 In the previous lecture video, we learned about what it means for a matrix to be in echelon form or row reduced echelon form was also discussed. And this is very important when one studies an augmented matrix such as the augmented matrix you see here on the screen, as we've also learned augmented matrices in code systems of linear equations. And so if a augmented matrix is in echelon form, I want to show you what that says about the associated system of linear equations. But let's first check, double check here to make sure that this matrix is in fact in echelon form like I talked about. Now to be in echelon form, for an augmented matrix, you only look at the coefficient matrix, you can care less about this side right here. Again, for determining whether it's in echelon form. If there's any row of zeros, they should be at the bottom, but there's no row of zeros in this matrix. So we go to the next one. Look for the leading entries. The leading entries are sometimes called the pivots are the first nonzero number in a row. And I'm going to draw a box around them to highlight them. So to be an echelon form, these pivots should be forming a downward staircase as you move from left to right here, which is where the name echelon comes from. It's French for staircase in that situation. And likewise, the numbers below the pivots should be zeros, which is exactly what happens. So this matrix is in fact in echelon form. Now it's not in row reduced echelon form, which remember to be in row reduced echelon form, you have to be in echelon form. But you have to also satisfy two other conditions. The pivots have to be one, which they are. The problem is the numbers above the pivots have to be zero, which they aren't in this situation. That's okay though, because if an augmented matrix is an echelon form, it turns out we can very easily solve the associated system of linear equations. So if we take this matrix and put it back into a system of equations, note that each row of the matrix represents an equation here. And our variables are x, y, and z here. So the first row gives us the equation that 1x plus 1y minus 3z is equal to 2. The second row then tells us that 0x plus 1y plus 3z is equal to negative 2. Now I'm not going to write the 0x right there, there's no reason to consider it. But then we get a y plus 3z is equal to negative 2. And then for the last row, you're going to get 0x plus 0y plus 1z is equal to 1. And so we get the following matrix right here. Notice that because of this staircase of zeros, there's a big gap that shows up in this system of equations. This gap is actually quite wonderful for us. Note, look at the very last equation, z equals 1. Hey, I know what z is supposed to equal, it's supposed to equal 1. Now if this coefficient was something other than 1, let's say it was like able to like 2. So you get a 2z equals 1. That's not really worse. I mean, could you divide both sides by 2? You get z equals 1 half. Sure, your answer is a fraction or a decimal, whichever you prefer there. But my point is, you can easily solve the equation if there's only one variable. It's just a regular linear equation. We've learned how to solve those before. Super simple, super nice. And so because of all the zeros inside that row, the last equation has only one variable and we can solve for it. In this case, because the coefficient was 1, it was solved for immediately. Now, thinking of the substitution method, if I know that z is equal to 1, look at the equation that's above it, y plus 3z is equal to negative 2. That equation only depends on the y and the z, but I know what z is. So if I were to substitute in z equals 1 into the second equation, I could solve for y. And let's do that here. Let's take the equation y plus 3z is equal to negative 2. We're going to move the z to the other side so we get that y is equal to negative 2 minus 3z. And then substituting the fact that z is equal to 1, we get that y is equal to negative 2 minus 3 times 1, which of course is negative 2 minus 3, which then gives us the negative 5. So because we knew z and we have an equation that depends only on y and z, we could substitute z into that equation and solve for y. So we get that y is negative 5 and z is equal to 1. Now look at the first equation, x plus y minus 3z equals 2. I know that z is 1. I know that y is negative 5. I could substitute those values into the first equation. Then I only have one equation, only one variable I don't know, which is x, and we can solve for x in that situation. So again, move the y to the other side, move the z to the other side. So we're going to get that x equals 2 minus y plus 3z. Now y is equal to negative 5. A negative negative 5 is in a positive 5. It's a double negative. And then here, z is equal to 1. 3 times 1 is equal just to 3. And then adding these together, 2 plus 5 is 7 plus 3 is 10. x is then equal to 10. So if I look at all the numbers I discovered, x was 10, y was negative 5, z was 1. That then gives me the unique solution to this system of equations. And so by having a matrix which is an echelon form, I could actually very quickly solve the linear system using a method which we call back substitution. So it's like the second half of the substitution method. The substitution method is you, once you start finding out values, you should start plugging them back in to figure out the other things. If you have a matrix with an echelon form, you're ready to do this back substitution, which is actually a fairly simple straightforward process. So that's actually pretty awesome. If a matrix is an echelon form, we can solve the system of linear equations. And while this example has three variables, there's nothing that requires it be 3. If I had five variables for which I knew one of them, and then I could substitute it into the next equation to find the second one, then I could substitute those into the next equation to find the third one. I could then substitute those into the next equation to find the fourth one. I could then substitute those into the next equation to find the fifth one. I could keep on going and going and going. This strategy of back substitution will solve any system of equations so long as the corresponding augmented matrix is an echelon form. Okay. This is an example of a linear system which has a unique solution. So this was consistent and independent. What about some of the other cases? Consider the following augmented matrix this time. So we have 1, 0, negative 5, augment 1, 0, 1, 1, augment 5, 0, 0, 0, augment 0. If you were to look at this matrix right here, okay, this matrix is in fact in row reduced echelon form. Look at the leading ones. They make a downward staircase. The row of 0 is at the bottom. The leading ones are actually ones. We have zeros above and below all of them. This matrix is in row reduced echelon form, RREF. If we were to translate this RREF into a system of linear equations, we get the following right here. The first equation will become that x minus 5z is equal to 1. The 0 means we can omit the y. The second equation then tells us that the second row tells us we're going to get 1y plus 1z is equal to 4. Again, we can ignore the 0 because 0x doesn't do anything. And then the last equation here says that 0x plus 0y plus 0z, which just simplifies to be 0, is equal to 0. This last statement here is an identity that our totality was the other word we used here. This is a statement that's universally true regardless of what's happened with the variables. It turns out it places no restriction on our variables whatsoever. This is exactly the dependent case that we were considering beforehand. That is, there's infinitely many solutions to this. And in fact, I can describe the general solution very easily. The general solution will come about by moving the non-pivot column to the other side. There was a pivot in the first and second columns, which actually translates to mean that x and y are going to be dependent variables. The pivots in your matrix give you the dependent variables. On the other hand, the column, which is not a pivot, it doesn't have a pivot, the z column, this turns out to be our free variable. And we can then describe the general solution using that free variable. So notice I can take these equations, rewrite them. The first equation, x minus 5z equals 1. This tells me that x equals 1 plus 5z. The second equation tells me, if I solve for y, that y equals 4 minus z. And therefore, the general solution to this system of equations then will be 1 plus 5z comma 4 minus z comma z. Like so, every solution to this linear system will have that form. You can choose any real number for z that you want, but then once you choose the free variable, the dependent variables will then become assigned based upon that choice. And because the matrix is in row reduced echelon form, the system of equations is already solved. And I can even describe the general solution in the dependent case very, very nice. So it's like, wow, row reduced echelon form is nice. If you have an augmented matrix in row reduced echelon form, you have solved the system of linear equations. So we've talked about the independent case. We've talked about the dependent case. What about the inconsistent case? Well, this look, this matrix here, it's basically identical to the matrix we just saw a moment ago. The first two rows are identical to what we saw here. But I made one small change in this last row where you have rows zeros. I actually changed this zero to a seven. So what does this tell you about your system? Well, the first equation is still x minus 5z is equal to one. The second equation is still y plus z is equal to four. But the last equation is now a contradiction zero equals seven. There's no assignment of the variables that can make zero equal to seven. And because of this contradiction, that means we actually have no solution. The system is in fact inconsistent. So if a matrix is in row reduced echelon form, and it's inconsistent, the system, we can detect that. If we're in the dependent case and the matrix is in row reduced echelon form, we can detect that and pull out the general solution. And if we have the independent case, that is, there's a unique solution. If the matrix is in row reduced echelon form, we can grab the solution with no effort whatsoever. But even if you're not in row reduced echelon form, if you're in echelon form, we can readily find the solution. So a matrix in echelon form or row reduced echelon form is basically solved. The row reduced echelon form is exactly solved. And so this is why we care about augmented matrices being in echelon form. echelon form, particularly row reduced echelon form means we've solved the problem. So if I can take a matrix and transform it into row reduced echelon form without changing the solution set, then that is a method that solves the system of linear equations. And that method is known as Gaussian elimination, for which we'll start to learn how to do that in the very next video.