Is the vector 14, -5, -41
in the set generated by or spanned by
the set containing the vectors 0, 5, -1 and -2, 5, 5?
If the vector is spanned by the vectors in the set,
then it must be a linear combination
of the vectors in the set,
which means there must exist scalars c sub one
and c sub two,
such that c sub one times the first vector in the set,
which is the vector 0, 5, -1, plus c sub two
times the second vector in the set,
which is the vector negative -2, 5, 5.
Which must equal the given vector 14, -5, - 41.
To solve this vector equation,
let's first perform scalar multiplication on the left,
and then add the two column matrices,
this will result in a three by one column matrix,
where the first entry is c sub one times zero,
plus c sub two times -2 or -2, c sub two.
The second entry will be c sub one times five,
plus c sub two times five,
or five c sub one plus five c sub two.
And the third entry is c sub one times negative one,
plus c sub two times five or negative c sub one
plus five c sub two.
And this is still equal to the vector on the right,
the vector 14, -5, -41.
And now let's write the corresponding system of equations
where the first equation is negative two,
c sub two equals 14.
The second equation is five c sub one,
plus five c sub two equals -5.
In the third equation is negative c sub one
plus five c sub two equals -41.
From here, we normally write augmented matrix,
and then write the augmented matrix
in reduced row echelon form,
in order to determine if we have a solution to the system.
However, in this case,
notice how the first equation only contains one variable.
So in this case,
let's go ahead and solve this first equation
for c sub two by dividing both sides by negative two,
which gives us c sub two equals -7.
Now substitute negative seven for c sub two
in the second equation and solve for c sub one.
This gives us five c sub one
plus five times negative seven equals negative five.
Simplifying, we have five c sub one
minus 35 equals -5, adding 35 to both sides
gives us five c sub one equals 30.
Dividing both sides by five,
and simplifying, we have c sub one equals six.
And we're not done here,
we need to make sure that c sub one equals six,
and c sub two equals -7 satisfies this last equation.
So let's go ahead do that now.
Negative c sub one is -6, plus five,
c sub two is five times -7 equals -41.
Simplifying, we do have -41 equals -41.
So we now know that when c sub one equals six,
and c sub two is equal to negative seven,
the vector equation is satisfied,
which verifies the vector 14, -5, -41
is a linear combination of the vectors in the set.
And therefore, the vector 14, -5, -41
is spanned by the set of vectors.
But just in case, your example doesn't have a zero
as one of the components of the vectors in the set,
let's also show the augmented matrix
and also show we get the same result.
So writing the augmented matrix
from the system of equations here
we would have a three by three augmented matrix
where the first row would be 0, -2, 14.
The second row would be 5, 5, -5.
And the third row would be -1, 5, -41.
Writing the augmented matrix in reduced row echelon form,
which I've already done to save time,
gives us 1, 0, 6 in the first row.
0, 1 and -7 in the second row.
In the third row, we have a row of zeros.
And notice how again, the result is the same,
where c sub one equals six
and c sub two equals negative seven.
So to finish the question, the answer is yes.
The Vector 14, -5 -41 is spanned by the set of vectors
and the coefficients are c sub one equals six,
and c sub two equals negative seven.
I hope you found this helpful.