 In this video, we're gonna define the notion of the span of a set of vectors. So if you have a set of vectors v1, v2 up to vn, we define the span of those vectors to be the set of all possible linear combinations of the vectors v1 up to vp there, vp that should probably be a vn. That's a typo there, sorry. So we take the combination of all the vectors and then we call that the span of the things. And in particular, as these vectors v1 up to vn live inside of fm, the span is gonna be a subset of fm. And we call this the subset span by v1, v2 up to vn, right? And this is denoted as span of these things. And this idea of span is gonna be much more important. It's gonna be very important in the future, right? And in which case, we'll talk a lot about the geometric consequences of what we need by span. But for the moment, just accept that the span is the set of linear combinations. So we could ask whether b belongs to the span of some vectors a1, a2 up to an. Now, asking if b is inside the span is the same thing as asking, is b a linear combination of a1, a2 up to an? And as we've seen in previous examples, determining whether b is a linear combination or not comes down to solving this vector equation. So if you're asked, is b in the span of these vectors? Then that means you have to solve this vector equation. Let's look at some examples. So given, we'll work over the real numbers here. So given the vectors a1, which is one, negative two, negative five, and a2, which is two, five, six, is the vector seven, four, negative three in the span of these two vectors? That just means is the equation x1, a1 plus x2, a2, equal b, is this equation consistent? That's really what we're trying to do here. And this equation can be solved with this augmented matrix where the first column is a1, the second column is a2, and the last column is b. Putting your pivot position in the one, one position, to go down, you're gonna have to take row two, plus actually two times row one, and row three plus five times row one. So we're going to get a two, a four, a 14. We're going to get a five, a 10, and a 35. Those will combine together to give us this matrix right here. Then I noticed that in the second column, a second row, excuse me, everything's divisible by nine. So we're gonna divide everything by nine in the second one there. And also this one, everything's divisible by two. It is by two, but I meant by 16. We're still gonna divide everything by 16 in the third row. That brings us down here to this matrix where our pivots in the two, two position. You'll notice that row two and row three are identical with each other. So when I take row three minus row two, you'll get a row of zeros. No contradiction though. Consistency appears to be the case. And I want to point out to you here that we have this row of zeros, but this system has a unique solution. You'll notice that there is a pivot in every single column. And this is a misconception students sometimes have. They think that when you get a row of zeros, that means you have a free variable. Nuh-uh, no sir. You get a free variable if you have a non-pivot column, which we don't actually have that here. The row of zero actually has no effect to it, right? It's like saying, okay, x has to equal three, y has to equal two, and the sky is blue. It doesn't matter. It has nothing to do with what we're trying to consider right now. And sure enough, you can see here when you solve this system, it's in, we're now in row reduced echelon form. We see that when x one equals three and x two equals two, we'll have a solution to the system of equations right here. Therefore, the vector b is a linear combination of a one and a three. It is, in fact, inside the span. And the evidence is here. If you take three times the first vector and two times the second vector, those will add up to be b. And just to verify that, right? The first vector was one, negative two, negative five. The second vector was two, five, six. If you take three times one, that's a three. Two times two is a four. Three plus two, three plus four is seven. Three times negative two is a negative six. Two times five is a 10. 10 minus six is a four. And then lastly, three times negative five is a negative 15. Two times six is a 12. 12 take away 15 is a negative three. We in fact proved it here. We had a unique solution. And so yes, the answer here is yes, b is inside the span of these vectors a one and a two. It just comes down to checking to be inside the span means that you're a linear combination of set vectors and that comes down to solving the associated linear system. Let's do another example of this. Again, we'll work over the real numbers in this situation. We have two vectors a one, which is given as one, negative two, three. A two is given as five, negative, which is, which a two is five, negative 13 and three. And then b is negative three, eight and one. Is b inside the span of these things? Same basic idea here. Let's see, set a one as the first column, a two as the second column, b as the augmented column. Your first pivot will be in the one, one position. So we'll take row two plus two times row one. Row three will be, you'll subtract from the three times row one. So we get two, 10, negative six. We're also gonna get negative three, negative 15 plus nine. The second row will become zero, negative three and two. The third row will become zero, negative 18 and 10. Now in this situation, your pivot will move to the two, two position. I'm okay with the negative three in the pivot position there. We have to get rid of the negative 18 below it, which as 18 is a multiple of three, I'm especially thrilled about that. We'll take our three and then we're gonna subtract from it six times our two. So that's gonna be positive 18 and negative 12. When you row reduce that last one, you're gonna get zero, zero, negative two. Now in this example, you will see that the last row gives us a contradiction. And since we got a contradiction, right, this is saying since like basically zero equals negative two, we're not working on two, we're working over the real numbers. This is a contradiction. So we have inconsistency here. If this linear system is inconsistent, that means that b is not inside the span. Consistency means you're in the span, you are a linear combination. Inconsistency means you're not in the span, you're not a linear combination. And that's all there is to determining whether a vector is a combination of others, whether it's inside the span of others. You can connect this vector equation to a system of linear equations and the solution of said linear equations will help you determine whether you're inside the span or not. That will bring us to the end of section 2.1. Thanks for watching everyone. If you feel like you learned anything, feel free to hit the like button. Feel free to subscribe if you wanna hear more about these videos in the future. Other than that, I will see you next time. Feel free to post questions in the comments below if you have any and I'll see you in section 2.2. Bye everyone.