Hey, for this problem, we're going to be examining the positions of people in a marching band. We're going to do this in a few steps. First, we have, uh, matrix that they've given that they're calling B. And as you can see, if you look at how be a set up, the third row is just once. There's one person in each place, but the top and the middle rows 1st and 2nd Rose show the positions of three different band members. If you kind of think of the football field is a grid, you've got one band member at 50 0 one at 50 15 and one at 45 23 non ca linear band members and we've been asked to start with We're gonna find the inverse of B. So if this is be that we're showing here, I've got a copy to be on the screen to find the inverse. We're going to remove that second bracket and we're going to augment it. We're on the let right hand side. We're going to copy the identity matrix, and we're going to use row manipulation to make the left hand side look like the identity matrix and when that's done, the right hand side will be the inverse. So in order to do this, the first step, we're gonna leave the first row exactly the way it iss, huh? And my goal is, since the identity matrix has an element in the first row. First column will leave that first row first column just as it is. And they will make everything else in that first column equal to zero. Well, one thing that makes our job a little easier. There's already a zero in the second row so I can leave those numbers alone. I don't have to do any manipulation on those, but what about the third row? I do have a one here, and I need to change it to a zero. So my row manipulation will be to take the opposite of the first row, plus 50 times the third row. And those will be my new third row entries all the way across. And when I do that, that's going to give me a new third row of 005 Negative 10 50. Hey, first row is done, or the first column now the second column. There is an element in the second row. Second column position of the identity, Uh, matrix. So I'm gonna leave the second row alone. Copy it just the way it is. And I want to have zeros everywhere Else in the second column again, we got kind of lucky. The third row already has a zero there, so no manipulation needed. We could just copy it. But what about the first row? I have a 50 here that I need to get rid off. So what I'm going to end up doing is I'm going to take negative 10 times the second row, and I'm gonna add three times the first row. Doing that gives me a new top row. And my new top row is going to be 150 0 negative. 65 three negative. 10 0. Hey, one Maurin oration should have all of our zeros in the proper place. Uh huh. There is a There is an element in the third row. Third column position of the identity matrix. So we're gonna leave the third row just the way it is and deal with that third column. Our goal is to make every element in the third column except for that third row third column spot to equal zero. So what do I need to dio? Well, first of all, let's look at the first row. I have a negative 65 so I'm gonna take 13 times the third row and add it to the first wrote. Doing that gives me a new top row. 150 00 negative. 10 Negative. 10 650. Hey, now, my second row, I have a 20 that I need to get rid of. So I will take the opposite of four times the third row, and I'll add that to the second row. That gives me a new second row of 0. 15 0 41 negative. 200. Okay, we are just about done. I'm going to just scroll a little, give us a little bit of space. It almost looks right. All the zeros in the right spot. I just need ones on that diagonal. So I'm going to divide everything in the top row by 150 everything in the second row by 15 and everything in the third row by five. And that will give me my new inverse matrix. So the left hand side is now the identity matrix and the right hand side is negative. 1/15 Negative. 1/15 13 3rd. Second row is 4/15 1/15 Negative, 43rd. And the bottom row is negative. 1/5 zero and 10. So this is the inverse of matrix B. Great. So we're almost ready to find our new band positions. But first, we need to compare this be have to multiply it with our grid. A and A is going to have the positions to which they will be moving. So this is gonna be a multiplication. I've got a times the inverse of be so first before we look at where it comes from. Let me just re copy this inverse matrix. You need to be able to see that top row. So it's negative. 1/15 Negative. 1/15 13 3rd for 15th, 1/15 Negative. 43rd negative. 1/5 0 10. Okay, just re copy. So a is our new positions. So the first column of B was the person at 50 0 at that point on the grid, and we're told that they were moving to the 00.0.40 10 and again we're going to have ones across the bottom. The person represented by column to the one at 50 15 is now moving to 55 10 and the third person who is at 45 20 is moving to 60 15. Okay, so we need to multiply this out. And once we do this will give us our movement matrix. It'll take where they are now and show us where they're going to be. So when we do our matrix multiplication, just a reminder. Here we take the first row times the first column, and that becomes our first row first column entry. So 40 times negative 1/15 plus 55 times for 15th, plus 60 times negative, 1/5. And when we do that, we get an entry of zero. Hey, second row sec. Oh, sorry. First row, second column that will give us our first row second column value doing that. We get 40 times negative 1/15 plus 55 times 1/15 plus 60 times zero. And that gives us a value of one. And we're going to do that for all nine, uh, entries. All the rows and all the columns and our result is 01 40 Negative. 10 60 and 001 Oh, on I apologize. I labeled this wrong. That is not a Our result is a They did not actually name this first matrix for us. I miss I miss read where that a was. My apologies on that. So this is a This is our movement matrix where you start from toe where you go. This matrix will help you determine that. So let's try it with some of the other. We had nine players on the field. We've already done three. So let's look at some of our other players. The one who is furthest on the left is starting at the 10.40 20 on the grid. So what's his new position? Well, we're gonna take that matrix a that we just found, and we're going to multiply it by a matrix that shows where this player started. They started at 40 20. We're gonna put a one in that third position, and when I multiply that out, I'm just gonna put this I'm gonna just put a little thing here over on the side. I'm gonna put our matrices to show where are new positions are multiplying. First row by first column. Gives me 60 second row by that column is 20 and then I get a one in my third position. So the person who starts at 40 20 moves to the position 60 20. What about the next person? Well, in order to do that, all we have to do is erase the current position that we're looking at. Put in the new one. The next person. Next Thio. Next person over. I'm just gonna go along all five of those at the top. The next one is one of the ones we already looked at, so I can ignore him. The next one over is going to be at 50 20. And when I multiply that out, this new person is going to end up at the 0.60 10. Okay, next one over. I could just erase these numbers. This next person is at 55. 20. Doing that multiplication gives me a final spot of 65. Okay, 60 across, five up. Hey, next person is at 60. I'm sorry. It's It's 60. 20 60 20. Multiplying that out gives me a new position of 60 0. Okay. Two more to go now. I'm gonna come down that t it's on the grid that we have. The next person that we haven't looked at yet is standing at the 0.50 10. And actually, when we do that multiplication, this person doesn't move. There must be a pivot point. So they stay at 50 10. No motion. And our last person is at 55 and they end up at the 0.45 10. Okay, so those air, all of our players Now, what does our final shape look like? Well, I took all nine points, the six over here with the Red Star, plus the three that we were given within the problem, and I've graphed them onto a graphic application. As you can see, we're still in a T, but it's gone sideways. It's fallen over to the right. So this tea is our new position. Based on the information given in the problem