 So let's continue our look into the null space. Now yesterday we had a few examples, but in some sense there were very special examples and I don't want you to think that that extrapolates to all examples. So let's have here three equations and four unknowns. We don't have enough equations and our matrix of coefficients would be 1, 1, 3, 1, 1, 3, 2, 4, 10, 3, 5 and 13. You can take the matrix of coefficients which would make our matrix A. You can make the augmented matrix. If it's all zeros as well, you're going to end up with the same thing. This is the row reduced form of the matrix of coefficients. And if we had done the augmented matrix, there'd be another 0, 0, 0 here. That would happen here. So what are we saying here? We're saying that x sub 1 plus x sub 2 plus x sub 4 equals 0. In the second row we have the fact that x sub 3 plus x sub 4 equals 0. And in the last row we just have zeros. We just have all zeros. So that doesn't help us at all. So what do we want to do from here? Now in these circumstances, let's just look at that first one. And there are four unknowns and I have three in that first one. So it's very easy for me to alternatively make this a 0 and this a 1 and then make this a 1 and this a 0 and see what happens. So if I make 2 a 1 and this is 0, that means x sub 1 is minus 1. I made this a 1. Now what happens to 3? Well if I made this a 0, this means x 3 must also be 0 and I made x 4 a 0. So all I'm doing here, I just solved this first equation for x sub 1. So that's minus x sub 2 minus x sub 4. And this second one we can also make that's minus x sub 4. So in the first instance I made this a 0 and this a 1. That made x sub 1 negative 1. I made this a 1, so it is there. If I made x sub 4 a 0, x sub 3 is a 0. Now I'm just going to swivel that around. I'm going to make x sub 4 1 this time. So I'm going to make, so this was in essence making x sub 2 into 1. But there's only a combination of that. If I could have made it a 2 or a 3, doesn't matter. This would be sort of the simplest form I can get to. And if I make x sub 4 a 1 in this instance, and that means x sub 3 is going to be a negative 1. And if I made this a 1, well if I made this a 1 and this a 0, x sub 1 is still minus 1. And I made this one a 0. And this was x sub 4 remember that I did. I made x sub 4 a 1. And what we are saying here is, now this is going to be different because there's not a 0 there. There's a 0 and there's not a 0. There's a 0 and there's a 0. But there are not two 0s there. So I can't just say any linear combination of this. It's got to go to our special, it worked out for us especially in our previous examples that there was a 0 there, for instance, and I could do linear combinations. But what we are saying here is linear combinations of these two. And in our instance here, making x 2 1 and there we made x 4 1 and x 2 0. There x 2 was 1, there x 4 was 1, x 2, x 4 was 0 and x 4 was 1 and x 2 was 0. So let's do linear combinations of those. So that is going to be minus x 2 minus x 4. There's going to be an x 2 there. I'm just adding them up. And there's going to be a minus x 4 and there's going to be an x 4. And there, that is linear combination of these two. What we are saying is that these two special cases, that special case and that special case, that is our null space. So it's our special cases in our null space, these special cases, special cases. So I can solve, I can plug this one in there, this one, I can plug this one in there. But if I want to look at the whole null space, it is linear combinations of these together with this, together with this. So in that instance, so now if I sit, let's do something very wild. I'm going to set x, let's make x 2 equal to, let's make that negative 3 and I'm going to make x 4 into 6. That means a linear combination of this is going to be minus, minus 3. That's 3 minus 6 and x 2 is minus 3 and x 4 is 6. So that's minus 6 and x 4 is 6. So I'm left with this, I'm left with minus 3, minus 3, minus 6, 6. That will also be, that would also be in my solution space. So that makes up this whole null space. So be very careful when you look at these. And sometimes this feels very tricky, you know, what do I do? Solve each of these lines for our pivot here, which is x 1, that one's x 3. And then this first one is very nice because I can just play around with these two, which means it gives me a solution to this. So alternatively make this 1 and 0 and make this 1 and that 1 0, which gives you, if you make x 1, x 2, if you make that x, if you make that equal to 1, this is your special case. If you make this one 1 and x 2 0, this will be your special case. But the linear combinations of these are the linear combination of this, which is that. So I can plug in values for this is my x sub 1, this is my x sub 2, this is my x sub 3, this is my x sub 4, so that I end up with this. If I make x sub 1, this and x sub 2, this and x sub 3, this and x sub 4, this, I have a solution to my, I have a solution. That will be a solution. Linear combinations of these will give me a solution to this problem that I have here. So very, very important for you to realize how to construct these two. And it is linear combinations of these two special cases in this form, in this form which will give me possible solutions. And that null space, that is a special, that is a subspace. That, this null space is a subspace in R4. If we took that, all these coefficients here and all the solutions here must be real numbers. This will be a subspace of R4 called the null space of our initial problem. So I hope you see that distinction from what we had yesterday. So here we are in Mathematica and let's have a look at this example that we just had on the board. I've created the augmented matrix here. I've called it uppercase a, lowercase a, could have called it anything. And let's just hold down shift and enter. It is there, let's just look at it in matrix form. And you can see, different from what I had done on the board, this is the augmented matrix. And let's get that in row reduced form. So row reduced a, a, and I want to see that in matrix form as well. And you see that all these zeros on the right hand side, it's going to work out like that. This is all the right hand sides are equal to zero. So in effect, if I were to read this off, it says one times X sub one plus one, one times X sub two, there's no X sub three is there. So X plus X sub four, that's zero. And there I have an X sub three and X sub four is zero. And I solve here, X sub one is then minus X sub two, minus X sub four, and X sub three is minus X sub four. And I'm going to just look at that first one, because you can see X sub four isn't the second one. So alternatively, I'm going to make X sub two one and X sub four zero to