 Sometimes it's convenient to go farther in our row reduction and to produce what's called reduced row echelon form. In reduced row echelon form, the leading entry of each row is a 1. Below and above the leading entry are 0s. And if you're one of those people who like working with fractions, we can deal with them and apply what's called Gauss-Jordan elimination. But for the rest of us, what we can do is we can extend our Fang-Chang Shi method upwards. So let's find our row echelon and reduced row echelon form of the following matrix. So our first row has pivot 2, so we'll multiply the other rows by 2, making them 6, 2, 8, 2, and 10, 4, 2, 6. If we multiply our first row by negative 3 and add it to the second row, that will eliminate the entry below the first row pivot, and that'll give us 0, negative 4, 5, negative 7. Likewise, if we multiply the first row by negative 5 and add it to the third row, that will eliminate the entry below the first row pivot, giving us 0, negative 6, negative 3, negative 9. And purely as a matter of convenience, we can multiply the second and third rows by negative 1 so we don't have to work with negative pivots. We also note that every entry in the last row is divisible by 3, so we can multiply everything by 1 third and get a smaller set of coefficients. Now we can continue the row reduction. Our second row pivot is 4, so we'll multiply the following rows by 4, which gives us a new second row, 0, 8, 4, 12. Now if we multiply our second row by negative 2 and add it to the third row, that will get us a 0 in the entry below the pivot. So multiplying that second row by negative 2 gives us 0, negative 8, 10, negative 14, and adding it to the third row gives us a new third row, 0, 0, 14, negative 2. And again, while it's not absolutely necessary, it's useful to take advantage of the fact that every entry in the last row is divisible by 2, so we can multiply everything by 1 half to get a smaller set of entries. And this produces our row echelon form of our augmented coefficient matrix. Now if we want to get the reduced row echelon form, we'll work from the bottom upwards. And to avoid fractions, we won't worry about getting the pivots equal to 1 until the very end of the problem. So now the third row will be the working row, which has pivot 7, so we'll multiply the other rows by 7. So multiplying the first row by 7 gives us 14, 14, 7, 21. And multiplying the second row by 7 gives us 0, 28, negative 35, 49. In the second row, we need to eliminate the coefficient above the third row pivot. So if we multiply the third row by 5 and add it to the second row, that will get us a 0 in this place. So multiplying the third row by 5 gives us 0, 0, 35, minus 5. And adding it to the second row gives us 0, 28, 0, 44. And that will be our new second row. And likewise, we see that if we multiply the third row by negative 1 and add it to the first row, we'll eliminate the coefficient above the third row pivot. So multiplying by negative 1 gives us 0, 0, negative 7, 1. And then adding to the first row gives us 14, 14, 0, 22 as our new first row. And as before, we can reduce the coefficients by removing any common factors in a row. So here we might see that everything in the first row has a common factor of 2. So we can multiply that first row by 1 half. So it becomes 7, 7, 0, 11. And likewise, the second row has a common factor of 4. So we can multiply everything in the second row by 1 fourth, which makes the second row 0, 7, 0, 11. Since the terms in our third row have no common factor, we'll just leave them as they are. So now all the coefficients above the third row pivot are 0, so we'll move on to the second row. And we'll want to clear out the coefficients above it. So if we multiply the second row by negative 1, it becomes 0, negative 7, 0, negative 11. And if we add that to the first row, our new first row is 7, 0, 0, 0. Finally, to get this into reduced row echelon form, we'll want to make sure that all of the pivots are going to be equal to 1. So we'll multiply the first row by 1 seventh, the second row by 1 seventh, and the third row also gets multiplied by 1 seventh. And we'll check, in every row, the leading entry is both above a column of 0s and below a column of 0s. So we've produced the reduced row echelon form of our original matrix.