Determine if vector X is a linear combination
of vector V sub one and vector V sub two.
If vector X is a linear combination of vector V sub one
and vector V sub two, then there must exist two scalars,
C sub one and C sub two such that C sub one
times vector V sub one plus C sub two
times vector V sub two must equal vector X.
It's also true that vector X is a linear combination
of vector V sub one and vector V sub two.
Vector X is in the span of vector V sub one
and vector V sub two.
Let's go ahead and set up this vector equation
where we have C sub one times vector V sub one
with components five, four, negative three,
plus C sub two times vector V sub two,
which has components four, one, three,
must equal vector X,
which has components seven, 10, negative 15.
And now from here on the left side,
we will perform scalar multiplication
and then add the two column matrices.
The result will be a three by one column matrix
where the first entry will be C sub one times five
plus C sub two times four,
or five C sub one plus four C sub two.
The second entry will be C sub one times four
plus C sub two times one, or four C sub one plus C sub two.
And the third entry is C sub one times negative three
plus C sub two times three, or negative three C sub one
plus three C sub two,
and this is equal to vector X on the right,
with components seven, 10, negative 15.
From here, we will write a system of equations
with three equations and two unknowns,
where the first equation is five C sub one
plus four C sub two equals seven.
The second equation is four C sub one
plus C sub two equals 10.
And the third equation is negative three C sub one
plus three C sub two equals negative 15.
And now to solve this system and determine if vector X
is a linear combination of vector V sub one
and vector V sub two, we will write an augmented matrix
and then write the augmented matrix
in reduced row echelon form.
Because we have three equations with two unknowns,
the augmented matrix will be a three by three matrix,
where the first row is five, four, seven.
The second row is four, one, 10.
And the third row is negative three, three, negative 15.
To save some time I've already found
the reduced row echelon form,
where the first row is one, zero, three.
The second row is zero, one, negative two.
And the third row is zero, zero, zero.
So we know because the last row is all zeros,
we have a solution.
Remember the two variables are C sub one and C sub two,
which are the first two columns.
So the first row indicates C sub one equals three.
And the second row indicates that C sub two
equals negative two.
So now we know the vector equation does have a solution,
and vector X is a linear combination
of vector V sub one and V sub two.
We now know that three times vector V sub one
plus negative two times vector V sub two,
or if you want, minus two times vector V sub two
does equal vector X.
And let's go ahead and write this out.
We have three times vector V sub one,
where the components are five, four, and negative three,
and then either plus negative two times vector V sub two
or just minus two times vector V sub two,
where the components are four, one, three.
This linear combination does give us vector X
with components seven, 10, negative 15.
And we can go ahead and check this.
Three times five minus two times four is seven.
Three times four minus two times one is 10,
and three times negative three minus two times three
is negative 15.
So vector X is a linear combination of vector V sub one
and vector V sub two.
Which also means vector X is in the span of vector V sub one
and vector V sub two.
I hope you found this helpful.