 In this video, we will provide the solution for question number 14 from practice exam number one for math 2270. In this question, we are given a set of vectors, four vectors are provided to you, and I'm just going to give them some names. We're going to call the first one A1, the second one A2, the third one will be A3, and then the fourth one will call it A4, just so we can talk about them. So we have this set of vectors A1, A2, A3, A4, and we have another vector B. We are supposed to determine with proof whether B belongs to the span of the set S. That is, is B a linear combination of A1, A2, A3, A4? I'm also going to do some matrix A whose column vectors are these four vectors in question A1, A2, A3, and A4, like so. And so what we have to do to determine this, if B is a linear combination of the A's that is B belongs to the span, that means the linear system associated to the augmented matrix A, augment B, must be consistent. That's what we're trying to do right now. So we want to solve that system of equations. So copying it down, we get the first vector A1 is just 1111, A2 is 1234, A3 is 1, negative 11, negative 1, and then A4 is 00001. We augment that with a B, 4, negative 201. We want to solve this linear system here. Our first pivot position would be in the 11 spot. I want to get rid of all the ones that are below it. So we're going to take row two, minus row one. We're going to take row three, minus row one. And we're going to take row four, minus row one. Make sure you do show all your steps on this question here for full credit. So you're going to get a minus one, a minus one, a minus one, and a minus four. I skipped when there was a zero. And you're just going to do this to all of the columns, all the rows, I mean. Minus one, minus one, minus one, and minus four. And so then write down the next matrix here. The first row stays the same. 11104, I'm going to give myself a little bit more space here. Just scoot you to the right. There we go. So for the second row, we're going to get 0, 1, negative 2, 0, negative 6. For the third row, we're going to get 0, 2, 0, 0, negative 4. And then for the fourth row, we're going to get 0, 3, negative 2, 1, and negative 3. So we get a matrix that looks something like the following. That takes care of the first column. So for the next column, focusing on the next pivot right here, we got to get rid of the, we have a one there, that's great. We got to get rid of the two and the three that are below it. So we're going to take row three, minus one, two times row two. And we're going to take row four, and subtract from it three times row two. So we get minus two, plus four, there's a zero right there. So I'm just going to skip that one. And then we're going to get plus 12 right there. So for row four, we're going to get a minus three, a plus six, a zero. And then we're going to get a plus 18. And so then I'm going to transcribe that over here. Again, nothing happens to the first row. So just copy down 11104. For the second row, we're going to get 0, 1, negative 2, 0, and negative 6. Again, nothing happened there. For the third row, do make the appropriate modifications. You're going to get 0, 0, 4, 0, and 8. And then for the fourth row, you should get 0, 0, 4, 1. And then we have 18 take away three. That should be a 15. Now we see right there. The next thing I want to do is looking at my third pivot position, which would be the 3, 3 spot. I'm going to subtract row four. I'm going to subtract from row four, row three. That'll cancel the four that's right here, minus an eight. And so this puts the matrix then in echelon form. Let me scooch over a little bit. 11104. We're going to get 0, 1, negative 2, 0, negative 6. Then we're going to get 0, 0, 4, 0, and 8. I also can't help but notice that everything in row three is actually divisible by four. So I'm actually going to divide row three by four as well, just to help me out here. 1, 0, 2. And then we're going to get 0, 0, 0, 1. And then 15 take away eight is going to be seven. So notice now what we have here. We have a matrix which is in echelon form. We have a pivot in each and every column right here. Because we get a pivot in every column, this actually does tell us that the system is consistent. We have no contradictions. This matrix represents a consistent linearism. And because the system is consistent, there is a solution. This tells us that, in fact, B is a linear combination of the As. Now, in the case that B is a linear combination, we have to now continue to find that linear combination. If this system turned out to be inconsistent, we could then report no, B is not inside the span. But we have to keep on going now. So starting with the backwards phase, the 1, 1, or the 4, 4 positions already a 1. That's great. And everything above it is already a 0. So then I'm going to start focusing on the 3, 3 spot. It's a 1. That's great. And so I got to go to this negative 2 right here. We're going to do row 2 plus 2 times row 3. So we get plus 2 and plus 4. Let me transcribe that down here. So again, first row stays the same. There's a lot of copying when you do these matrix operations. We're going to get 0, 1, 0, 0, negative 2, 0, 0, 1, 0, 2, and 0, 0, 0, 1, 7. And so then the last thing to deal with when you're looking at the 2, 2, we got to get, that's already a 1. That's great. Get rid of the 1 that's above it. So we're going to take row 1 minus row 2. So we have to take a minus 1 and then a plus 2. And so this then gives us our final form. We're going to have 1. Oh, did I forget to do a row operation? It looks like I did. So let me go back in time a little bit because I still need to do, I have this 1 right here. How did I forget that? So yeah, let's make sure we get rid of that one too. So we're actually going to do row 1 minus row 3. So we get a minus 1 and a minus 2 right there. That's fairly easy to switch to fix it here. So we get rid of this one, this 4. So that should be a 0. And then we had 4 minus 2, which is a 2. Sorry about that mistake there. Then we can go to R1 minus R2. So we're going to minus a 1 and we're going to plus a 2 right here. And so in the end, despite that little hiccup there, we have just a bunch of 1's and 0's on the other side and we should have a 4, a negative 2, a 2, and a 7. And so the solution to this linear system gives us the linear combination. So we'd say something like the following. Yes, B does belong to the span of S. And so in fact, we see that B is equal to 4 times a 1 minus 2 times a 2 plus 2 times a 3 plus 7 times a 4. And since these vectors turned out to be independent, it turned out this is the only way you can express this linear combination like so. So this is the evidence that we wanted.