 In this video, we present the solution to question number 11 from the practice midterm exam number two for math 2270, we're given a matrix A which is three by five and it's presented right here on the screen. We're also told that a is row equivalent to this matrix right here, which we can see very quickly that this matrix is in echelon form. It's in fact, it's in fact row reduced echelon form. And so we're asked to compute a basis for the null space and a basis for the row space of this matrix right here. Both of these we can get from the echelon form. So since we have this row reduced echelon form, we notice there is a pivot position in the first and third column. That tells us that the basis, that tells us that the variables two, four and five are the free variables to the associated homogeneous system of equations. If you prefer, you can write this as a system of equations, right? x1 minus 2x2 plus 9x4 minus 16x5 equals zero. You're going to get x3 minus 3x4 plus 5x5 equals zero and then zero equals zero. We don't need that last equation honestly. It doesn't help us all at all. And so we could pull these things apart, right? So we solve for the dependent variables. These are going to be the variables in the pivot positions. x1 equals negative x2 minus 9x4 plus 16x5. And then we also get that x3 equals 3x4 minus 5x5 like so. And so then we can think of the general solution to the homogeneous system. It's this vector with three coordinates, well, four, excuse me, five coordinates there. x1, 2, 3, 4, 5. And so then we substitute in these relationships we found between the free variables and the dependent variables. So the dependent variables were one and three. So we're going to get negative x2 minus 9x4 plus 16x to the fifth. x2 is free, so we'll leave it alone. x3 is dependent, so we get 3x4 minus 5x5 and then we get x to the fourth and then we get x to the fifth. Like so. In which case if we peel this apart, we're going to get x2 times, we're going to get a negative 1, 1, 0, 0, 0. Like so. Then we're going to get an x4 times, just decomposing these vectors, negative 9, 0, 3, 1, and 0. Like so. And then the last one we're going to get is x5, which would be 16, 0, negative 5, 0, and 1. Like so. And so then these three vectors then form a basis for the null space and we should then record as much. What I'm going to do at this moment though is actually rewind it a little bit. You know, back when we had VCRs, that's the sounds they made. Yep, that's exactly the sound they made. So just again, just kind of erasing what's happening here because I want to mention that the basis of the null space, we can actually gather straight from the REF. We don't really need to show any of our work whatsoever in this situation. We get that the null space of A is going to equal the span of the following three vectors. There are three non-pivot columns. That gives me the nullity of the matrix. So we see that the nullity here is equal to 3. So we're going to get three vectors associated to these three non-pivot positions, the three free variables. Kind of like Dr. Seuss right there. The three free variables. And so what we're going to do is I'm going to put a 1 and then zeros for all of the free variables. Like so. So our free variables correspond to X2, X4 and X5. So basically, I'm just going to write down the vectors E1, E2, E3, except I'm going to leave a gap when there is a spot for the dependent variable. So I left an open spot for the first and third variables. That's what we're doing there. And then when you look at the column associated to that free variable, so the second column gives us X2, look at the numbers in the pivot rows and write their inverses. So when I see a negative two, I'm going to write a positive two. When I see a zero, I'm going to write a zero. Then for the next one, you look at the fourth column right there and we're going to write the opposite of the numbers we see in the pivot rows. We get a negative nine and we get a positive three. And then lastly, you're going to look at the coefficients there and write the opposite. You're going to get 16 for X1 and you're going to get negative five for X3, which was the exact same three vectors we did earlier. But you can see that we are able to do that without having to show all of that stuff. That's perfectly acceptable on this one. You can get it from the RREF pretty quickly. How about the row space? How does one compute a basis for the row space of A? This is actually, in fact, even easier. When you have a matrix right here, it turns out that the pivot rows of an echelon form, particularly the RREF, give us a basis for the row space. So I'm just going to record these vectors right here. So we're going to get the vector one, negative two, zero, nine and negative 16. I'm going to deliberately write them as if they were rows because it is the row space. It makes sense to write them as rows. If you wrote them as columns, that would be just fine. And then the second pivot, we're going to get zero, zero, one, negative three and five. So to find a basis for the row space, you just grab the pivot rows of the RREF. Now, hypothetically, let's say this question asks us to find a basis for the column space. If you want to find a basis for the column space, what you do is you recognize where are the pivot columns in the RREF? That was the first and third. That then means that the first and third columns of the original matrix A form a basis for this column space right here. So the column space is actually the span of the vectors one, two, three, and then the other vector three, seven, eight. So this one didn't actually call for the column space basis, but a variant of this question does ask for the column space. So I just want to show you how would you compute a basis for the column space if it were asked of us?