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3) Determine the matrixP =3 2is the transition matrir from what basis B to the basis B' = {(1,1,1), (1,1,0), (1,0,0)} for R: ?...

Question

3) Determine the matrixP =3 2is the transition matrir from what basis B to the basis B' = {(1,1,1), (1,1,0), (1,0,0)} for R: ?

3) Determine the matrix P = 3 2 is the transition matrir from what basis B to the basis B' = {(1,1,1), (1,1,0), (1,0,0)} for R: ?



Answers

The matrix $$ P=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 3 & 2 \\ 0 & 1 & 1 \end{array}\right] $$ is the transition matrix from what basis $B$ to the basis \{(1,1,1),(1,1,0),(1,0,0)\} for $R^{3} ?$

Hello there. Okay, so here we have the phone matrix speed given by 11010 to 9021. Okay. The point is that this let's suppose that this p is the transition matrix from some basis, B to the standard basis. So what's going to be these bases? Okay, so if we are in this scenario, let's remember how we can construct this kind of matrices. Well, if we're going to construct using using the the standard procedure here, we need to put the new basis and here on the right, we need the old basis. In this case the new basis will be the standard matrix. That is already the identity. Okay. All right. 2111. So here the new basis is already the identity. So what is on the right is already these these transition matrix from bits. And this is it the case? That is our matrix speed one zero on 02 & 0: one. Okay, so here we have already the transition matrix from P. But not is not only that but this is also the old basis in the extent the extended matrix, that means that the basis B will be composed by the victors. We need to pick the columns of this matrix. So the first Vector will be 110. The second vector will be one 02. And the third vector will be 02. What? Then I'm going to go up here again. The matrix speed 10-0-1. Okay, so now what's going to what happened if p is the transition matrix from the standard basis to the basis? B wow, it's going to be the base is beef. Okay, so here we're going to assume that P corresponds to this transition matrix from the standard basis to be So first. What? So in this case what we need to do is first attained what's going to be the transition matrix from businesses order to find the And that means taking the members of pain or the members of these matrix. And that is the matrix, 1/50 times 4. -2. 1 -1, 2 And -2: one. Okay, so this is the inverse matrix. Okay, And this is the matrix the transition matrix from the basis B to the basis. S as is the standard matrix. So that means that the basis B will be just the columns of this matrix to the basis in this case will be there will be the vectors 1/5 times 4, -2. 1 50 time of the vector, 1 -1, 2 and 1/5 Times The Victor -2: one.

