 In this video, we will present the solution to question number six from the practice and interim exam number two for math 2270. We're given a three by three matrix A right here, one, zero, negative two, negative three, one, four, two, negative three, four. And we're given the sequence of row operations that reduces A to the identity I3 in this situation. We're asked then to compute the LU factorization of A. So the LU factorization is going to be two matrices, L is going to be three by three and U will also be three by three. U will always have the same shape that A does. So since A is three by three, U is going to be three by three. Now, L is always a square matrix. And so its dimensions will always be the same as the number of rows in A. So this would also be three by three. And so this is our matrix L and this is our matrix U. Now to find U, it turns out this sequence of row operations gives us way more information than we need. To find U, we have to grab the first echelon form we find in this sequence. So A is not an echelon form. The first echelon form is going to be right here. This is echelon form. It's not row reduced echelon form, but that's fine. To get the LU factorization, we don't do any row operations except for forward phase replacements. No interchange, no scaling, no backwards phase replacements. Just forward phase, lower triangular matrices. That's all we want there. And so then we can just copy down U. So there's no computation there. We just have to find U, negative two, zero one, negative two, zero zero two. And then L is always going to be a unit lower triangular matrix. So we get ones along the diagonals, zeros above the diagonals. And so now to figure out the remaining terms in U right here. So we have these numbers to determine. We go back and look at the row operations in place. We have to see the replacements. So with our first pivot right here, how do we go from here to here? Well, we look like we took row two and we add it to a three times row one. And so that tells me I'm going to write the inverse number here. So we're going to get a negative three in the box right there. Then the next one, we still have our pivot in the one one spot. We got rid of this two right here. How are you going to get rid of the two? We're going to take row three. It's a track from a two times row one. And so we're going to put a positive two right here. And then lastly, to get into echelon form, we have our pivot in the two position. We got to get rid of this negative three. And to do that, we're going to take row three, add to that three times row one, row two, excuse me. So in the three two position, we're going to put a negative three. And so that would then, this would then give us L, this then gives us U, and we have the LU factorization.