 What if we want to consider a system of equations with three variables? Again, for geometric purposes, we're going to call them x, y, and z. Although for our applied linear algebra course, we would typically call this x1, x2, and x3. And so when you graph one of these things, I'm not going to put a picture of this on the screen. But when you graph one of these things, it would give you a plane. So the linear flat object in three dimensional space would then be a plane, in which case we're then looking for the intersection of the planes. And there's a lot of ways that planes can intersect. But graphically, that's what we're looking for. We're trying to intersect these three planes together. Now, I claim that a solution to this system would be 234. And so let's see what happens when we plug it into the first equation. So you get 2 plus 2 times 3 plus 3 times 4. And then simplifying on the left-hand side, you have 2 plus 6 plus 12. Let's see, 6 and 12 gives you 18 plus 2 is 20. So that works out. If we put 234 into the second equation, you'll notice the second equation doesn't have a z. That just means whatever. When you plug in the 4, well, there's nothing to plug in. Not a big deal. You get negative 2 times 2 for x. You're going to get 3 for y. And then trying to simplify that, negative 2 times 2 is negative 4. Add 3 to that. You end up with a negative 1. That's what we were expecting. And then if you try the third one here, we plug in 2 for x, 3 for y, and 4 for z. Negative 3 times 2 is negative 6. Negative 6 times 3 is negative 18. And 5 times 4 is 20. If we do 20 and negative 18, that gives you a 2. 2 plus negative 6 gives you a negative 4. And that's exactly what we expected. So this point 234 is, in fact, a solution to this system of three equations, three unknowns. We call it a 3 by 3 system, much like how we describe matrices. You always count the rows first. That's going to be your number of equations. Then you count the columns, which in this case, will give you the number of variables. Now, try, try, try as you might. You're not going to find another solution to this system. This system has a unique solution, and it's exactly 2, 3, and 4. In this situation, the three planes that come together, they do intersect in three-dimensional space at a unique point. And this geometric principle does continue into higher dimensions as well. If we took some linear equations in a four-dimensional space, it's kind of hard to visualize, so you don't see a picture of that on the screen right now. But each equation forms what we call a hyperplane. A hyperplane is then a flat object with one dimension smaller than the ambient space, so like a line in the plane is a hyperplane, a plane in three-dimensional space, that's a hyperplane. And then the hyperplane would be a, if we had four variables, the hyperplane would then be something that looks three-dimensionally. It's flat, but it lives in this four-dimensional space. It's kind of hard to visualize, but geometrically, that's what's going on here. And so still, as we try to solve systems of equations, we're looking for the intersections of hyperplanes, for which we can get a unique solution. It's also possible, of course, you can get no solution, like if you take two lines and up the plane, it could very well be that two lines are parallel to each other, so you don't get a solution. In higher dimensions, you get a little bit more variety, but you can get multiple solutions to a system of equations. Consider the following example right here. So we take this system, 2x minus y plus z equals 11, x plus 3y minus 10z equals two, negative 3x plus 2y minus 3z equals negative 17. And so I claim I can give you multiple solutions to this. So let's first consider the solution, at least I say it's a solution, right? And one should be careful, it's like, what do you mean is everything I keep on checking is a solution? I should also warn you that some things aren't solutions, right? Like if I try one, one, one, you put it into the first equation, you get two minus one plus one, that's gonna give you two, that's not 11, right? So it's not a solution. It's not a solution to the first equation, so it's not a solution to the system. To be a solution, you have to be satisfied every equation. If you violate any of the equations, you're not a solution. But a real authentic solution here is gonna be five, negative one, zero. And we can check this, right? Put it into the first equation. You're gonna get two times five, which is 10. I'm just gonna write over it if that's okay. So you're gonna get a 10 right here. Then you're gonna get negative, negative one. That's a plus one, and then you're gonna get zero. Well, that's 11, all right, that's great. The second equation, you're gonna get five plus three times negative one is a negative three, and then another zero here. So you get five minus three, which is two, all right? And then lastly, you put five in for X, you're gonna get negative 15 minus two, and then zero, again, that's gonna be negative 17. So that is, in fact, a solution to the system. So that one checks out. What about, again, I'm gonna give you another one. What if we do six, two, one? Interesting, what's happening here? Well, if we were to plug this in the system, see what happens? You're gonna get 12, you're gonna get minus two, and you're gonna get one right here, right? So 12 minus two is 10 plus one is 11. That checks out. Try the next one here. You're gonna get six plus two times three is another six, and then you're gonna get a minus 10 right here. Six plus six is 12, minus 10 is two. That one works. And then lastly, if you plug in six, you get negative three times six, which is negative 18. Two and two is four, and then you get a negative three like so. So notice what happens here. Four minus three is one, one minus 18 is 17, negative 17. So we found another solution. So in this case, we actually found two solutions. Two, I should say distinct solutions. Two distinct solutions to the system. Are there any more? Well, it turns out that if you have two different solutions, you actually have infinitely many solutions. It's always, that's always the case, because if you take any affine combination of two distinct solutions, that always gives you another solution. So consider the following. So consider the system of equations AX equals B, all right? For some generic AX and B. And let's suppose, let's say that AU equals B. So U is a solution. And let's say that AV is also a solution. So here, X is being treated as a variable. In this situation, U and V are specific