Hello there. Okay, so in this exercise we have two bases the basis. B define it here by these two vectors to one minute four on the basis. S which corresponds to the standard base? No, these two are our bases in there too, of course. And the first thing that we needed to find these transition matrix from the basis beat to the basis. S This can be done easily because basically the new the new basis will correspond to the standard basis. So in that case This transition matrix from B two s. It's just a matrix that the columns are just pictures of the basis, peak 2 to 1 minus three. The reason behind this is the the standard procedure to obtain these decision matrix, is that you need to construct extended matrix? Well, first we're going to put the new basis and in the extended part you're going to put the old ways one that you have constructed these extended matrix, you need to perform some row operations in order to reduce it to the asian form. So that in the left part you're going to the identity matrix. And in the right part you will obtain the transition matrix from the old basis To the new one. So as you can observe in this case, the new basis corresponds to the standard basis. And this is already the identity matrix on the left part. And on the right part which corresponds to the old basis, we're going to put the vectors from the basis be That is just what we have done here. So whenever you have this kind of transition matrices From some arbitrary basis to the standard one then the only thing that you need to do is build the vectors as the columns of the matrix. The next thing that we need to do is to find the transition matrix from the standard basis to the basis. So this is kind of different in this case we need to do the whole procedure. So first we're going to put the the basis beef. So that means here bring the you're 21 minus or and here on the right, we're going to put the standard basis. So that means the identity matrix. Okay, so then we need to perform some role operations to this. And that means First we need to put the number one in this first entry of the matrix. And that means Dividing the Freeze Row by two. And in that case we obtained here one minus. Call how zero on the other really be the same. Okay, the next step will be putting a zero below the first one. So to do that we just need to assign to the second row. This obstruction of the second row with the first row and we obtained here juan minus three hall how zero and here zero 11. Health -1/2 and one. Then we need to transform this 11. How? 2, 1. So that means we need to multiply the second row by two, divided the left So that we obtain here one minus three health one. How was it this part will 01 -1/11. I'm too over finally we need to eliminate where we need to put a zero in the position of the meeting. And that means to Assigned to the 1st Row. The first road lost three health times the second row. And then finally here on the left part. These extended matrix, we obtain the identity matrix. And in this right part we're going to the transition matrix that we need. That is for 8 11 three, 11 -1 divided 11. And to really and of course we can rewrite this better way. So the transition matrix from the basis as to the basis will be one over 11 times the matrix forward. Three mine is one great. The next will be showing that these to transition matrix B from from us to be. And the transition matrix from the basis to ask Our member says of one another. And in order to do that, the only thing that we need to do is multiply two matrices and show that they are the identity. So what we need to do is multiply these matrix with the other matrix that we obtain it at the beginning. Okay, so this first matrix when people deal with that color. So this this first matrix is the one that we just competed in the previous part. So it's 1/11 or three minus 12. And the other matrix is the one that we obtain it at the beginning. That is just putting the vectors of the basis be as columns of the matrix. So to one minus 34. And then we just need to multiply them. Okay, so we have one over 11. The kind of factor and then the winding to multiply the So there's as equals to eight those three The next term will be -12. I was 12. The next term will be -2 plus And the last one. Yes. Three. So you can observe that the off diagonal terms are going to cancel out. So we're going to have zero in both parts. And here eight plus 3 is an A plus three. Of course. Mhm. And therefore here we obtain the identity matrix. So that means that these two matrices Are the member says of one another. The next step is exercise. Is to now consider the following vector. The flowing vector is 5 -3. It's important that this vector is already in the standard basis of are too So you need to keep that in. Okay, basically the the vector in this thunder base. Okay, so first we need to write this factor in the basis being so to do this. Usually what you need to do is take the picture here. That is where in the standard basis of the space that you're working these cases are you? This vector life and minus. And you need two. Great. This as a linear combination of the elements in the basis b. Okay, so let's say that the basis B Is composed by the Vectors B one and B two. I'm not great in the victors because we're not using this procedure. I want to show you what's the standard procedure reading. Okay. So happened that you need to find too ah Coefficients of for one and over two that multiplies these elements of the basis. B one and B two. Such that they are equals to the vector in the standard based. Okay. And this implies that your picture in the basis B Will be Alpha one, Alpha 2. The basis. Okay, that's the standard procedure. However, in this case we can avoid this to doing this because basically, as you can observe this victory is in the standard basis and we have a transformation matrix from the standard basis. The basis B. So we can use this. So the pictures of you in the basis B will be equal two. The transition matrix from the basis. As to the basis B. In the vector W. In the basis. S That is just the The victory in the standard basis of five. Okay, so these matrix is the one that we obtain it. The last one That we obtain it is one over 11 or 3 -1 and two. And times the vector W. That in this case, In the Basis S is 5 -3. Just in the standard base. And the result of this is the vector one minus it. The basis. The next one is to find the vector W in the basis as producing the transition meat. So in this case we need to use the other metrics. The vector W in the basis, as will be the transition matrix from the basis B to the basis. As times the vector W in the basis B. This matrix was 21 minus 34. And here Times A Picture, one -1. That is the one that we obtained in the previous part. The base is B. And the results this matrix multiplication Is the Victor three minus in the standard five minus three. Five minus. Okay. And then we need to do something similar. No, with a vector post three means in this case the first thing that we need to do is to find the vector job in the standard basis, but we have said before this is a red in the standard basis of this vector, in the standard basis is just three. This and then we need to find the vector in the basis. Be using the transition made. So this will be equals to using the transition matrix from the standard basis to the basis. B times the vector W In the standard, This matrix was one over 11 time is a matric 4, -1, 2, nine times here. The Vector three lines. This is equal to one over 11 minus 1/11 times the victor, three, 13.

