In this example we're given three vectors,
V sub one, V sub two, and V sub three.
We want to determine if they
are linearly independent
or linearly dependent.
If they are dependent,
we want to find a dependence
relation between the vectors
to express vector V sub
three as a linear combination
of V sub one and V sub two.
Looking at the notes below,
a set of two or more vectors
are linearly independent,
when this vector equation here,
where we have a constant times each vector
equals a zero vector,
only has a solution
where all the C values are equal to zero,
and because we have three vectors,
we'll have C sub one, C
sub two, and C sub three.
So we'll set up this vector equation,
solve it to determine if they are
linearly independent
or linearly dependent.
So our vector equation is going to be
C sub one times vector V sub one,
plus C sub two times vector V sub two,
plus C sub three times vector V sub three,
equals the zero vector.
So we have C sub one times V sub one,
plus C sub two times V sub two,
plus C sub three times V sub three,
equals the zero vector.
Let's write this as an augmented matrix,
and then we'll reduce row echelon form.
So we'll have a three by
four augmented matrix,
where the first row would be five,
14 negative three, zero,
the second row would be two,
negative 13, five, zero,
and the third row would also be two,
negative 13, five, zero.
So this kind of way row two
and row three are the same,
let's obtain a row of zeroes by replacing
row three with negative one,
times row three, plus row two.
Let's also get a leading
entry of one in the first row.
Notice a negative two times
two, plus five would be one.
Let's replace row one with negative two,
times row two, plus row one.
So in the first row
we'd have negative two,
times two, plus five, that's one,
and then we'd have negative
two times, negative 13,
that's 26, plus 14, that's 40.
Negative two times five,
that's negative 10,
plus negative three, that's negative 13,
and then we'd have zero.
Second row stays the same.
So because we have a row of zeroes,
we already know we're going to have
an infinite number of solutions,
and therefore the vectors
are linearly dependent,
but we still need to express
the relationship between
C sub one, C sub two, and C sub three,
in order to write vector V sub three,
as a linear combination of
V sub one and V sub two.
For the next step let's get a zero here,
by replacing row two with negative two,
times row one, plus row two,
and we'll keep row one and row
two the same for right now.
So the first row stays the same.
Last row stays the same.
Now for row two we have
negative two times one,
plus two, that's zero,
negative two times 40, is negative 80,
plus negative 13, that's negative 93,
negative two times 13,
is 26, plus five is 31.
Then we have zero.
Now let's get a leading entry here of one
by replacing row two
with negative one over 93
times row two.
So row one and row three stay the same,
and then for row two we have zero, one,
and then negative one over 93 times 31,
actually simplifies to negative one third,
and then we have zero.
Again, we multiply negative
one over 93 times 31,
and we get negative 31 over negative 93,
but 93 is equal to three times 31.
The 31's simplify out,
giving us negative one third.
Let's continue on the next slide.
Our last step, let's get a zero here
by replacing row one with negative 40,
times row two, plus row one.
So the second and third
rows stay the same,
and now for row one,
we'll have negative 40,
times zero plus one, that's one,
negative 40 times one
plus 40, that's zero.
Negative 40 times negative
eight and one third,
that's forty thirds,
and then plus negative 13,
or minus 13 over one,
common denominator of three.
So the denominator is
three the numerator is 40,
minus 39, which is one.
So we have one third and then zero.
So now the augmented matrix is
in reduced row echelon form.
We already know there is nonzero solution.
Notice the first row
tells us that C sub one,
plus zero C sub two,
plus one third C sub three, equals zero,
and the second row tells us
that one C sub two, or C sub two
minus one third, C sub three equals zero.
So here we know C sub one must equal,
negative one third C sub three,
and we know that C sub two
must equal positive one third,
C sub three.
So we can use these relationships
to express vector V sub
three as a linear combination
of vector V sub one and V sub two.
Remember our vector
equation was C sub one,
times vector V sub one, plus C sub two,
times vector V sub two, plus C sub three,
times vector V sub three,
equals the zero vector.
So if we solve this for C sub three,
times vector V sub three,
we'd have C sub three,
times vector V sub three,
equals, we can leave off the zero vector,
and this would be equal
to negative C sub one,
times vector V sub one, minus C sub two,
times vector V sub two.
So if we want to write V sub
three as a linear combination
of V sub one and V sub
two, notice how C sub three
would have to be equal to one,
and if we know C sub three equals one,
we have the equations here we need
to find C sub one and C sub two.
So going back to our first slide,
we now know the three vectors
are linearly dependent,
because the solution to
vector equation was not just
C sub one, C sub two, and
C sub three equals zero,
and again, now to express
vector V sub three
as a linear combination of
V sub one and V sub two,
if we want V sub three here,
notice that C sub three is equal to one.
So if C sub three is equal to one,
notice how this tells us
that C sub one is equal
to negative one third,
and C sub two is equal
to positive one third,
but we need to be careful
here because notice how,
the right side of this
equation is negative C sub one,
times vector V sub one,
minus C sub two, times vector V sub two.
So because C sub one
is negative one third,
the scalar here is going
to be positive one third.
So V sub three equals one
third times V sub one,
and then because we have minus C sub two,
times vector V sub two, and C
sub two is positive one third,
this would be minus one third,
because we have a plus sign here,
the scalar here is going
to be negative one third.
I hope you found this helpful.