solve for X sub one. And because I made X sub four zero, I also have a solution for X sub three, and then I'm going to swap it around. I'm going to make X sub two a zero and X sub four one, which also gives me a solution for X sub three, whilst giving me a solution to X sub one. Okay, so let's go carry on. I've just now made the matrix of coefficients. I've left out the zero, zero, zero. I want to do that because there is a null space function inside of the Mathematica, but you can only enter the matrix. Remember, we are writing AX equals B. We are really interested in that matrix of coefficients, not really the augmented matrix. I just did the augmented matrix here just to show you how we get. It's just an easy way to see how we get then from here to these equations that we have here. So let's just have A, and now I can call null space, null space of A. It's going to give me two rows, but I actually want the column view. So I'm just going to transpose it. Remember to always transpose it when you print it out to the screen with the null space, and I want it in matrix form. And there we go, I should say the null space of A. And there we go. And we see what we had on the board, the negative one, zero, negative one, one. It's just in a different order than we had. Here we can see X two is zero. And in this one, we see X four is zero. And here in the first one, X four, X up four is one. And here X up two is one. So it's just in the reverse order. But it gives me that exact null space. So just as we did on the board, the Mathematica is going to give you this exact same solution. Now let me show you how we can just test what we did on the board. So I'm going to show you how to construct a function in Mathematica. I'm going to call my function F1. And I'm going to, inside of square brackets, give it all the variables that we are dealing with. Just as you would have done in algebra or in calculus, you say the F of X, or in multivariable calculus, you would say the F of X comma Y. So let's put in X one. But to tell Mathematica that this is a placeholder variable, I'm going to put an underscore after that one. Comma, there's an X two. I'm not going to do the subscript notation. I'll just keep it easy like this. Underscore, I have X three underscore. And I have X four underscore. And I underscore, there we go. And I close my square brackets. Now I don't want any immediate execution here, or any immediate equation of the right-hand side to the left-hand side. So I'm going to use this notation, colon, forward slash. And look what happened. These turned like a greenish color. And the colon means don't execute now. This is just a placeholder function that I'm creating. So just as null space row reduced was a function, I am creating my own function. And one of the functions I can create is just normal mathematical functions. And remember our initial thing we had here? It was a system of linear equations. So that said it was X sub one plus X sub two plus two times X sub three. And instead of times, I'm leaving a space between the two and X sub three plus two times plus, what did we have three times? Times is a space X sub four. And I'm going to put a semicolon because I don't want that expressed to the screen. So let's press Enter. And I'm just going to copy and paste here from my second screen, the other two. So all I'm doing here basically is recreating my linear system that I had initially. And they're all equal to zero. And the trivial solution is, let me show you that. Let's shift and Enter. Let's do the trivial solution. I'm going to say if one and the placeholder see that it's just like a mathematical function now. I have all these arguments that I have to pass to my function. It just so happens to be this time around a mathematical function. So zero comma zero comma zero comma zero. If I were to do that, I get the solution zero. The zero vector is a solution. But let's do something special here. Let me say a new style. And I'm going to choose, I'll bring it in for you. I'm going to choose display formula numbered. Let's do that. And remember, this first one was X sub four and the second one was X sub two. So let's make X sub four. Let's redo that. Let's make X sub four. Let's make that equal to, let's say, six. And we're going to make X sub three. X sub three, we're going to make that equal, say, negative three. And remember, this was in reversed order. So this first one was X four. And this one, that should be X sub two. The second one was X sub two. So let's just do that and see where we get to. So I'm going to say minus, this minus negative three, times this minus one, that's three, minus six. So I'm going to end up with, I'm just going to do some placeholder text here. So that first one is going to be a negative three. And if I do minus, or six times zero, zero, and a minus three, so that's also going to be a minus three. And if I do six times minus one, that's minus six. So this is going to be a negative six. And if I do six times one is six. And the negative three times zero is zero. So then I'm going to end up with a positive six. I'm just doing a little placeholder work for myself there. And let's see what happens if I take F one. And I pass this now. It's negative three, comma, negative three, comma, negative six, comma, six. And if I pass that, it is zero. That is a solution. So remember that you have to multiply this first null space here, that's x four times this, plus x two times this second one. And that's going to give you a new solution in the subspace. And you can check any one of these. Let's just make sure. Let's just do F two there. F two was let's do a negative three, comma, negative three, comma, a negative six, comma, six. If I were to do that, also zero. So that is a legitimate solution. Let's just check F three as well. And it's negative three, comma, negative three, comma, negative six, negative six, and a six. Close there. It's a solution. That linear combination of x four times this first one and x two times the second one gives me a new vector. And I've just wrote it here just for convenience sake, because I don't want to remember negative three, negative three, negative six, negative six. But I've created this function called F one, F two, F three to represent my system of linear equations. I pop in this new solution there. Pop in these new there and I do get zeros. So I know that this is a solution. It is in this null space, which is a subspace of R four. It's R four because we're dealing with four dimensions here, four unknowns. And you see how beautifully this works out. And you can also see my two special solutions here. And linear combinations of those, remember where the x up four and x up two there, those linear combinations make up the full null space.