 So, let's see if I can find a basis for a vector space spanned by some set of vectors. So, for example, let's say I want to find the basis for the vector space spanned by vector 1, 1, 0, 3, 2, 0, 1, 1, and 2, negative 3, 3. And so I have 1, 2, 3 vectors, and I want to find the basis of the vector space spanned by these three vectors. Now, the key approach that we're going to be using is that if I can find a linear combination of these equal to the zero vector, any non-trivial solutions to that linear combination equal to zero correspond to a way of expressing some of the vectors in terms of the others. Let's set up that system of equations. So, what I'll do is I'll see if I can find a linear combination of these three vectors that gives us the zero vector. So, there's my equation, some linear combination equal to the zero vector, and I am looking to see if this particular equation has non-trivial solutions. So, I can set down my augmented coefficient matrix, and again, I don't really need those constants because they're all going to be zero and they're going to remain zero. But it's worth keeping in mind we are actually trying to solve a system of equations. So, here's my augmented coefficient matrix. I'll perform row reduction, and that will get me my reduced matrix. And so, remember, each row of this matrix corresponds to an equation. This is 1x1 plus 0x2 plus x3 equals zero. This is 1x2 minus 3x3 equals zero, and this last row is eliminated entirely. So, this is a rank two matrix, which means that two of my variables are dependent on the third variable. And so, that means I can parameterize a solution. So, I'll make that third variable. So, this equation here, this row corresponds to the equation 2 minus 3x3 equals zero. So, I'll let x3 be t, x2 is going to be 3t, and x1 is going to be minus 2t. And so, that gives me a solution. And one solution, if I let t equals 1, then x1 is negative 2, x2 is 3, x3 is 1. And that gives me this as an equation in my three vectors. And I can solve this for fact v1, v2, or v3. It's easiest to solve for v3. And so, I get the vector v3 expressed in terms of v1 and v2 only, which says that v3 is superfluous. I don't need it, and I only need v1 and v2. So, my basis, this vector and this vector only.