In this video, we are going to discuss linear models and systems of linear equations. Our focus is going to be on solving a system of linear equations by discussing what we call the Gauss-Jordan row operations. This will cover pages 48 and 49 of your lecture notes. When we talk about the Gauss-Jordan row operations, these will be the steps or the procedures that we will use on a matrix in order to simplify and solve a system of equations that we have put into an augmented matrix. It is a step-by-step process when we talk about the Gauss-Jordan process or the Gauss-Jordan method but we want to discuss the row operations first. There are three types of row operations that we can look at. The goal of using these Gauss-Jordan row operations is that at every step along the way your system in that matrix is equal to the previous system. This will be similar to the addition method that was discussed in section 2.3. So first of all, it's very important that you start with your original augmented matrix, and use the row operations that we're about to give you, in order to algebraically manipulate that matrix to get our solution set. The row operations that we are going to use are called elementary row operations and the notation discussed in this video will be the same notation that you will see in your online homework. There are three elementary row operations that we would like to focus on in this class. The first is is that we can interchange two rows. When you interchange two rows that means you literally take every entry and swap it with the different row's entries. The notation that we will use is say we want to place R one's entries into R two's entries and vice versa. We will use the R to tell us which row we're operating on. The one tells us that we are going to swap one with two. The double arrow means that it goes in both directions. Our second row operation is is that you can multiply any one row by a non-zero constant and replace that original row with the new row being multiplied by a non-zero constant. And it does not change anything about the system that you have. The notation that we will use is k times our row so say we wanted row three we wanted to multiply by k. The one directional arrow the right arrow tells us that we will place our result back in row three. So unlike when we were interchanging, our goal is to replace our row with a new row that every entry has been multiplied by some non-zero constant. The third operation and the final operation that we will use in this class is to add the multiple of one row to a different row and write the result in that second different row. So what we mean is if I take a row and multiply it by a constant and add that multiple of the row entries to the entries of a different row, I can write my result in that second row. So what's important here is that wherever you're adding to is where the result will end up. So for this class whenever you add the second row will be where the result goes. And again this is a one directional right arrow to indicate where my results will be. The important thing to remember though is that no matter which row operation you are computing, that all operations are performed on rows not columns because the rows represent the different equations in your system. Let's take a look at an example. In this example, example four, we would like to use the matrix given and the following row operations. Our first row operation is we want to take row one and multiply by one-third and put the result in row one again. I'm not going to touch row two or row three so I can when I write down my new matrix immediately write down row two and row three. So as I said in my new matrix I can write down my original row two and row three, because those will remain unchanged. So my original row two was negative two, negative three, for my vertical line to represent my equal and then seven. My row three was zero, three, one, and 18. The only row that's going to change is my row one. The operation that I was to perform to transform my matrix was I was to multiply by one-third, row one and place the result in row one. You can see that in the directions the ones were subscripts but when I write this on my paper I'm writing it with a large one so that it is similar to what you would see in your online homework. Below this I'm going to do the calculations so that it's easy to keep track of. So I'm going to take every entry of my first row and multiply by one-third. I can see three times one-third will give me one, negative 18 times one-third will give me negative six, the negative 24 times one-third gives me negative eight, and negative 66 times one-third gives me negative 22. Those will be the new entries that I place in my first row. One, negative six, negative eight, negative 22. and while I have completed my row operation, my two matrices are still equivalent. My two matrices are still equivalent. Now then, we would like to perform the second operation which is to take twice row one and add it to row two and write the result in row two based on the new matrix that I now have. That original matrix that I started with no longer exists in my head. The only thing I have is this new matrix here. So I would like twice row one plus row two and the result will go to row two. Since the only row that we'll be changing is my second row, I'm going to write down the first and third row. Those will remain unchanged even though I'm multiplying row one by two, that's not where my result is. So the one, negative six, negative eight, and negative 22 remain as well as the zero, three, one and eight. So the only row that will change is row two. I'm going to draw an arrow to indicate that that's what my result is going to. And let's see. Underneath it we're going to put the algebra. So twice row one would give me two, negative 12, negative 16, and negative 44. and then to that I'm adding row two which was, above, negative two, negative three, four, and seven. If I add each set of entries, I have zero, negative 15, negative 12, and negative 37. So row two becomes zero, negative 15, negative 12, and negative thirty seven. And that's because my result was supposed to go into row two. We have now seen two steps in a process in order to simplify an augmented matrix in order to get the solution to the system of equations that the original augmented matrix represented.