 Like a homogeneous system, a non-homogeneous system of linear equations with multiple solutions can have its solutions expressed in parametric form, where we're going to see that the solution set is a line, a plane, or some other flat, which can be expressed as a vector equation or with parametric equations. So consider the matrix A that's 3 by 3 given right here, 3 5 3 4 5 4 6 1 6, and consider the linear equation AX equals B, where B equals 0 6 3, and this is actually the exact same coefficient matrix we used in a previous example, and recall that we actually are working mod 7 in this example here. So without going through all the details, if you augment the matrix A with the vector B right here, so we have this matrix A augment B right here, when you reduce this, I would actually invite you to compare this example to the previous one we did. When you reduce this matrix A augment B, you're going to do the exact same row operations to put this matrix into its row reduced echelon form right here. The fact that the system is non-homogeneous compared to homogeneous makes no difference. The row operations you do will be the exact same, although the consequence on this final column will change as you switch from one B to another B. But the row operations necessary to reduce this matrix depends entirely on the coefficient matrix A. And so because of that, I'm going to skip over the details, and we're going to see that the RREF of this system will look like 1 0 1 6 0 1 0 2 and 0 0 0 0, and I invite the viewer to double check these calculations on their own to verify them. Well, when you take this RREF right here and switch it back to a linear system, you would get that x1 plus x3 equals 6, x2 equals 2, and then 0 equals 0, which that last equation is non-consequential whatsoever in this conversation. So because of this, we still identify that x3 is going to be a free variable, and x1 and x2 are dependent variables. x1 is going to look like x1, well it's going to look like 6 minus x3, but as we're working mod 7, we're really going to write this as 6 plus 6x3, and again, x2 is just going to be given as a specific number, in this case it's a 2. So using those substitutions, x1 is 6 plus 6x3, x2 is just 2, and then as x3 is free, we haven't really made any assignment there. Now what we're going to do is we're going to separate this vector into two separate vectors. One of the vectors is only going to depend on the multiples of x3, so we get a 6x3, a 0x3, and a 1x3. When you factor out the x3, like we did here, that leaves the vector 6, 0, 1, like so, and then if you then focus on the constant terms, right, there's this constant 6, there's this constant 2, and then in the last one there's a constant 0. If you take away the free variable portion, what's left as a constant, you get this 6 to 0 right here. Now if you did go and watch the previous video when we worked with this, the same matrix as a homogeneous system here, you'll notice this 6, 0, 1 might look a little bit familiar. 6, 0, 1 was the exact same spany vector we had before. We noticed that the null space of this matrix, that is its solution set to the homogeneous system, is just going to be the span of this vector V given as 6, 0, 1. So this wasn't a coincidence. This spany vector is based solely on the matrix A. It has nothing to do with B whatsoever. On the other hand, this vector 6, 0, 2, we're going to call it X null right here because this vector 6, 0, 2 is a solution to the equation AX equals B. If you take the matrix A and you multiply it by the vector 6, 2, 0, you'll see that that actually is equal to our vector B which was 0, 6, 3. I invite you to double check that calculation yourself. Pause the video if you need to. And so what we've now discovered is that the solutions to this, so because X here is just the general solution, the solutions to this linear system can be expressed in the following form. You have a particular solution, X naught, plus any scalar multiple of this directional vector 6, 1, 6, 0, 1 which actually spans the null space of this matrix. And so when you write this equation again, you see the following. X is equal to X naught plus Tv, right? And this right here looks like the vector equation of a line. You have a specific vector that's on the flat, and you have this spanning vector which tells you the direction, the slope, you could say, of the associated line. And in particular, this Tv is just a generic vector from the null space. Tv is the general solution to the homogeneous system right here. And so what then, when you compare these things, you see that you get the general solution for the homogeneous system, right? And then this right here is what we call a particular solution. So it's a vector that does solve this non-homogeneous system. And if you combine a particular solution with the general solution of the homogeneous system, that actually gives you the general solution to the non-homogeneous system. And as we can see that this is a line, X equals X naught plus Tv, using the specific points on this line, we could describe this solution set as an affine span, in which case you have X naught right here, G2 0, 620, excuse me. And then this other one, like if you take the T value right here to be 1, X naught plus V, you would get 521. That gives you two points on this line, two points to terminate a line. And so we would actually see that there are seven points on this line. And it's exactly these seven points that belong to the affine set spanned by 620 and 521. And this example demonstrates a much more general property. If we have any system of linear equations, AX equals B, where we can, I mean B could be the zero vector, or we could also think of it as any vectory, right? So it could be a non-homogeneous system here. So suppose the equation AX equals B is consistent for some given B. So B belongs to the column space of the matrix A here. And we're going to let X naught be a particular solution. That is, we're not talking about the general solution. We're saying this is a specific vector that solves this system of equations. Then the solution set of AX equals B is going to be the set of all vectors of the form X equals X0 plus Xn, where X0 is the particular solution before. And Xn is any solution to the homogeneous system of equations. And this vector equation gives us the solution to this linear equation. So in general, what I'm saying is the solution to a non-homogeneous system of equations is you take a particular solution, which we will actually find this as we roll over to the matrix, plus the solution to the homogeneous system. Now this is a pattern that shows up all over in linear algebra and has its consequences outside of linear algebra. For example, when one tries to solve differential equations, which one can recognize that differential equations have something to do with linear algebra, but I won't go into the details of that right here. But once one tries to solve a non-homogeneous differential equation, we relate this to a homogeneous differential equation. We solve it, we find the general solution. But then we go back to our non-homogeneous differential equation, we find a particular solution to the differential equation, and we squish the things together. We take a particular solution, plus the general solution of the homogeneous system, and those together form this general solution for this non-homogeneous system. And so every solution set to a linear system can be broken up in this way, a particular solution, plus the general solution to the homogeneous system. And therefore, as the solution to the homogeneous system is always a subspace, we are taking the translation of a subspace, which is in general going to be this affine span. It's this affine set, these flats, we have all these different names that describe all the same things, that affine sets and solutions to linear systems are one and the same thing. And so that brings us to the end of section 2.6 about solution sets, and kind of summarizes this three-part lecture, you could say, as we've talked about affine sets, subspaces, and solution sets, how the things are all very related to each other. And this is a geometric interpretation of these algebraic problems of solving linear systems that we're trying to do here. So as usual, if you feel like you've learned something in these videos, feel free to give it a thumbs up, a like, click, that'd be great. If you want to learn some more stuff about linear algebra in the future, please subscribe, or if you want to see other math videos as well, click the subscribe button. And as always, if you have any comments, questions, or concerns as you're watching these videos, feel free to post them in the comments below, and I will gladly reply to you when I can. Have a great day, everyone. I'll see you next time. Bye.