Welcome back to our lecture series linear algebra done openly. As usual, I'll be your professor today, Dr. Andrew Missildine. In section 2.6, we're going to talk about solution sets of linear systems, which admittedly we've solved many linear systems throughout this series already. So why are we finally talking about solution sets? Well, given the last two lectures we've done from the textbook 2.4 and 2.5, we've really dissected the geometry of affine sets and subspaces. We're starting to realize that solution sets, aka affine sets, aka flats, these are geometric structures and so we want to revisit these ideas of solving a solution, solving a linear system with this new geometric perspective we've taken on. Now some things to remember, so we say that a linear system is homogeneous. If the right-hand side of all the equations are all zero, homogeneous you're meaning that everything's of the same family, right? And so the right-hand side of every equation in the linear system is a zero. If you have something non-zero, of course, we call it a non-homogeneous system there. Homogeneous systems are always consistent because they contain the trivial solution. That is when x is itself the zero vector that always solves a homogeneous system. And so when it comes to homogeneous systems, consistency isn't the concern. The concern is whether it has a unique solution, the zero vector, or if it has multiple solutions. And if it has multiple solutions then it has a non-trivial solution. That is a non-zero solution to the homogeneous system. And so again, summarizing what we've learned about homogeneous systems so far is that a homogeneous system AX equals zero has a non-trivial solution as something other than zero as its solution if and only if it has at least one free variable in the system. Because if you have any free variable, you could set that free variable equal to one. And that would then create a non-trivial solution. And of course, if you have lots of free variables, you're gonna get lots and lots of solutions. And so we get multiple solutions if and only if we have a non-trivial, if we have a free variable in the system, excuse me. So let's look at an example of such a thing. Let's consider the three by three linear system. 3x1 plus 5x2 plus 3x3 equals zero, 4x1 plus 5x2 plus 4x3 equals zero. And 6x1 plus 6, plus x2 plus 6x3 equals zero. This is clearly a homogeneous system like we've seen before. And as most of our examples seem to be over the real number system, let's try something working mod seven. You notice our coefficients are only chosen as numbers between zero and six right here. Well, when it comes to solving the homogeneous system, we're gonna take our coefficient matrix and augment it with zero right here, right? So we get 35304540 and 6160, like we can see. And so we'd have a pivot position in the first spot right here. And so to get rid of the four that's underneath it, we're gonna take R2, subtract from it, 4 over 3, R1. Now, when you're working mod seven or mod jillow anything, right? We don't really want fractions in the system as much as possible. So how can we deal with this four thirds, right? So when we think of four thirds, really we should be thinking of, okay, since I'm working mod seven, the numerator four can be replaced with anything congruent to four. So we could take four plus seven, that would end up with being 11 thirds. That's not divisible by three. We can add seven one more time, we get 18 thirds right there. 18 is divisible by three. So we can reduce the fraction, that situation, to be a six. And so with that perspective in mind, we actually erase the four thirds right here. And we think of it as we're actually taking R2 minus six times R1. So we're gonna get a minus four right here, but then for the next one, you're gonna get a minus 30. And for the next one, you're gonna get a minus 18. And then the last one's just a zero, right? We'll come back to those in just a moment. For the third row, we need to take row two minus. Well, in this situation, I already know that if I take three times two in the usual sense, I can get a minus, I can get, excuse me, if I take two times row one, two times three is six, that's gonna cancel out. So no fractions are necessary because I can get the usual or integer arithmetic to work out for us right here.  And then finally, we're gonna get a zero right there, okay? And so now simplifying these things, we're gonna get four minus four, which is gonna be a zero for the first line.  Which if we add 28 to that, that gives us a positive three, which is what we get right here. And although I wrote negative 18 earlier, notice that should just, I mean, three times negative six, that's just gonna give us a negative four. And then four minus four gives us a zero again, so that cancels out. And then you'll notice zero minus zero is gonna be a zero. Looking at the third row, the six should cancel out. You're gonna get one minus ten, which is a negative nine. If you add seven to that, that's a negative two. Add seven one more time, that's a positive five. And then you'll notice you're gonna get six minus six, which is zero. And then lastly, zero minus zero, it's a zero. Nothing changed in that last column, interesting. So then looking at the next pivot, I should say here. So we have a pivot in the two, two position. In that situation, we wanna get rid of the five that's below it. You'll notice that we have zeros, zeros, and zeros, right? So to get rid of the five that's below it, we're gonna have to do row three minus five times three row two. But it's the only non-zero entry in row two is just the three right here. We actually don't need to simplify five thirds as a fraction mod seven. I know this is gonna cancel out to be a zero. And this gets us a matrix in echelon form. The reason this is important is that when we reach echelon form, we can see the nature of the solution set