 In this video, we're gonna introduce the notion of the column space of a matrix. So suppose we have a matrix A, which is M by N. We define the column space of that matrix, which we'll denote as C-O-L of A for short. We define the column space to be the set of all linear combinations of the column vectors of A. So for example, if A has the form where the first column is A1, the second column vector is A2, up to the last column vector, which is AN. Well, the set of linear combinations of some vectors is just what we call the span of those vectors. And so the column space, be aware, is just the span of the column vectors of a matrix A1, A2, up to AN. So we've seen previously that the matrix equation AX equals B is consistent exactly when B is a linear combination of the column vectors of A. So with the introduction of the column space here, we then see that the matrix equation AX equals B is exactly consistent when B is inside the column space of A. And so if we want to determine whether a matrix equation is consistent or not, it really just comes down to determining when is B inside the span of a set of vectors, which is something we've done before. But we often did it in such a way that if we had a specific vector, like B is the vector one, two, three, then we checked, oh, okay, B was inside the column space. Then we take a different vector, like, okay, this time B can be 105. And then we checked, oh, it's inconsistent that time. It seems a little bit haphazard to have to check for each individual vector. Is this one in the span? Is this one in the span? Is this one in the span? Or in this case, is this one in the column space? Is this in the column space? Is this in column space? Is there some way of sort of doing this universally? And so in this example, I do wanna give you a method of determining what is in the column space and what is not in the column space, which of course generally shows us if vectors are in the span or not. So let's say we take a matrix. Here we have a three by three matrix, one, three, one, two, four, four, three, five, seven. Now be aware that me giving you a matrix is really no different than me giving you a set of vectors. We can really identify the things together, right? Because a matrix is just a list of the column vectors. So a set of vectors and a matrix essentially are the same thing, right? At least how we're thinking about them right now. And so what we're gonna do is we're gonna decide is the vector B inside the column space of A or not. And by doing this generically, that is we don't know what B is. B is just B1, B2, B3. We're gonna solve this linear system generically and see what happens, right? So to solve the matrix equation, we would have to then solve the matrix equation AX equals B. We solve the associated system of linear equations, which is represented by this augmented matrix. It's very nice to go from the matrix equation to the augmented matrix because the matrix A is just the coefficient matrix. And then the vector given is just the augmented column. So it's really nice to do it that way. So we wanna solve the linear system associated to this augmented matrix where our pivot position will be one one, which is already a one, which is nice. So I'm gonna take the second row and subtract from it two times row one. And we're gonna take the third row and subtract from it three times row one. So we're gonna get a minus two, minus six, minus two, and then minus two B1. We can't really simplify that one better because we don't know what B1 is. And then for the next one, we're gonna get minus three, minus nine, minus three, and then minus three B1 again. And so simplifying the coefficients, of course, two minus two is gonna cancel out. Four minus six is a negative two. Four minus two is a two. And then you're just gonna get B2 minus two B1. Can't do much more about that right now. Next year, we're gonna get three minus three. So those cancel out, giving us a zero. We're gonna get five minus nine, which is negative four, and then seven minus three, which is a positive four. And then you're gonna get B3 minus three B1. Again, without knowing what the Bs are, that's the best we can do. We can't simplify it, but that's kind of the point. We want the Bs to be generic so that we can determine universally which vectors are gonna be inside the column space or not. And so now with the first column done, we're gonna transition our pivot position to the two-two spot. I noticed that everyone in the first row is, sorry, the second row, at least from the coefficient side, those are all divisible by two. We could divide both sides by a negative two. So we're gonna take negative one-half R2. But also, when you look at the third row, I also noticed that the coefficients are both divisible by four. So we're gonna divide by negative one-fourth in that matrix, thus bringing us down here. Now it's gonna be easy to take negative two and two divided by negative two, right? Negative two divided by negative two is a one. Two divided by negative two, of course, is going to be a negative one. So there's a mistake there, sorry about that. So that should be a negative one. But then when we take the B2 minus two B1, when we divide that by negative two, we have to do that one as well. Don't forget the other side. That then becomes B1 minus one-half B2. So we get some fractions involved there. When we do this to the third row as well, you divide everything by negative four. The negative four will become a one. The four will become a negative one. And then here, we're gonna have B3 minus 3B1. When we divide that by negative four, you're gonna get one-fourth 3B1 minus B3. Notice I switched the signs on the negatives because we times by negative one there. So we did that. And so again, focusing on the pivot position right here of a one, we gotta get rid of the one below it. So we're gonna take row three, and we're just gonna subtract from it row two. So you're gonna get a minus one, a plus one. And then we're just gonna subtract, we take this entry right here, we're just gonna subtract the entry that's above it. And in which case, then we're gonna get a row of zeros, which is fine. But then when you take this entry minus this entry, you'll have to combine like terms for the Bs, right? You're gonna have a three-fourth B1 minus B1. So that gives us a negative one-fourth B1. In terms of the B2s, you'll have a negative negative one-half B2, so that's a positive one-half B2. And then you're gonna just have a negative one-fourth B3, nothing to combine there. And so we get this entry right here. And this is significant because notice it is next to a row of zeros. The only way that this system of equations can be consistent is that the row of zeros on the left must match up with a zero on the right. And so this tells us that will be consistent. This system, this matrix equation is consistent if and only if we have that negative one-fourth B1 plus one-half B2 minus one-fourth B3 is equal to zero. That has to follow in order for this to be a consistent system. Now, if you don't like fractions, right? We can times both sides by, we'll say negative four. Negative four, that will clean up the fractions a little bit. That way you get B1 minus two B2 plus B3 is equal to zero. And so this one's a little bit more tame to use. But what this tells us is that the only way that our matrix equation can be consistent is if B satisfies this relationship. And so we could pick something like, well, what do we wanna do? We could take the vector B, it could be the vector 111, that would work because notice one minus two plus one is zero. That's a vector that we could do. We could also do as another alternative, we could do like one zero negative one. That's another vector that would be inside of the column space of the matrix A. On the other hand, if we did something like, say this time you said B equal to B211, that doesn't work there. You notice you'll get two minus two plus one, which is not zero. This is not inside the column space of A. So by solving the system of equations generically and thus finding this equation right here, we then could describe the, we could describe the column space using this equation. And in some respect, we think of this as an equation with three variables where these two variables, you could then make as free variables. And then this B1 depends on that choice. You know, that's a way of considering this, right? But this is a way of determining whether we have a vector inside the column space, whether we have a vector inside the span or not.