z equals z.
From here we want to write the basic variables in terms
of the free variable z.
So the first equation remains the same, x equals zero.
For the second equation, we add z to both sides,
y is equal to z.
Third equation remains z equals z.
So this would be one way to express
the nontrivial solutions to this system,
but we often parameterize the solutions
by setting the free variables equal to different variables.
So in this case,
we can parameterize the solution
by letting the free variable z equal t.
So if we let z equal t, we have x equals zero,
y equals t
and z equals t.
This is another common way to express nontrivial solutions.
Another way is to write a matrix equation,
which could also be referred to as a vector equation.
To write the matrix equation,
we first have a column matrix that contains the variables,
in this case, x, y, and z,
equal to another column matrix
indicated by the parametric equations,
x is equal to zero, y is equal to t, and z is equal to t,
but we often factor out the variable here on the right.
So if we factor out t,
we have the column matrix where the elements are x, y, z,
equals t,
and then times the column matrix,
that will be the coefficients of t,
which are zero, 1, and 1.
So this is another common way
to express the nontrivial solutions
to a homogeneous system of equations.
Notice how, when t is equal to zero,
we would have the trivial solution,
zero comma zero comma zero
where it expresses a column matrix,
as we see here on the left.
And the column matrix on the right being multiplied by t
is called a basic solution.
And once again,
these column matrices can be referred to as vectors.
Let's take a look at a second example.
Again we want to find the nontrivial solutions
to the given homogeneous system of equations.
So first we write the augmented matrix
that the first equation gives us the row 1, 4, 3, zero.
The second equation gives us the row 3, 12, 9, zero.
Next we write the augmented matrix
in reduced row echelon form,
which I've already done here on the right.
Notice, now we only have one leading entry
or one pivot position here.
So x is the only basic variable or leading variable,
both y and z are free variables.
The first row indicates
that x plus 4y plus 3z equals zero.
Because both y and z are free variables,
we also have y equals y and z equals z.
Because both x and y are free variables,
it's common to rewrite the first equation solving for x
in terms of y and z.
So that would give us x is equal to negative 4y minus 3z.
And we still have y equals y and z equals z.
This would be one way to express the nontrivial solutions,
but, again, it's more common to parameterize the solution.
We're here because we have two free variables, both y and z.
We can let y equal s and z equal t.
And therefore the nontrivial solutions can be expressed
as x equals -4s minus 3t,
and y equals s, z equals t.
Let's also write the solution as a matrix equation,
which again, can also be referred to as a vector equation.
We have the variable column matrix
equals
the column matrix
where the first element is -4s minus 3t.
The second element is s, the third element is t.
But notice, here we do have two variables.
We have s-terms and t-terms.
So now we're going to write this
as a sum of two column matrices,
where we have the s-terms in the first column matrix
and the t-terms in the second column matrix.
The elements
in the first column matrix are -4s, s, and zero.
The elements
in the second column matrix are -3t, zero, and t.
But again, we typically factor out the variables.
So we'll factor out s from this first column matrix
and t from the second column matrix.
So again we have the variable matrix
equals,
if we factor out s,
we have the column matrix with elements -4, 1, and zero.
Notice, these are the coefficients of s,
and then plus, if we factor out t,
the coefficients of t are -3, zero, and 1.
So this is another way to express the nontrivial solutions
to a homogeneous system.
And once we have the solution in this form,
notice when s and t are both zero,
we have the trivial solution,
which can be expressed as a column matrix or vector.
And then the column matrices on the right
are called the basic solutions.
And there's one more thing I do want to mention
before we go:
If x sub 1 through x sub n and V are column matrices,
then V is said to be a linear combination
of the column matrices x sub 1 through x sub n
if there exists scalars a sub 1 through a sub n
such that column matrix V can be written in the form shown
here on the right,
where in our case, a sub 1 and a sub 2 would be s and t.
And x sub 1 and x sub 2 are the two column matrices,
which we know are called basic solutions.
So we can say the solutions to this system
are linear combinations
of the two column matrices or vectors,
which are the basic solutions.
I hope you found this helpful.