 In this video I want to show you a simple way to calculate the specific solutions for the null space of a matrix. So I've already typed a few things here, but that's just to explain to you again, should you have forgotten what a null space is, remember you can also watch some of my playlists here on YouTube. I'll link to them below where you can watch videos on the null space. So I've just created a title here, Null Space of a Matrix. You'll see I've named the notebook Null Space of a Matrix from SMPI. We're importing matrix and init printing. Let's run that and we're just going to initialize init printing with open and closed parentheses. So we do have large tech printing to the screen. You'll note that this next cell is already a markdown cell. So what I want to show you is these three equations in four unknowns. So we have the four unknowns x sub 1, x sub 2, x sub 3 and x sub 4. We see three equations, so not really enough and we see that they're all equal to zero. I want to know the solution to this and you see that I've printed it here in matrix form. So I have my matrix of coefficients here. I have this column vector here for which I want solutions and that all equals this zero vector. Now if you're into spaces and subspaces, the null space is always a subspace of a larger space and we want to know what the null space is, that what solutions here of this four dimensional space here will give us these zeros. So what we can do is we can just take this matrix of coefficients and do gas elimination on that and we get to this row echelon form. So this is this matrix and we're going to do that just in sympi to see that we are correct and we see this reduced form here and what we have at the bottom here is a row of all zeros. Now look at it, we don't have enough equations for unknowns. Another little thing to look at here if you notice, if we look at the column space, so what this is saying I want x sub 1 times this first column plus x sub 2 times the second column plus x sub 3 for the third column and x sub 4 times the fourth column should equal the zero vector. That's what we're doing here with a column space. But you'd notice that column 2 is just twice column 1. So if I take column 1, 1, 2 and 3 and I multiply by 2 a scalar times this column vector 1, 2, 3, it gives me 2, 4, 6. So it gives me that second column vector and that's a problem. These are not independent of each other. This is just a linear combination, at least column vector 2 here. It's a linear combination. It's twice this first column vector and that's going to leave us with issues here. We see the 0, 0, 0. If I were to try and scribe this though just into normal equations that we are used to just from this row echelon form, it would say 1 times x sub 1 plus 2 times x sub 2 minus 2 times x sub 4 equals 0 and 1 times x sub 3 plus 2 times x sub 4 is going to equal 0. If I just simplify that by just getting one element on the left hand side, this is all I'm doing here, it would say x sub 1 equals minus or negative twice x sub 2 plus 2 times x sub 4 and x sub 3 equals minus 2 times x sub 4. This is what this row echelon form is showing us if I just simplify it. Now to get to those special solutions, remember you're just going to let all the unknowns here on the right hand side equal 0 except 1 of them, which will make 1. So let's start off by making x sub 4 0. Fortunately for us that means x sub 3 would also be 0 as you see here and make x sub 2 1. So if x sub 2 is 1 and x sub 4 0 that means x sub 1 is negative 2 and that leaves us with this first particular solution minus 2 1 0 0 and as far as our null space is concerned. Now we're going to let everything be 0 except x 4. If we make x 4 1 and all the other 0, well if x 4 is 1 that does mean that x 3, x sub 3 is negative 2 except 2 we kept that at 0. That means x sub 1 from these two equations here is going to be 2 and there we have a 2 0 negative 2 1 and that is going to be the second special solution, particular solution in our null space and any linear combination of these two would still be in the null space and that would solve our system of equations here. So if that was too quick for you, if you didn't fully understand please watch some of my videos explaining this in detail. Let's create this matrix though in some pie so we can see how very quickly we can get to these two particular or special solutions. So there's our matrix of coefficients and I can just print this into reduced row echelon form. Remember we've done that very easily and that's exactly what we saw there and I can just say a dot null space, a dot null space, open close parentheses and there we go the two special solutions that we saw there. There we go negative 2 1 0, negative 2 1 0 0 and 2 0 negative 2 1 2 0 negative 2 1 there they are as simple as that you could do all your hard work come and check it in some pie very very easy to do to get the null space and remember you can plug this into your system of linear equations and you'll get to the zero vector on the right hand side either of them and any linear combination of them for instance three times this plus four times that is going to give you a new column vector and that will also be a solution and all of these solutions together form the null space of this matrix here that we started with.