Hello there. Okay. So for this exercise we have these two basis of our three. So the first basis is B. But this composed with these vectors 1-1 250 and 338. And then there is a basis S which corresponds to the standard basis for our three. We need to find the transition matrix from B. Two S. Okay, so this can be performed really easy because whenever you have and arbitrary basis in the transition matrix here but you're going to change from an arbitrary basis to the standard basis. Then that transition matrix corresponds to put the vectors from the basis B as the columns of the meeting. Mm In that sense we obtain the following metrics. 21 what? 1, 2, 1, 250 and three 38. The reason behind this is that the standard procedure is to construct an extended matrix or first you're going to put the new basis. And in the extended part you're going to put the old Yeah. Then you perform some row operations in order to reduce these extended matrix to the asian reform. So that in the left part you obtain an identity matrix. And in the right part you will obtain the transition matrix from the old basis to the new. Okay. And that's what we need. But you can observe that in our case the new basis is this standard? The standard basis. So we have here already the identity midge. And the old basis will respond to put the vectors of the basis as the columns of the matrix. That is exactly what we have done here. So in in those cases the the you can obtain this immediately. Okay, just write the vectors of the basis as it comes. Great. Then we need to find the transition matrix. But now from the standard basis to the basis peak here, we need to do the whole procedure. Okay, so first let's good here on the the new basis that are the vectors from the basis B want to one 250 338. And then here the standard basis, that is just identity matrix. Okay so we need to perform some reparations first. We're going to eliminate these these two elements here, we're going to Mhm. There's some zeros. So in order to do that we're going to take the second row and we're going to sign the second row -2 times the first one. And for the third one which are we're just going to subtract the third and the first road of the matrix. So we obtain here the matrix the first road remained the same. But the last two will change the 00, 1 -3 10 here minus 25, one minus one. Then we need to in the next step we need to put zeros in this position and then this position of the matrix and that can be done by taking the first row. And so obstructing two times the second one and for the third one we're going to do almost the same. But instead of something we're going to instead of obstructing we're going to some This two times the 2nd row. So we obtained here The Matrix 109 5 -20. The second role remained the same. 01 minus three minus 210 0, -1 -5, 2 and one. Right. Then we need to change the sign of this. So to do that, we're just going to Multiply by -1. The 3rd room. Okay? So the third row we're going to multiply by -1. So here we obtain one here, 15 5 -2 and-. And the last thing to do is just eliminating these two numbers here. Just put in zero and disposition. So for the first one we need to take the first row and obstructing nine times the third row. For the second one we need to do the same. So 2nd row plus three times good. And then we obtain finally here on the left part, the alien deity matrix. And on the right part we obtained the transition matrix. That is the one that we need. So that is -40 here, 16 nine, 13 minus five minus three, five minus two. And minus. And that this is the transition matrix that we need. So let me look right here will be transition matrix from the basis as to the basis B is just this matrix here copy and great. Now for the next part, I'm going to copy here again. The matrix from the basis B to the basis. Yes, that is composed by the vectors In the basis be so 1-1, 250 and 33. Okay, so now in this part we need to show that these two matrices from it. S I'm going to change to define communication bit to us. The prime going to be the transition matrix from us to be these two. Our member says we need to show that they are in versus of one another. Okay, so how to show this? Well, basically if they are embassies of one another, the multiplication of these two matrices will give us the identity matrix. Okay, and we can observe that if we multiply these two matrices here minus 40 16 9, 13 -5 -3. Bye -2 -1. Yeah, I was 14 fine. I mean minus five from the Yeah. Been killed here is on two three three. 108 times the matrix -40-16-9, 18 minus My name is three. 5 -2 -1. If you multiply these two matrices, it did you obtain the identity matrix three basically. So with this you show that they are in verses of one another. That Okay. In the next part We need to consider the following factors. So the vector w is fine by five minus three here, we need to remark something. And is that whenever you have this made this some arbitrary vector This given in the standard basis of our three. Okay, so that means that the gov the vector W in the basis s will be just the picture well. And For two great in the annotation. So 5 -3. 1 does the same. That Okay. And that's going to help us a lot because the next thing that we need to do is we'll write this vector W. But in the basis. B. And we have already here the vector in the basis. S and even more, we have a transition matrix from us to be So we're going to use these matrix in order to obtain these factories. So this vector will be equal to the transition matrix from the basis as to the basis B times the victor. W In the basis, this vector this this matrix here was minus 40 16 nine. 13 -5 -3. My minus two minus one. And then we need to multiply of course this Bye vector five minus three. And um the result of this matrix multiplication by a picture is equal to the victor -239, 77 and 30. So there's the victory to all of you. But in the basis. Great. Now for the next part we need to write the backyard W. But in the basis. S. And for that we're going to use the other transition that means the transition matrix from the basis, B to the basis. S. And here of course the lecture W. But in the basis. So these matrix course one one 250 three 38 times the victor -203. 77. There there's multiplication result into the victor, bye -3 and one in the basis as others. Great. And now we need to repeat this procedure. But now with the specter that is three -50. Again, this vector is given already in the basis in the standard basis. So the first thing that we need to do is write this vector in the standard basis. But because it is already written in the standard basis, we just need to copy the vector and now we need to transform it two. The vector in the basis beat. And to the days we're going to use the transition matrix, the transition matrix from the basis. As to the basis B. Times the vector W. In the basis. Yes. Okay, this matrix equals two -40-16-9, 13 -5 -3. 5 -2 -1. And the victor is three minus five. The result of this is the vector minus 200 64. In the basis. The

Yeah. Okay, so in this case we have considered that we have the flowing basis for some vector space, let's say the prime B and C are the basis for some vectors. And we have that the transition matrix from the basis be primed to be is the matrix B The transition matrix from the basis beat to the basis, say is the matrix cube. So the first question, it's what's going to be the transformation matrix from the basis? Be prime to the basis seat. So basically you can observe that from this. Is this this basis B? We can take the transformation matrix Q. And we're going to send here to the basis see so we can define just want transformation matrix from the prime to see. That will be given by multiplying 1st B times the matrix Q. Okay, so let's say that here this matrix, the transition matrix from big prime to see will be multiplying the matrix B times the matrix. Yeah. Now what happened? We want the inverse. That means from the basis see to the basis be prime then what's going to be this transfer transfer transition matrix? Well, first we're going to need here this same diagram and going to raise this because we're not going to need it. So the first so we have already define this transition matrix from the prime to see. But now here we need to map from Z to be Okay and that will be taking the embers of cuba. And then we need to take the members of B to B from b to b prime. Okay, so this inverse transformation that's define it here, from B from set to be will be defined by first taking cute in verse times P inverse. Okay, so this transition matrix from the basis see to the basis the prime be cute reimbursed times P inverse.


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