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Question

Let $\mathcal{B}=\left\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\}$ and $\mathcal{C}=\left\{\mathbf{c}_{1}, \mathbf{c}_{2}\right\}$ be bases for a vector space $V,$ and suppose $\mathbf{b}_{1}=-\mathbf{c}_{1}+4 \mathbf{c}_{2}$ and $\mathbf{b}_{2}=5 \mathbf{c}_{1}-3 \mathbf{c}_{2}$
a. Find the change-of-coordinates matrix from $\mathcal{B}$ to $\mathcal{C}$ .
b. Find $[\mathbf{x}]_{\mathcal{C}}$ for $\mathbf{x}=5 \mathbf{b}_{1}+3 \mathbf{b}_{2}$

Runpeng Li · Accepted Answer

call them too. So we need to find Thea Person Need to find the change of courting the matrix from B to C and your work even be one is elective C one us or I&#x27;ll see to it. Sorry, just or I&#x27;ll see to. And you too is five C one minus three. Upset you. So by over your own given in the 15 given in the chapter 4.7 we I know that the Matrix changeup quote in the Matrix is you want record C and B two bracket See? And this is my ovary questions here. Active one and four and five and negative three. So this is our change of coordinates matrix. So the next part I need to find ex breakfast, see two giving question. There&#x27;s X equals five, one less three. You too. So the way to do that is to notice, except C is Come, I don&#x27;t need to see times. The vector here is a five three. So this will be our previous meetings. Negative one. Why? For negative Three times I three, which iss 10. He left

Runpeng Li · Answer

So in this problem, we&#x27;re given three questions as one ISS two of the one and minus you too. Minus d two plus G three and two iss three up. D too. Us the three and at three iss negative three. Do you want plus two of the three. So by these three questions, we can immediately right down the change of coordinated milkshakes. That is from after D. So this gives two to negative one one and zero 31 and lack of 30 and two. So this is our Beatrix from back to t change of corn images from that day at that stops today are eight. So for a part B, are we We need to find x 50 for Coolidge X is F one minus two, uh, to us three three. So the way to do that, it&#x27;s just Ah, so that&#x27;s, uh, already down. Ex expected d first. So we used our change of Courtney and Matrix here, so that&#x27;s XY equals P. You&#x27;re mad to be times one negative two and three. So we have how write it down in the next page. So that is 20 active. Three next one three and zero and 112 times one negative, too. So during the Matrix multiplication, the first entry two times one class zero minus nine. So that&#x27;s selective. Seven. Or the second entry connecting one times one active, one minus six. So think of seven. And that&#x27;s zero Still negative. Seven. Last entry, one times one and negative. Two times. One that&#x27;s connected to So we have an active one and plus two times three at 65 So this is our ex. Break it. Hey!

Runpeng Li · Answer

Okay, so for problem to we have the basis di at his d one and you too, and are so basis be, which is B one B two. So we we need to find out the matrix t relative to D and B. So by the giving permission ti of the one is to be one minus three b two and t of d to its inactive for B one plus five b two. So by using that relation by observing their coefficient, you can have a transformation off the one and we should have a deep coordinate. Sorry, the be coordinate here. So the first thing that the 100 people made by our first equation it&#x27;s okay to be one Sorry. Step two. We should be two and three and would be other. You too. And under ordinate b, this will be by our second equation, like a born by now the meet our metrics t I would just be the span of these two vectors to elective three next four and five

Runpeng Li · Answer

problem. Eight. So we first right down the vectors. The one is connective. 18 He too is one negative five and be three iss. Sorry. I see one C one ISS 14 and see to is one. So we let&#x27;s first find the, uh, change of coded matrix from me to see. So in order to find the change according the Matrix from B to C, we put column vector See, on the right hand side 1411 and I call Alexis would be electable. 18 one, activate and goodbye. So we need to turn the left inside to be the ad in the Matrix. So excuse me. We performed the guttering ammunition. First thing we do is too used a second role, minus four times out. First role. So 11 if you want one. Four times. Four times four. You know, one time, one minus four would be negative. Three, eight miners. Negative for So that is 12 and negative. Five minus four. That&#x27;s a negative. Naughty. Okay, so next we used the first roll. Plus three. Happy about the second roll. We can do the three having a second row purse because we to turn, um, this entry to be one. So Israel, we&#x27;ll keep the same 11 and negative one and one negative one. And the one and zero Nothing. One, four and negative three. So we add first row by the second rule. So that would be one, you know, And three and negative too. And zero, we divided. Divide both or multiply both sides by tactical one or two by a second row bite into one. That&#x27;s one before. And so our change of corny matrix from you to see is the right hand part 32 naked before and three. All right, so to find the numbers, C two b. That&#x27;s, uh that&#x27;s to say, um, we need to find it Inverse off this matrix greed and to do next for Nick on three. So do this naked, too, Before and three and 100 one. So first thing I want to do is too. Divide the first roll by three. This one Now the negative to over three and 1/3 and zero. And we are So, um, have second role next, four, three, zero and one. So that is should be, uh, a narrow here. So we divide the second row by four. So we have next one new quarter and zero on 1/4. So one night get to third and 1/3 and zero. Okay, so So we add the first roll by the second roll awake, we add the second anniversary. So that&#x27;s zero and 3/4 3/4 minus to third. This will be nine divided by 12 minus. So that&#x27;s eight, if I did by 12. So that&#x27;s 1 12 So 12 right on DDE, we still have 1/3 and 1/4 and first role is one negative to over three. 1/3 and zero. So as first turn turned the second entry, that&#x27;s money. Sorry. It&#x27;s just multiply the second roll by 12. So we have 01 and for and three and one to over three and 1/3 and zero. So we we use the first roll had negative, too at 2 32 3rd off the second row. So we have one zero 01 and 1/3. So that&#x27;s so 1/3 plus to third up. Four. So that&#x27;s eight there. Yeah, so that&#x27;s 1/3 and we have three. Oh, it should be on the first rule. We have three and then we have say to third, all three will be too. And that by a zero said, That&#x27;s too. We have 0143 So this is our change of holding the Matrix and from Did you see? And that is 32 or three.

