 An even better form of a system of equations is known as row echelon form. We say that a system of equations is in row echelon form if, first of all, the equations are in standard form. So this system of equations is not in row echelon form because it's not in standard form. Also, if we've agreed to the order x, y, z of the variables, then this system of equations is also not in standard form because the variables are out of order. The next requirement is that for each equation, the first non-zero coefficient occurs later in the variable sequence. If we've written the equations in standard form, this means our system of equations will have a stepped appearance. So in this last system, the first equation starts off with an x term. The next equation starts off with a y term, as does the last equation. And so this cliff that we front into indicates that we are not in row echelon form. So none of these systems of equations is in row echelon form. What about these systems of equations? Well, in our first set, we see that the equations are in standard form, and the first non-zero term of each equation occurs later in the variable sequence. So this system is in row echelon form. Similarly, the second system of equations is in standard form, and the first non-zero term of each equation occurs later in the variable sequence. So this system of equations is also in row echelon form. So what's the big deal about row echelon form? To answer this question, we might consider the following. Even in a system of linear equations, it might have a unique solution. So for example, this system of linear equations has solution x equals 5, y equals 4. And you know this is true because I said it's true and I'm on the internet. But you might want to verify for yourself that those are the actual solutions. After all, this isn't like politics, where you can trust everything anybody says. Another possibility is we may have no solutions to a system of linear equations. For example, this system has no solution. And finally, there's another possibility, which is that we may have many solutions to a system of equations. So for example, this system of equations has solution x equals 0, y equals 5. It also has solutions x equals 1, y equals 4, x equals 2, y equals 3, x equals 3, y equals 2, x equals 4, y equals 1, and so on. And this situation suggests the following goal. Given a system of linear equations, we should try to find all solutions to that system of linear equations. Row echelon form becomes important through something called back substitution. Given a system of linear equations in row echelon form, we can write every variable in terms of later variables and then use back substitution to find the values of all of the variables. Let's see how that might work. So we'll take our system of linear equations in row echelon form. And since our variables are listed in the form x, y, z, w, we'll assume that's the ordering of the variables. It's convenient to start with the last variable w, and we see from our last equation, w equals 3. Now if we go to the next to last equation, this equation has variables z and w, and we can solve it for z in terms of w. And if we do that, we find z is equal to 3w plus 1. Moving up to the next equation, we see that it has variables y and w, so we'll solve for y, finding that y is equal to 2w plus 5. Finally, the first equation has variables x, y, z, and w. We'll solve it for x and find that x is 2y minus 3z minus w plus 4. So now we'll work our way backwards through these equations. So our last equation gives us w equals 3. We'll make a note of that. Next, we know z is equal to 3w plus 1. Since we already know the value of w, we can substitute this in and find z. And so now we know that z is equal to 10. Our next equation tells us that y is equal to 2w plus 5. So we'll use this and our value for w to find y. So we know y is equal to 11. And finally, our first equation gives us the value of x in terms of y, z, and w. So we'll use that, substituting our values for y, z, and w, and find x. It's convenient to write this solution in vector form. Because we've decided in order of the variables, we also have an order of the vector components, x, y, z, w. And so we can compactly express our solution as the vector negative 7, 11, 10, 3. Rho echelon form is particularly useful when we have to parameterize our solutions. In this case, what we can do is set any variable that is not the leading variable of an equation, equal to some arbitrarily chosen parameter. The variables we set equal to a parameter are called the free variables. We can then solve for the remaining variables in terms of these parameters. And these other variables are called the basic variables. And as before, we can try to express our solutions in vector form. So I suppose I have a system of equations in Rho echelon form, and I want to express all solutions to this system. So the first thing we should do is to identify the free variables, those that will be set equal to parameters. So remember, these are going to be the variables that are not the leading variable of any equation. So if we take a look at our equations, we see that we have four variables, x, y, z, and w. And while x and z are leading variables, y and w are not. So we'll let w equal t and y equal s to parameters. Now we can use the equations to solve for our basic variables. And we're going to work our way backwards through the equations. So in the equation z plus 3w equals 1, we'll substitute our value w equals t, giving us the equation z plus 3t equal to 1, and we can solve that for z equal to negative 3t plus 1. And we'll write that down as a part of our solution. Next, we'll go to the preceding equation x minus 3y plus z minus 4w equals 5. And so we'll substitute in the values for y, z, and w into our equation, and then solve that equation for x. And when we do that, we find that x is equal to 3s plus 7t plus 4, and we'll write that down. What about a vector form of the solution? To find that, we might begin by writing our solution as a vector. And here's a useful idea in mathematics and in life. Consistency counts. When we've been working with equations in standard form, we've made sure we've included all variables and even used coefficients of 1, 0, or negative numbers so that we could express things as sums. If we look at the components of our vector corresponding to our solution, we see that some of the components include s, t, and a constant. Others only include s, some only include t, and so on. So let's try to be consistent and write every component so that it includes s, t, and a constant. Our first component already has s, t, and a constant, so we'll leave it alone. Our second component is just s, so we'll write it as 1s plus 0t plus 0. Our third component is minus 3t plus 1, so we'll write that so that it's now 0s plus negative 3t plus 1. And our final component, t, will be rewritten as 0s plus 1t plus 0. Since we add vectors component-wise, what this means is we can split this vector up into a vector consisting of the s terms, a vector consisting of the t terms, and a vector consisting of the constant terms. But wait, there's still more. Because of scalar multiplication, I can remove the s from each component of this vector and place it as a scalar multiple. And likewise, this t can become a scalar multiple of a constant vector. And this gives me a vector form of our solution.