 Alright, so let's try to find another set of basis vectors and see if we can try to generalize our process. So, again, we'll take a vector space span by the vectors v1, v2, v3, v4, and v5, and, again, what we want to do, the vectors will be linearly dependent if I can find a linear combination equal to the zero vector with a non-trivial solution. So, I'll set up the corresponding coefficient matrix, so, again, here's all of my vectors as column vectors, and I want to augment that with the zero column vector. Again, we don't really need this, but it's useful to keep it in there to remind us that we actually have an equation. I'll reduce it, and after all the dust settles, I end up with this. And so, notice that my leading variable is here, x1, x2, x4, and so, let's see. Well, I can take this and find the parametrization of the solutions. So, my solutions, x5, I'll make that 9t, x4 equals minus t, x3 equals s, and so on. So, I can parametrize my solutions. I don't need that matrix anymore, I'll slide my solutions up there. And, well, if I let, I have two free variables, t and s. If I let s be zero and t be one, then I get the solutions, x1 equals three, xt equals negative five, and so on. And, I get this equation here, and that means that I can find v4 in terms of v1, v2, and v5. Likewise, same parametrization, I'm going to let s equal one, t equals zero, that wipes these two out, and I get a different solution, minus v1, v2, plus v3, and so that allows me to solve for v3, and I have vector v3 as the linear combination of v1 and v2, I have v4 as a linear combination of vectors v1, v2, and v5. Now, putting this together, that means v4 and v3 can be expressed in terms of vectors v1, v2, and v5. So, the vectors I need, the basis, this vector, this vector, and this vector, and these two are superfluous. Now, you might notice that the dependent vectors v3 and v4 are the ones that are directly expressible in terms of the free variables. Vector v3 corresponds to the variable x3, and that corresponds to one of my free variables. Vector v4 corresponds to minus t, and that's another one of our free variables. And, well, this is a coincidence. Or is it? Well, let's think about that a little bit more. So, suppose I have my set of vectors, and I have a set of linearly dependent vectors. So, again, when I go to solve the corresponding coefficient matrix, what I'm going to get is a non-trivial solution, and it'll be parametrized using p-free variables, where p is to be determined based on what our reduction turns out to be. So, if I think about this a little bit, we should come to the following conclusions. If, as we just did, set all but one of the free variables to zero, and let the remaining free variable be equal to actually any non-zero value that we want, we can write the corresponding vector as a linear combination of the vectors corresponding to the basic variables. So, anything that used any of the other free variables is gone, and the only thing that's left over are vectors that can be expressed using our one free variable that we set equal to our non-zero value. And so, what that means is that none of the vectors corresponding to the free variables have to be included in our basis, because they have all been zeroed out. With a little bit more thought, it should also be clear that the remaining variables have to be linearly independent. And again, that goes back to the idea that if this has any non-trivial solutions, then what we have is a set of linearly dependent variables. And that gives us a really nice result. The basic variables to the matrix whose columns are the vectors of v correspond to the basic vectors of the vector space spanned by v. So, our basic variables correspond to basis vectors.