Runpeng Li · Answer

get. So in this problem, we present you to find the changeup Coordinate matrix from you did see so you to the scene. And this will be this can be achieved by first food. The column vectors of be on right heart, its matrix seven nectar to two active one and column column vectors of sea or one. And why two? So the way to do that we do the, um, reduction. So we first divided first roll by poor 5/4 7/4 and to quitters. And second of all, would still be one too negative. Two and one next one. Okay, so this gives. So we used the second row diva minus the first roll. We have you to minus 4/4. That is 3/4 and 7/4 minus 7/4 minus two. That is 1/4. Well, yes. So he&#x27;s actually we used one minus one and two minus five over four. That&#x27;s 3/4. And next to minus 7/4. That is negative. 15 over four. Right, Because selective to minus. Just for 60 minus 7/4 that&#x27;s Ah, eight rito plus seven. I could sit 15 over four. Yeah, and the last one is 91 minus two or four. So that is. See, um, keep six. 04 And we still have the first roll. Libel before seven over four and tour. Okay, now we first to turn the second row. We multiply second row by for over three so we can turn this entry to be one. So we have a life. Local traitors, 7/4 two porters and 01 and three over four. Yeah, it&#x27;s one and or third times negative. 15. What? Hers that is 5 95 and negative. Six or four times where? Oh, where Three. That is connected to. Okay, So for the first rule, look, I use 5/4 off the second reel ad. First role. Sorry. Minus first, So we can cancel the entry of the entry here. So that&#x27;s 10 and 544 times negative. Five used. Seven over four minus is value. So it&#x27;s seven over four plus 25. She&#x27;s 32 or four. So that&#x27;s eight. And we also have to over four. Finest. I&#x27;m over four times negative too. So that is 10 plus two or four. So that&#x27;s three. We have three here. 01 and active, applied negative too. So I were change. According the Matrix from B to C is eight three negative five and negative too. And the inverse C two B. It&#x27;s the inverse off the you to see and remember how we find it. Find the inverse up two by two matrix First the best A determinant. Uh, this matrix, which is? Uh um you to see times we have the changes have to entertain you The diagnosis that selective too. And eight No, we have to take the negative 23 And what? So what&#x27;s the determinant determined then? Days negative 16 plus 15. That&#x27;s negative one. So multiply by multiplied by wants us to read negative five and activate.

Ernest Castorena · Answer

I get for this question? They give us the Matrix A is equal to 000 001000 And they ask us what is that I mentioned of the null space. Uh, well, we know that this is a two by three matrix three? I think so. It&#x27;s trying to map three dimensional that there&#x27;s 23 dimensional electors, so it&#x27;s trying to map it are three. However, we know that this matrix has a rank of one is the only non zero. So the rain, uh, is one. We also know that the knowledge e of a plus the ring okay equals that I mention of the record space were trying to map Teoh, which is three police, that with one bring that over to the other side and we have the banality of it just to so that I mentioned of the North bases, too.

Problem

Let $\mathcal{U}=\left\{\mathbf{u}_{1}, \mathbf{u…

Problem 2 Easy Difficulty

Answer

A. $\underset{C \leftarrow B}{P}=\left[\begin{array}{cc}{-1} & {5} \\ {4} & {-3}\end{array}\right]$
B. $[X]_{C}=P[X]_{B}=\left[\begin{array}{cc}{-1} & {5} \\ {4} & {-3}\end{array}\right]\left[\begin{array}{l}{5} \\ {3}\end{array}\right]=\left[\begin{array}{c}{10} \\ {11}\end{array}\right]$

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Kristen K.

Samuel H.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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A. $\underset{C \leftarrow B}{P}=\left[\begin{array}{cc}{-1} & {5} \\ {4} & {-3}\end{array}\right]$B. $[X]_{C}=P[X]_{B}=\left[\begin{array}{cc}{-1} & {5} \\ {4} & {-3}\end{array}\right]\left[\begin{array}{l}{5} \\ {3}\end{array}\right]=\left[\begin{array}{c}{10} \\ {11}\end{array}\right]$

Linear Algebra and Its Applications

Kristen K.

Samuel H.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Kristen K.

Samuel H.

Michael J.

Joseph L.

A. $\underset{C \leftarrow B}{P}=\left[\begin{array}{cc}{-1} & {5} \\ {4} & {-3}\end{array}\right]$
B. $[X]_{C}=P[X]_{B}=\left[\begin{array}{cc}{-1} & {5} \\ {4} & {-3}\end{array}\right]\left[\begin{array}{l}{5} \\ {3}\end{array}\right]=\left[\begin{array}{c}{10} \\ {11}\end{array}\right]$

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications