 In this video, we will provide the solution to question number nine from the practice exam number three for math 2270 for which we're given a linear transformation, let T be a map from R2 to R3 given by the formula T of x1 x2 is x1 plus 4x2 for the first coordinate, 2x1 plus 5x2 for the second coordinate, and 3x1 plus 6x2 for the third coordinate. We're given a basis for R2B, which is given as the vectors 1 and 2, and then also 3 negative 1. We're given a basis for R3, a non-standard basis of which we get 1, 2, 3, 1, 0, negative 1, and 0, negative 1, 2. And then we're asked to compute the matrix representation of the linear transformation C, T, B. So we're going to represent the domain using B coordinates and we're going to represent the co-domain using C coordinates. Now this is not the standard matrix representation. This is the matrix representation with respect to C and B coordinates here. So recall that the formula that we need to do here, let's call this matrix A that we're trying to compute right here. So A is going to be given by the formula, which we're going to compute the image of B1 within, we put it inside of C coordinates. We're then going to take the image of B2 and put that also into C coordinates. And now if basis B was bigger, where of course B here is the basis for the domain. If we had more, we could keep on going here, but that's going to be sufficient for this one right here. So we need to find the C coordinates for T of B1 and T of B2, which what are those vectors? Let's stick B1 and B2 into the transformation. So you'll notice if we take the vector B1, which is 1 and 2, we plug it into the formula, we're going to get 1 plus 8, which is 9. That's the first one there. And then we're going to stick in for the next one, we're going to get 2 plus 10, which is 12. And then for the third coordinate, we're going to get 3 plus 12, which is 15. So we need to find this vector in C coordinates. And then for the next one, we're going to stick in B2, for which we're going to get 3 minus 4, which is negative 1. Then we'll stick this 3 negative 1 to the second corner, we're going to get 6 minus 5, which is 1. And then for the last one, we're going to stick it in, we're going to get 9 minus 6, which is 3. So this right here, we need to put this into C coordinates, like so. For which in order to then compute these, to find these coordinate vectors, what we're going to do is we're going to take the matrix, which we take C, augment these vectors we just found a moment ago. So we're going to take 1, 2, 3, 1, 0, negative 1, 0, negative 1, 2, augment that with 9, 12 and 15, and then negative 1, 1 and 3, like so. So we didn't need to now re-reduce this matrix, for which this is a short response question, if you use your calculator to re-reduce it that would be perfectly acceptable. I'm just going to do the details right here, just to show you everything that goes on, although these steps are not necessary for the calculation. For which I want to get pivots below the 1 there in the first column. So we're going to take row 2 minus 2 times row 1, and we're going to take row 3 minus 3 times row 1. So we get a minus 2, minus 2, 0, minus 18, plus 2 for the second row. For the third row, we're going to get a minus 3, minus 3, 0, we're going to get minus 27, and then we're going to get plus 3. For which then the next matrix in consideration here, we're going to get 1, 0, 0, that's the first column. Then the second column, well I'm just going to write down the rows here, we get 1, 0, bring that down, 9, negative 1. So then for the second row, you get 2 minus 2, which is 0. You get 0 minus 2, which is going to be a negative 2, negative 1 there. You're going to get 12 minus 18, which is a negative 6, and then 1 plus 2, which is a 3. For the third row, we got 0, negative 4, 2. We're going to get 15 minus 27, which that's going to give us, what do we get there? We're going to get negative 12, and then lastly, we get 3 plus 3, which is 6. That's the first step, moving our pivot position now to the second row there. I want to get rid of the negative 4 below, so I'm going to take row 3. We're just going to subtract from it 2 times row 2, so we get a positive 4, positive 2 right there. We're then going to get a positive 12, and then lastly, we're going to get a negative 6, like so. For which then you can see what happens with that row. So leaving the first row alone, 1, 1, 0, 9, and negative 1. Leaving the second row alone, we get 0, negative 2, negative 1, negative 6 and 3. Then the third row you see turns all into 0s, except for the 3, 3 position for which we then get 2 plus 2, which is 4. We're going to get negative 12 plus 12, which is 0, and 6 minus 6, which is 0 as well. Now when you look at that last row, there's no concern going on there whatsoever, the fact that you have a 4 and a bunch of 0s everywhere else. This would be saying something like 4x3 equals 0. There is a solution to that equation. It's not a contradiction. Just be x equals 3 is 0 there. No big deal. The next thing I want to do is I would then take the third row, excuse me, and I'm going to scale it by 1 fourth. So we're going to replace that with a 1, and then we can get rid of the number above it pretty easily because you have these 0s everywhere. It's not going to affect anything. So we're going to take row 2, add to it row 3. This is going to give us a 0 right there. It won't affect anything else. That's why I'm doing all these raw operations at once. Then the next thing to do is you move your pivot position back to the negative 2 right there, for which then we need to divide everything by negative 2. So we're going to get negative 1 half row 2, for which again, that would then replace that with a positive 1. We're going to get here a positive 3, and then this number right here becomes a negative 3 halves. And then lastly, to get rid of the 1 that's above it, we're going to take row 1, subtract from a row 2, for which case that'll go to 0. We didn't change anything. There we get 0. You get 9 minus 3, which is going to be 6. And then this last one is the one that's going to give us a little bit of concern because of the fractions, right? This right here needs to become negative 1 plus 3 halves. So that's just a 1 half, nothing too complicated there, but still, you know, fractions sometimes confuse us. In which case then our matrix, when we write it in this RREF, we end up with 1, 0, 0, 0, 1, 0, 0, 1. So the basis should have reduced to be the identity matrix. Now the numbers we really care about, we're going to get 6, 1 half, 3, negative 3 halves, and then 0, 0. In which case then our final answer is going to be this matrix right here. This matrix is our matrix A. You do need to indicate that this is the matrix. If you just stop with this augmented matrix, then you won't get full credit because you haven't indicated that what the answer is to the question. You should not leave it up to the inference of the grader. You need to specify yourself. In which case then we record down this change of basis matrix, which was 6, 1 half, 3, negative 3 halves, 0 and 0. This is not the change of basis matrix. Excuse me. This is the matrix representation using both B and C coordinates for this linear transformation.