 In this video, we'll present the solution to question number 14 from practice exam number two for math 2270. We're given two bases, B and C as bases of R3, and we're asked to compute the change of basis matrix P changing from, changing to B from C. So that's what we're going to do first. Then we have to compute the coordinate vector of X with respect to B if we know the coordinate vector with respect to C. We're going to have to show all of our steps. There's going to be some real operations in play here. So to compute the change of basis matrix, we're going to augment the set B with the set C and make sure you go in this order, right? You go to B from C. So this will look like we plug in the vectors 113, 142, 216. So we just take the vectors from B and then they take the vectors from C, 101, 1-30 and 212. We need to reduce these, but show all of our steps here. So for the first pivot position, we have a one right there. We've got to get rid of the numbers below it. We're going to take row two minus row one. We're also going to take row three minus three times row one. That'll look like minus one, minus one, minus two, minus one, minus one, minus two. For the third row, we're going to get minus three, minus three, minus six, minus three, minus three, minus six. Like so. And so then recording what we discovered from those row reductions, the first row stays the same, 112, 112. Then for the next row, we're going to get 0, 3, negative one, negative one, negative four and negative one. For the third row, we end up with zero, negative one, zero, that's a zero there, negative two, negative three and negative four. All right, so with that in mind, notice my new pivot position comes to two spot right here. I love a one there, I could divide by three, that give me fractions I don't want. I can notice that hey, this row right here, there's a negative one right there. I could swap the rows around and multiply the new second row by two. If I kind of did that at the same time, that would actually put me in a pretty nice situation for my pivot position, I'd get a one there. So don't do anything to row one, so you get 112, 112. I'm going to take the old row three and put it on top, but I'm also going to times everything by negative one. So you get zero, one, zero, two, three, and four, which is nice. Everything in that row is negative. And then the old row two becomes row three, zero, three, negative one, negative one, negative four, negative one. It's never a good idea to do too many row operations at once, but if you label it, I think we can handle it. So now with this pivot position right here, we've got to give the three that's below it. So we're going to take row three minus three times row two, minus three, minus six, minus nine, minus 12. So the first row stays the same, it's just along for the ride here. Second row is also going to stay the same this time, one, or zero, one, zero, two, three, and four. For the third row, we end up with zero, zero, negative one. We get negative seven, negative 13, and negative 13, like so. So for this pivot position right here, again, I'm going to do a double whammy on this one, I'm going to times negative one by the third row. I'm also going to take the current third row, and I'm going to take row one and add to it two times the current row three. So that would look like minus two, minus 14, minus 26, and then minus 26 again, like so. So what does my matrix look like in that situation? Draw it over here. So we get one, one, zero, augment that with negative 13, negative 25, and then negative 24. For the second row, we didn't do anything to that, so we get zero, one, zero, two, three, four, and for the third row, we times everything by negative one, remember. So we get zero, zero, one, and then seven, 13, and 13. And so the last thing to do is looking at the pivot position again into two, two spot, we got to get rid of the one above it. So we're going to take row one subtract from it row two. So we get a minus one right there. We're going to get a minus two, minus three, and a minus four. So in the end, this would reduce to be one, zero, zero, zero, one, zero, zero, zero, one. That's the identity. That's what we want right there. Then you're going to get negative 15, negative 28, negative 28 for the first row, two, three, four for the second row, and then seven, 13, 13 for the last row. Now this matrix right here is the change of basis matrix, but don't leave that up for me to guess. You should be specific about it. So the change of basis matrix that we changed to be from C. This is the matrix, negative 15, negative 28, negative 28, two, three, and four, and then seven, oops, seven, 13, and 13. So that's the first part of the problem. The second part of the problem is we then had to compute the change of the change of coordinates there, right? So we had the vector X, C, if you multiply it by the change of basis matrix, that'll convert it into B coordinates, which is what we're looking for. So let's multiply this matrix, negative 15, negative 28, negative 28, two, three, four, seven, 13, 13. Let's multiply it by the coordinate vector, which the coordinate vector with the coordinate to C coordinates was one, negative one, one, remember. And so in that situation, you're going to end up with negative 15, plus 28, minus 28. Those will get each other. You're going to get two, minus three, plus four, and then you're going to get seven, minus 13, plus 13. Those will get each other. So we end up with negative 15. We're going to have three and then seven as the B coordinate vector. And maybe write a little bit smaller so we don't wrap on to the next problem, but that solves question number 14.