 As long as we can get our augmented coefficient matrix into a row echelon form, we can solve it for the unknowns. But it's sometimes convenient to go a little bit further and to obtain what's called Gauss-Jordan form. And so in Gaussian reduction, we want to get a row echelon matrix in Gauss-Jordan reduction. What we want to do is we want to produce a reduced row echelon matrix where our leading non-zero term is 1 and there are zeros both below it and also above it. For example, we can use Gauss-Jordan reduction to solve our system of linear equations and first of all, we'll write down our augmented coefficient matrix. And as we did before, we'll avoid fractions by using the Chinese method, but we'll do one more thing to really delay getting fractions. We'll delay getting leading ones until the very end of our reduction and we'll see why that's actually useful to do. So we'll start with our augmented coefficient matrix. I like the first rows. I'll copy that over. I'll multiply the equation corresponding to the first row by 3. That's my next coefficient, so that gets me my equation there. I'll multiply the equation corresponding to the second row by negative 2. So again, my first equation becomes this. My second equation becomes this. And by that process, my x variables have now become equal but opposite. When I add them, I get rid of that x coefficient and my row operations, first row times 3, second row times negative 2, add and store the values in row 2. So that's 0, 11, negative 7, 13 as my new second row. Again, I'll take my first row corresponding to my equation, multiply that by the coefficient 5. That's my first leading coefficient of my last row. That's 10x plus 15y minus 5z equals 25. I'll multiply the third row, the equation corresponding to the third row by negative 2, that's my leading coefficient there. And I'll add and that eliminates the x terms and so I get a new third equation. And again, what did I do? First row times 5, third row times negative 2, add, store the results in row 3. So that's 5r1 plus negative 2r3 stored in r3 and I now have a new third row. So now I have a new coefficient matrix. I like that first row. I'll keep it. I will not make that leading coefficient of 1 because that introduces fractions and I don't want to do that. I will work with that second row. So I'll keep it and I'll use it to eliminate the third row. So we could multiply our second row by 11 and our third row by negative 11 and that's exactly what we do if we were machines following an algorithm. However, because we are human beings, we might notice that the coefficients are actually the same. So I don't have to do that. What I can do is I can just take that equation corresponding to the second row. I can take the equation corresponding to the third row and multiply it by negative 1. That changes those coefficients and I can add which will eliminate the y variable. And again, what did I do? Well, I took the second row, second equation. I multiplied the third row, the third equation by negative 1. I added them and I stored the results in row 3. So here's my third equation. I'll store those in my third row. And there's my row operation. So I don't have row echelon form, but I have a, you might call this a quasi row echelon form. Leading coefficient is above a bunch of zeros. Leading coefficient is above a bunch of zeros. And so we're good so far. Now we want to get zeros above those leading coefficients as well. Now notice that we worked our way down our matrix to get those zeros. And now we'll work up our matrix to get the zeros above the leading coefficients. So remember each row corresponds to an equation. So let's take a look at these last two rows and my equation 0x plus 11y minus 7 and so on. And if I want to eliminate the next coefficient, I'll do almost the same thing I did before. If I multiply this by 7 and this equation by negative, negative 8, then I'm going to get equal but opposite z coefficients. So I'll multiply the equation corresponding to the last row by 7. I'll multiply the equation corresponding to the second row by negative 8. And so that gives me equal but opposite z coefficients. And when I add the 2, z drops out and the only thing I'm left with is y. That gives me a new second equation. So what did we do? Well, we multiplied the third equation by 7, r3 times 7. We multiplied the second equation by negative 8. That's minus 8, r2. We added them and we're storing the results as our second row. So there's my original third row, which I was working with. I'll keep that there. And here's my modified second row. And you'll notice that now I have an 8. My leading coefficient now has a 0 above it. So we can do that also for the first row. So now I'm going to multiply the equation corresponding to the first row by 8. That gives me this. I'll multiply the equation corresponding to the second row by negative 1. So there's my next equation there. And when I add those 2, my z term drops out. So again, that's my first equation, multiplied by 8. My third equation, multiplied by negative 1. And added drops out the z term, because the coefficients of z are equal and opposite. And again, what I did, 8 times the first, plus negative 1 times the third, store the results in the first row. And I have new equation, new row. I'll continue to reduce that. So I have my new coefficient augmented matrix looking something like this. And now my leading coefficient of the third row is above it. My leading coefficient of the second row, we want to put 0s above it. So let's take a look at those first two rows. So again, the first equation, first and second rows correspond to two equations. And so I can take the first equation, multiply by the negative of the coefficient in the row below it. That's going to be positive 88. So I'm going to multiply the first row by 88. I'll get a new equation. I'll multiply these equation corresponding to the second row by 24. And that gives me equal but opposite y coefficients. So when I add the two equations, y drops out and I'm left with a simpler equation. And again, what I did, 88 times the first row, plus 24 times the second row, and store the results in the first row. And so there's my new first row. And my leading coefficient of the second row has 0s above it, which is exactly what we want. And now we're ready to complete our steps. So we want to get the leading coefficient of each row equal to 1. So remember, every row corresponds to an equation. This is 1400 x equals negative 736. If I want to get 1x, I'm going to multiply by 1 over 1408 and store the results in row 1. I can get that. Likewise for my second row, this is negative 88y equals negative 258. If I want to get leading coefficient 1, I'll multiply by negative 188. And I'll store that in row 2. And then finally, my third row, I'm going to get negative 8z equals negative 22. If I want a leading coefficient 1, I'll multiply by negative 18. And store that in row 3. And here's my Gauss-Jordan reduction. And the minor advantage to this is that each row corresponds to an equation. So I can read each of these as 1x equals negative 23 fourths, 1y equals 129 fourths, and 1z equals negative 11 fourths. And here's my solution to the system of equations.