 In this video, we're going to provide the solution for question number 10 from practice exam number 1 for math 2270. We have a 3 by 3 linear system that we have to solve. I would begin by writing the augmented matrix 1, negative 2, 3, 7, 2, 1, 1, 4, negative 3, 2, negative 2 and negative 10. Looking at this matrix, if I want to row reduce it, I'm going to see there's a pivot in the first position. So I'm going to have, that's a 1, that's great. So we're going to take row 2 minus 2 times row 1. And we're also going to take row 3 plus 3 times row 1. So we're going to get a minus 2, a positive 4, a minus 6, and a minus 14. And then for row 3, we're going to get 3, negative 6, 9, and 21. So then copying down what we did to that matrix, the first row stays the same, 1, negative 2, 3, and 7. The second row will become 0, 5, negative 5, and negative 10. The third row will become 0, negative 4, positive 7, and then 11, like so. The next thing I want to do, moving my pivot position to the 2, 2 spot, I recognize that the third, the second row, everything zooms by 5. So I'm going to divide everything by 5 in row 2. This gives me the matrix where the first row stays the same. In the second row, everything divided by 5, 0, 1, negative 1, and negative 2. This gives you 0, negative 4, 7, and 11. And so again, still with my pivot in the 2, 2 spot, I want to give it a negative 4 that's below. So I'm going to take row 3, and I'm going to add to it 4 times row 2. This will give us plus 4, minus 4, and minus 8. So the first row stays the same. The second row stays the same. And the third row will become 0, 0. We're going to get 7 minus 4, which is 3, and we're going to get 11 minus 8, which is likewise 3. And so then my pivot will move to the 3, 3 spot that I want to think about. I want that to be a 1, so I'm going to divide the third row by 3. Scroll things up so I have some more space. So we get 1, negative 2, 3, and 7. We're going to get 0, 1, negative 1, negative 2, and we're going to get 0, 0, 1, and 1. So still now we're in the backwards phase, right? We want to get zeros above the 1 right there. I'm going to take row 2 and add to it row 3. This will give me a plus 1 and a plus 1. We're going to then get take row 1 and subtract from it 3 times row 3. So we get minus 3, minus 3, perform the row operations. So you get 1, negative 2, 0, slash 4, and then we're going to get 0, 1, 0, negative 1, and then we get 0, 0, 1, 1, like so. So now looking at the 2, 2 pivot again, we want to get a 0 above that because the pivot's already a 1, right? So we're going to take row 1 and add to it 2 times row 2. So we're going to get a plus 2 right here and we're going to get a minus 2 right here, which I need a little bit more space. So let me just squeeze it in just right here. I wrote too big, apparently. So you can get 1, 0, 0, 2. You're going to get 0, 1, 0, negative 1, and 0, 0, 1, 1. And so this would indicate to us that the solution, our solution x is going to be the vector 2, negative 1, 1, like so. And so this is how we showed how we could solve this system of equations using Gauss Jordan elimination.