Given vector u and v and R2
were asked to show vector w with components [4,5]
is in the span of the vectors u and v.
If vector w is in the span of vectors u and v,
it must be a linear combination of vectors u and v.
Which means there must exist two scalars,
which we will call C sub 1 and C sub 2,
such that C sub 1 times vector u plus C sub 2 times vector v
must equal vector w.
Now let's write the vector equation,
and we'll also write the vectors as column matrices,
which gives us C sub 1 times the vector [1,1]
plus C sub 2 times the vector [3,2]
must equal the vector [4,5].
And now lets perform the scalar multiplication on the left,
and also add the two matrices
which will result in a 2 by 1 column matrix
where the first entry will be
C sub 1 times 1 plus C sub 2 times 3
or C sub 1 plus 3 C sub 2.
The second entry will be
C sub 1 times 1 plus C sub 2 times 2
or C sub 1 plus 2 C sub 2.
This is still equal to the vector [4,5] on the right.
And now from here, we'll write a system of equations
where the first equation is C sub 1 plus 3 C sub 2 equals 4
and the second equation is C sub 1 plus 2 C sub 2 equals 5.
And now we will solve the system
by running an augmented matrix,
and then writing the augmented matrix
in reduced row echelon form.
We will have a 2 by 3 augmented matrix
where the first row is 1 3 4,
and the second row is 1 2 5.
And now let's write the augmented matrix
in reduced row echelon form.
Let's get a 0 in row 1, column 2.
Notice, the least common multiple of 2 and 3 is 6.
Let's replace row 1 with
3 times row 2 plus 2 times row 1.
Notice -3 times 2 is -6
plus 2 times 3 would give us +6
which is equal to 0.
And lets also get a 0 in row 2 column 1
by replacing row 2 with -1 times row 1 plus row 2.
Notice -1 times 1 is -1 plus 1 is 0.
So again,
replacing row 1 with -3 times row 2 plus 2 times row 1
we have -3 times 1 plus 2 times 1,
which is -1.
And then we have -3 times 2,
which is -6 plus 2 times 3,
which is 6, -6 plus 6 is 0.
And then we have -3 times 5,
which is - 15 plus 2 times 4 which is 8.
15 plus 8 is -7.
And now we replace row 2 with -1 times row 1 plus row 2.
We have -1 times 1 plus 1 which is 0.
And then we have -1 times 3,
which is -3 plus 2, which is -1.
Next -1 times 4 is -4 plus 5 which is +1.
And now to make the leading entries positive 1, not -1,
we will replace row 1 with -1 times row 1
and replace row 2 with -1 times row 2.
Gives us 1 0 7 0 1 -1.
Remember the variables are C sub 1,
and C sub 2 where the coefficients of C sub 1
are in the first column.
And the coefficients of C sub 2 are in the second column.
So now we know that C sub 1 is equal to 7,
and C sub 2 is equal to -1.
Going back up to our vector equation,
since C sub 1 is 7, and C sub 2 is -1,
we now know that 7 times vector u
plus -1 times vector v
or just -1 times vector v is equal to vector w.
This verifies that vector w is a linear combination of
vectors u and v,
which means vector w is in the span of the vectors u and v.
Course, we can check this if we want.
7 times vector u,
which is the vector [1,1] -1 times vector v,
which is the vector [3.2] must equal vector w
which is the vector [4,5].
Notice 7 times 1 minus 1 times 3 is 4
and 7 times 1 minus 1 times 2 is 5.
Let's finish by saying that vector w is in the span
of vectors that u and v,
since vector w is a linear combination of vectors u and v
where vector w is equal to
7 times vector u minus 1 times vector v.
I hope you found this helpful.