 A useful aspect of matrix algebra is our ability to solve multi-linear systems through the use of what are called elementary row operations. So remember that if we have a system of equations in row echelon form, we can solve it easily through the use of back substitution. But what if it's not in row echelon form? In that case, we need to put it into row echelon form, using either the algebra of solving multi-linear system using the standard written forms of the equations, or the algebra of solving multi-linear systems, wait for it, using the augmented coefficient matrix. The advantage to this is that we can take a system of equations and write them down more compactly as an augmented coefficient matrix. Since every row of a matrix corresponds to the coefficient and constant of an equation in standard form, then anything you can do to an equation in standard form can be done to the row of a matrix. This produces what are called the elementary row operations. So suppose I have a system of equations in standard form. I can write down the augmented coefficient matrix for the system with each row of the matrix corresponding to one equation. So what can we do to an equation in standard form? Well, one of the things we can do is we can take the equation and multiply both sides by a constant. This has the effect of multiplying every coefficient and the constant by the same number. Since every row of the augmented coefficient matrix corresponds to an equation, then what we've done corresponds to multiplying every term of a row of the matrix by a constant. This means we have a new row with the new coefficients and constants. What about the rest of the system? Since we only changed the first equation, the second equation of the system remains the same, and so the second row is also unchanged. Now it is sometimes convenient to record our process. So in this case, we've multiplied the first row by two and replaced that first row. So we'll indicate this by two R1 being sent to R1. And so one of our elementary row operations includes multiplying every term in a row by a constant C, and we'll indicate that by CRI being sent to RI. Another thing we can do when we have a system of equations in standard form is to add one equation to another. So again, let's take our equation in standard form and the corresponding augmented coefficient matrix, and this time we'll add the two equations. Again, every row of the augmented coefficient matrix corresponds to the coefficients and constant of an equation. Now, strictly speaking, we now have a system of three equations and should have a coefficient matrix with three rows. But if you think about it, we got this third equation by adding the first two equations together, which means that we don't actually need all of the equations, because if we wanted to, we can recover one of the equations from the other two. Now, presumably, we added the two equations together for a reason, so we actually want to keep this third row, and if we keep the first row, we notice that we can recover the middle row because it's the difference between the entries. So this first entry one is the difference between two and three. This next entry negative one is the difference between one and zero and so on. So we can get rid of this second row we no longer need it. And again, it's convenient to record our process, so in this case, we've taken rows one and two and added them together, and we've replaced row two with the result. And so this gives us another elementary row operation. We can add one row to another and replace it. So we write this as ri plus rj stored as rj. Another thing we can do when we have a system of linear equations is we can change the order of the equations. So again, we have our system of equations and corresponding augmented coefficient matrix, and maybe I'll change the order of the equations, which is going to also change the order of the rows. And to record this, I'll indicate that I've switched row one and row two. Actually, for reasons that will become apparent later, it'll be better if we don't do this. And so we can also switch to rows, but it will be convenient not to do this as an elementary row operation. To make it a little easier to talk about elementary row operations, we'll introduce a couple of terms. The pivot is the first non-zero entry of a row. So if I have a system of equations with augmented coefficient matrix, then the first non-zero entry of a row is going to be the pivot of that row. So for this matrix, our first row pivot is three, our second row pivot is one, our third row pivot is three, and our fourth row pivot is going to be two. When we reduce our system of equations to row echelon form, then all entries below the pivot are going to be zero. So in row echelon form, three is the first row pivot, and everything below the three is a zero. Likewise, one is our second row pivot, and everything below the one is a zero, and so on. This leads to a process known as Gaussian elimination. We'll have a working row, we'll multiply that working row by a constant to make the pivot one, and then we'll add multiples of this row to the rows below so that we can get zeroes below the pivot, and we'll repeat this process until we're done, and our matrix is in row echelon form. And again, remember the whole point of doing this is so that we can then use back substitution to get all of our solutions. Now it's possible that after we've done this, some of our rows are going to consist of all zeros, and so we'll introduce another term. The number of non-zero rows in the row echelon form of the matrix is called the rank of the coefficient matrix. So in this case, we have three non-zero rows, and so this matrix is of rank three. It's worth emphasizing that while we determine the rank based on the row echelon form of the matrix, rank also applies to the original augmented coefficient matrix. So this matrix has rank three, and this matrix, which gave us this row echelon matrix, also has rank three.