 So let's look at a new space called the null space of our matrix A. And what we really have here is a system of linear equations such that all the right-hand side values equals 0, 0, 0, 0 so that my B vector equals the 0 vector appropriate to the size or the dimensionality that I'm interested in. So let's look at this A 1, 3, 2 and 6. You can already see that one is just a linear combination of the other. 1 times 3 is 3, 2 times 3 is 6. Even 1 times 2 is 2, 3 times 2, that's 6. So you can already see that there is a problem here. This matrix will not be invertible. But let's have a look at what happens if we do elementary row operations on this. So there we go. I can multiply this by negative 2. So I'm going to get negative 2, negative 6 and 0. I'm left with 1, 3, 0. 2 and negative 2 is 0. 6 and negative is 0, 0. So this is what I'm left with. In essence, I'm left with this. If this was x sub 1, I'm left with x sub 1 plus 3 times x sub 2 equals 0. And this is the same. 2 plus 3 equals 0. If I divide by 2, divide by 2, I get the same equation. I have this, I actually only had one equation as opposed to two equations. And if I were to solve for x sub 1, that gives me minus 3 times x sub 2. And I'm free to choose anything for that. And the choice is usually that we make this choice 1. So if that's 1, this means this is negative 3. So my x vector here really is negative 3, 1. Negative 3, 1. And remember though that we are talking about linear combinations of these. So a linear combination just of this one vector will mean that my solution, my null space, if I go negative 3, negative 1, negative 2, negative 3, let's go there. So I go negative 3 down and I go 1 down there. So this is my vector here, negative 3, 1. So any line, any vector on this line, which is a constant multiple of this vector, is in the solution space, is in the solution space. So I can plug anything on this line, any vector on this line in there to give me 0. So this becomes the null space of this matrix. And as such it is a subspace, this line that is in there. Let's have a look at 1. So here's a weird system of linear equations in as much as there's only 1. I only have 1. So it would be 1, 2, 3, 0. And there's 1 as my pivot. I can't really do anything else. Let's just solve as far as x1 is concerned. So that's going to be negative x2 sub 2 minus 3 times x sub 3. And what I can do here now is create two special solutions if you can look at it. Remember this is just, you know, I can simplify this as well. I can normalize this so that the norm of this vector is 1. It will just be a vector with a length of 1 so it will just lie there. So there's infinitely many solutions here. That is what we're saying, but they will all lie on this line. So that was just on that one line there. Here though we're looking at three dimensions. So, you know, something slightly different is going on here. Let's create two solutions. I'm going to call the 1 just s. Let's just call it s. And what I'm going to do is let this be 0. This be 0 and this be 1. So that's 1 that makes this one negative 2. So that lies on that line. And let's make another one. Or shall we make the other one? Let's make it t. It doesn't matter what it is. So I'm going to make this 0 and this one 1. So this is 0 and this one 1. So that means this is negative 3. So I have these two special cases as far as my solution is concerned. And remember any linear combination of those then should also be, that is my null space of this single equation system of linear equations. Let's look at a, we'll get deeper into this null space. There's a few things to understand about this null space. Let's do this. I'm just going to clear the board. Okay, so we've cleaned the board. I've just moved it up. I can multiply by negative 3, minus 3, minus 6, minus 6, minus 12 and 0. That's 1, 2, 2, 4, 0. That gives me 0. That gives me a 2. That gives me a 0. That gives me a 4. That gives me a 0. That's fine. I can multiply this by minus 2. So it's 0, minus 2, 0, minus 8 and 0. And minus 2, minus 4 should be there of course. But I 1, let's just leave that one out. So I multiply this by negative 1. So 0, minus 2, 0, minus 4, 0. Add that 1 plus 0 is 1. That gives me a 0. That gives me a 2. That gives me a 0. That gives me a 0. And here I'm left with 0, 1, 0, 2, 0. So what I'm saying here is that I have x sub 1 plus 2 times x sub 3 equals 0. And here I have x sub 2 plus 2 times x sub 4 equals 0. So I can solve for x sub 1 here. x sub 1 equals minus 2 times x sub 3. And x sub 2 equals minus 2 times x sub 4. So that's my solution that I have there. Once again, as we did with the previous one, let's make a couple of solutions here, special solutions. Remember I still have 0, 0, 0, 0, 0 as the solution to any of these. But let's do this. Let's make this a 0 and we make this a 1. So we're making x4 a 0 and we're making x3 a 1. That makes x2 a 0 and then makes x3, that's a 1, makes that a negative 2. Let's do another solution. Let's make this a 1 and that is 0. And this x sub 4 is a 1 that makes x sub 2 a minus 2 and that makes x sub 1 of that's 0 a 0. So you can see those special solutions that we can create once we get to this part. And these solutions are all, they all form part in the linear combination form part of the null space of my matrix A that we started with. Let's make this a matrix A. Remember it was the augmented matrix so we were only looking at that was one matrix A. So I'm creating solutions that are in the null space of my matrix. This is a bit tricky when you know you'll see different answers but remember the whole idea here is that the solutions are in this null space here forming this line, here forming a linear combination of these. So the solutions all fall within the solution that we are looking for. They all fall in this and the linear combination of these are in the null space of my original matrix.