vectors that we say solve it. So we'd say something like, so U and V, these are solutions. These are solutions to the system, all right? So next, what you're gonna do is we're going to take some coefficient C1 and C2. So these are some real numbers, such that C1 plus C2 is equal to one. So that's the condition we have when we want to combine together for an affine combination. So notice what's gonna happen in this situation. If I take A times C1U plus C2V, like so, well, this is matrix multiplication, so we can distribute this thing. So this becomes C1AU plus C2AV, which because U is a solution, AU is the same thing as B. And since AV was also a solution, you're gonna get that AV is the same thing as B. There's a common factor of B, so you factor out the scalars, C1 plus C2. This gives you a B. But remember by assumptions, C1 plus C2 equals one, this is just gonna give you a B, like so. And so in conclusion here, I just wanna mention that any affine combination, any affine combination of solutions is also a solution. So if you ever find two distinct solutions to a system of linear equations, any affine combination will also give you a solution. And the only requirement is that C1 plus C2, yeah, it's up to be one, there's infinitely many possibilities for the C's and therefore you're gonna get infinitely many solutions. And so in general, in general here, what you see is that there are basically three options. If you're counting the number of solutions to a linear system, you can have no solution. This is when we have no solution, we say that the system of equations is inconsistent. Another option is you can have a unique solution. We saw such an example of that or you can have infinitely many solutions. Cause again, if you have two, you actually get infinitely many. And so if a system of equations has a solution, we call it consistent. If it doesn't have a solution, we call it inconsistent. Now in this lecture, we're not gonna talk about how one finds the solution to a linear system. We're just talking about the nature of the problem, the nature of linear systems. And so a few other things I wanna mention here is that suppose that a consistent linear system has M equations and N variables. So it's consistent means it has a solution. If N is greater than M, so we have more variables than equations, then that means you must have at least N minus M free variables. So a free variable for our purposes would interpret that to mean there are multiple solutions. So there's two types of variables in play. There's so-called free variables and there's so-called dependent variables. Free variables basically mean that you could assign anything you want to them. You can assign any real numbers to those free variables, no restrictions whatsoever. And then the dependent variables, once the free variables have been assigned, then they will be computed as linear combinations of the free variables. So the dependent variables depend on the choice you have for the free variables. In such a situation, this is called an under-determined system. We have more variables than equations. There's not enough information to define what all the variables have to be since there's some free variables. Another possibility here is if M is greater than N, if you have more equations than variables, then that actually means that if it's consistent, that N minus N of the equations could have been removed, right? You have all these extra equations you don't need. They're redundant. This is called the over-determined system. You have more information than is necessary. And so the sweet spot is when you have N equations and N variables, the corresponding coefficient matrix would then be a square matrix. That's where we want. In the over-determined case, honestly speaking, it's usually inconsistent because you have too much information and most likely that information will conflict with other information you have. So we should always be cautious of over-determined systems. Over-determined systems will be sort of a fruitful field when we talk about the least squares problem later on because as they're generally inconsistent, we can then find an approximate solution since there's no solution at all. Under-determined systems are gonna have free variables. They show up in practice. They certainly do. But like I said, that sweet spot where you have the same number of rows as columns, that's where we want to be most of the time. And just a little bit more vocabulary before we end this video here. When we talk about systems of equations, we often talk about homogeneous versus non-homogeneous systems. So if the right-hand side of every linear equation is zero, we call it homogeneous. Homogeneous here, of course, means one family. That is to say it's, everything's the same. And then we're talking about the right-hand side of these equations. They're all zeros. Non-homogeneous would mean that they're not necessarily all zero. And when I say they're homogeneous, that means because all the constant terms have to be moved over. We're talking about linear systems, not affine systems. An affine system might have some term over here. But the good news is that if you have an affine system, you can always move all the constants to the right-hand side. So solving an affine system is really the same as solving a linear system. So we don't need to differentiate between the two. One thing that's, and homogeneous systems show up all the time in practice. One very special thing about a homogeneous system is that they're always consistent. And why is that? Well, it's because you can always take all of the variables to be zero, okay? So if you have a system, ax equals the zero vector, this is the homogeneous system. If you plug in zero for x, now these x's, the zeros might be different lengths because a is in my end. But if you take, if you take a times the n zero vector, this will always be the m zero vector. So that's a solution, great. This is what's commonly referred to as the trivial solution. And so homogeneous systems are always consistent because they have a solution. But often we're interested in, is there another solution to this system of equations, right? Is there a non-trivial solution? And so when the homogeneous system has a non-trivial solution, that's because the columns of the coefficient matrix are linearly dependent. But if ax equals zero, if the homogeneous system only has the trivial solution, that actually implies that the column vectors of the matrix A are necessarily linearly independent, connecting us to a topic that we've seen before.