Let's take a look at another similar example in the last video.
We were given a set of three vectors, and asked to find if these vectors form a basis
for R3.
And they did, because they were three linearly independent vectors.
In this example, we're going to change things just slightly.
We're going to work with a set of three vectors, but the final vector this time is minus 1,
1, 5.
So these are the three vectors.
And we're asked the same question, is this set of vectors a basis for R3?
Since R3 is three dimensions, we know that we need three linearly independent vectors
if we're going to form a basis for R3.
So really, this question boils down to finding out if the three vectors that are provided
are linearly independent or not.
So our approach to answering this question, are these three vectors are linearly independent?
We can take the same strategy as the last video.
We can basically set up a matrix, and perform reproductions to find out how many pivot columns
there are, and that tells us how many linearly independent vectors we have.
And then the second part of the video, we'll compute the determinant the same as we did
in the first video to see if that tells us anything about the vectors being independent
or not independent.
So here in part a, let's go ahead and set up our matrix.
So there is the matrix formed by making each column, one of the vectors that we were provided.
And then let's go ahead and do some row operations, our replace equation 3 with equation 3 minus
equation 1.
So the first two equations remain unchanged.
1 minus 1 and 0.
4 minus minus 2 is 6.
5 minus and negative 1 will give me 6 as well.
I'll go ahead and replace the third equation with the third equation minus 6 times the
second equation.
So I'm not touching the first two equations at all.
And this will modify the third one.
So I'll have zero minus 6 times 0 is 0.
6 minus 6 times 1 is 0, and 6 minus 6 times 1 is zero as well.
So at this point right here I have a row of all zeros.
So it's very different in the first time.
I'm missing a pivot.
I don't have a pivot in anywhere in this third row.
I only have pivots in the first two columns.
There's no pivot in the third column.
So that means these are not all linearly independent.
I only have two that are independent from this set.
So since these three vectors are not linearly independent, I only got two pivot not three.
There's no way that these vectors can form a basis for R3, because really I only have
two dimensions to work with here, and I need three dimensions.
So these vectors are not a basis for R3.
Let's go ahead and do the determinant way as well.
We're not going to work through all the intricate determinant computations.
I have videos that show you how to do that.
We're just going to use the information about the determinant to draw our conclusion.
So the determinant here is the determinant of 1 minus 2 and negative 1.
0 1 1 1 4 5, so that's just our starting matrix.
And actually, I misspoke we are going to go ahead and work through the computations here
to see how things cancel out.
So I said determinant I need to do.
And I'll go ahead and do my cofactor expansion.
So that's going to be 1 times those 2 by 2 determinant minus a minus 2 gives me plus
2 times this 2 by 2 matrix, and then minus 1 times this 2 by 2 matrix.
So again, we're just doing cofactor expansion.
If you don't know how to compute determinants by hand, I have lots of videos that show you
to do that.
Then go look at other places on the playlist to figure out how to do this.
As we continue the computation, now I'm doing the determinants of these 2 by 2 matrices,
it's very simple.
You just take 1 times 5 minus 1 times 4 plus 2 times 0 times 5 minus 1 times 1 minus 1
times 0 times 4 minus 1 times 1.
And then if we multiply this out, we have 1 minus 2 times negative 1 gives me a negative
2.
And then a negative 1 times a negative 1 gives me 1, and this is equal to 2 minus 2, which
is 0.
So I've computed the determinant of this matrix whose columns are the vectors that we are
provided.
And since I got 0, that tells me that the columns are not literally independent.
It actually tells me that they are linearly dependent.
Anytime you get a 0 in your determinant computation, that means that you have linearly dependent
columns.
So that makes it impossible for these three vectors to form a basis for R3.
And we've essentially answered the question twice now.
We've done it the determinant way.
We've shown that these vectors are not linearly independent.
They can't form a basis.
And in part a, we did answer the exact same question just using a different methodology.