Hi there in this occasion. We're going to work in the vector space of poor animals of utmost degree one. That means the space P one. We're going to consider to vacuum to basis for this space. The basis be that's going to be generated by the polynomial P. One and P two that you can find here. And the basis be prime. That is going to be composed by the point of Mills Q one and Q two that are defined below. We need to find first. The transition matrix from the basis be primed to be. What is the procedure? Will we need to consider next and extended matrix or first? We're going to put the new basis in this case the bases beat. And in the extended part we're going to put what corresponds to the old vases in this case the prime. Okay. The idea here is that This we need to write this point on those in the standard basis for P one. That means that in this case P one that is equal to six plus three X. Well corresponds to the picture 6, 3 In the standard basis for P one. Similarly, P two will be the victor. That is the .9 of 10 plus two X. Is the picture 10 to The victim. The point on your Q one in the more space here. Q one. Well, that is equal to two Is the vector just 2. 0. And cute too. That is a polynomial three plus two. X Will correspond to the Vector 3. 2. Okay. And we're going to use this back to representations in the basis. This coordinate factors to construct this extended matrix, that I define it here. So the new basis will corresponds to be. So we need to put here 6, 3 and 10 to in the extended part we need to put the vectors to zero on 32. Okay, so we have now or extended matrix. And what we need to do is reduce this matrix to the asian form in such a way that you obtain in the left part, the identity matrix. And in the extended part you will obtain the transition matrix from the old basis to the new basis in this case, from the prime to be. So that's what we're going to do. We're going to reduce this this matrix to the action form and to do that first we need to put a number one in the 11 entry of this matrix. That Can be achieved easily but multiplying the first row by 1/6. So one five thirds, one third on one half 3- zero and 2. Now we need to put a zero in the entry 1, 2 of these matrix. That means that we need to assign to the 2nd row. The second row minus 23 times the first row. So the first row will remain the same. And here we're going to have a zero, Then here -3 -1 and one Health. The next step will be put one in this position. That can be done easily. But my multiplying the second row by -1 3rd. So our matrix will become 110 here, five thirds, one third, one health one third And -16. Finally, we need to put zero in this position of the matrix and that can be done by multiplying the first row by taking the first road and obstruct him 5/3 times this second row. And now we have here on the left part, the identity matrix, that is what we wanted and in the right part we will obtain the transition matrix. That in this case is it was to to over nine, 7/9, 1/3 and -1 6th. So this is our transition matrix. We're going to regret this as from B prime to be. Yes. The transition matrix from Victory prime we're going to take Is a common factor. Why not? 19. So that we obtain here More writable matrix that these -2 seven three and I'm in here. Well, let me check yes minus three health's Or well, I think that this better 1/18. And in that way we will obtain the matrix here minus four, 14, 6 and -3. They think that this looks better. Okay, so this is our transition matrix from the basis. Be primed to be for the next part of this exercise we need to calculate now the transition matrix from B to B prime, but that can be obtained easily by taking the members of the previous matrix. That means the members from the transition matrix from from B prime to be will be this transition matrix from the basis B two B prime. So we just need to take the members of this matrix 1/18 times minus 46, 14 minus three. The members of this matrix, what's going to be that members will? The members is equal two, one over four times three, 14 six on four. And this will be the transition matrix from B to B prime. Then For the next part we're going to consider the following polynomial. The polynomial is equal to -4 plus x. We need to grate this in a coordinate factor in the basis beat. So what's going to be? Well, first, Well, to to to obtain this, we're going to write this first as a coordinate vector in the standard basis of the polynomial. That means the vector -4, 1. And we need to create this vector as a linear combination of the element basis in B. That means this is equals to all for one times the first element in the basis B. That is the vector 63 Plus α two times the vector 10 to when it to solve For Alpha one. Enough for two and we will obtain that of one is equal to one And that Alpha two is equals 2 -1. So the coordinate lecture of this point of melty in the basis B Will be the Victor 1 -1. And in order to write this this polynomial in the basis be prime. We need to multiply the previous vector by the maid transition matrix from B to b prime. Okay. Times deep coordinate lecture pete in the base is big. So these matrix was equal to 1 4th times three 14 64. And times in this case by the picture one minus one. That is the this polynomial in the basis be the coordinate vector of of the paranormal in the basement. Okay so we just need to multiply these pictures this picture by this matrix and we obtain that the polynomial the recording intellectual of the polynomial in the basis. Be prime as equals two. 1 4th times the vector minus 11 and two. And of course we need we can verify this directly. This is using the uh the transition matrix from the basis B two B prime. But we can do we can perform this directly. So let's regret. People was equal to minus four plus X. And the element basis in big prime. We're the we're the plan our meals too On 3-plus 2 x. Okay so again we can represent this as vectors to make the things easier we have here for one and we need to grate as a linear combination of the element of the element basis in the prime. That means you're all for one Times A Factory to zero. Let's put it be better. Better one plus B to to. And the victor here. 3, 2. And the solution here for B to one is minus 11/4 and forbidden to east one half. That implies that the the coordinate vector for P. In the basis be prime is the vector -11/4 And one half. Or what we obtain here, 1 4th Times The Vector -11. And here, too, just what we obtained in the previous part.