right here. So you'll see that because there is a pivot in the first position, variable one, x1 will be a dependent variable. Because there's a pivot in the second column, x2 will be a dependent variable. But there's no pivot in the third column. And as such, this indicates that we have a free variable in our system. Now, one should also be concerned about consistency, right? But with a homogeneous system, there's never gonna be a consistency problem. No contradictions will ever get here. Cuz even if you get a row of zeros since the system is homogeneous, it's always gonna be matched up with a zero. So no inconsistency is possible here. But we do have a free variable which tells us there's multiple solutions to this system of equations right here. So let's continue to investigate those multiple solutions here. And I should mention, because we have multiple solutions, because we have a free variable, there will be a non-trivial solution to this homogeneous system. Let's find such a non-trivial solution. We have to continue to solve it. So working where we left off with our echelon form, I don't want a three right there. So I'm gonna take one-third divided by a row two. But again, since there's only one non-zero entry in that row, I don't have to worry about simplifying the fraction one-third mod seven. I'm just gonna get a one right here. And then we have to get rid of the five that's above it. Well, that's easy enough. Take row one minus five times row two. That'll get rid of the five giving us a zero. And then the last thing to do is, well, I need a one in this position. So again, we're gonna do one-third times row one. But the only non-zero entries are actually divisible by three the usual integer sense. So divide by three, we're gonna get the following RREF. We see right here where our pivots are still in the first and second columns, but we do get this non-zero entry right here that's important to pay attention to. So if we rewrite this matrix as a system of linear equations, notice what happens here. The first equation looks like x one plus x three equals zero. The second equation looks like x two equals zero. And then the third equation would be zero equals zero, which we really don't even need it because that's not telling us much at all. And so we can see the exact same thing right here. And so as we try to solve for our dependent variables because you should solve for the dependent variables, you're gonna get the x one equals negative x three. But as we're working mod seven, I'm gonna rewrite that as a six x three because negative one and six are the same number when you work mod seven, right? And then x two was just equal to zero, right? Now this is something that's important to realize that a variable is dependent with meaning you cannot choose what it's going to be. Now in terms of x one, we don't choose what x one's gonna be because it depends on the choice of x three. On the other hand, x two is still considered a dependent variable even though it doesn't depend on x three. The thing is you don't have a choice when it comes to the dependent variable, x two has to be zero irrelevant of what you choose for x three. But x one is very much dependent on that choice there. So if you look at the general solution for x, well in general we have these variables x one, x two, x three but using the equations that we have above right here, we see that x three is whatever it is, right? But then depending on that choice of x three, we're gonna see that x one is six times x three and x two is zero. In which case when you look at these three entries right here, you notice that all of them are divisible by x three, right? You might be wondering what about x two but zero is just zero times x three. So each and every one of these entries has an x three as a common factor. You can factor it out using scalar multiplication. And so our solution x, our general solution x is just gonna look like some multiple of x three times the vector six zero one. This vector is pretty important here. Let's call it V for the rest of this exercise here, six zero one. And so with V in consideration, then we've now seen that the solutions, the solutions to the system AX equals zero, this right here is just x three times V right here whereas x three is kind of just some general, you know, it's just some number, right? We really are just taking scalar multiples of this vector V that in fact, the solution to this homogeneous system is a line that passes through the origin, a line whose slope is determined by this directional slope vector V. And so in fact, the solution set to the system is just gonna be the span of the vector V which will have the form T times V for any T that's in Z seven which as there's only seven scalars when you're working on seven, we can actually list every single solution. There's gonna be multiple solutions but in particular there's gonna be seven solutions and this is gonna come from when T equals zero and you get the zero vector. When T equals one, you just get back V 601. When T equals two, you're gonna get 507. When T equals three, you get this one. When T equals four, you get 305. When T equals five itself, you get 205. I think I said that incorrectly before. When T equals four, you get 304. And when T equals five, you get 205. And then lastly, when T equals six, you're gonna get 106. And those are the six solutions to this system of equations. You can check each and every one of them. I sorry, there's seven solutions because we do include the zero vector itself. And so solving a homogeneous system really comes down to this. And so the trivial solution is included here, 000. But then we get these six other non-trivial solutions which come about by setting your parameter, your free variable to something other than zero. So we found these non-trivial solutions to the system.


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