in-exercises-9-18-construct-the-general-solution-of-mathbfxprimea-mathbfx-involving-complex-eigen-10

Question

In Exercises $9-18$ , construct the general solution of $\mathbf{x}^{\prime}=A \mathbf{x}$ involving complex eigenfunctions and then obtain the general real solution. Describe the shapes of typical trajectories. 
$$
A=\left[\begin{array}{rr}{-3} & {-9} \ {2} & {3}\end{array}ight]
$$

Ibrahima Barry · Accepted Answer

you&#x27;re asked if I d generous ex prime clues e X and whereas to draw feats poacher describe the feast of dissolution. So now you know you should get when you take it to really be minus land I is to be I notice there&#x27;s no real part No determine how the feast scorch it looks like And I get vector You should get to be my street mystery I to solution It&#x27;s gonna be even be I may be useful to zero when you use always forego times V which is my screen must be I to So we get So what was Formula toes is that he has e he I be not just equal to it would be times co sign a Finally. So you know the group. The real court. Use your we get each with zero times suicide 15 plus I sine et times a director. Maestri you No, it&#x27;s just a place to get my three co sign Creepy. My I re sign creepy or stream. I co signed Clukey and then point we I squared said it&#x27;s minus any three and second world. We have to co side plus two. I sign spoken to zero and let&#x27;s just split the real imaginary part. First solution to get my speak Ozai taking my 5 30 and to co star with a freaky three pp My plus three co sign +33 and second row We had 22 sides off. Creepy. There are general solutions. One comes this solution What&#x27;s your time each year and are you gonna be? Is he I Euro rial part of you that implied our face portrait It&#x27;s gonna be look sees around and origin

Bobby Barnes · Answer

So they want to find the general solution for this using both complex and riel numbers and then also described its trajectory. So remember, our solution is going to look something in the form of X is equal to So it be C one b one will be one is going to be our icon Vector our first Eigen vector. And then it will be e to the land of one T and then plus C two second I conductor and then e to the second Eigen value times. So this is what our general solution is gonna look like. So let&#x27;s go ahead and find our Eigen vectors first. So we&#x27;re going to do a minus slammed. I is equal to negative 7 10 negative four, uh, minus lambda, then five minus lambda. And so I remember we take the determinant of this now and then set this equal to zero because this should have been able to the zero director and doing that. It will give us negative seven minus Lambda five minus lambda I then plus 40 Is it zero? Now let&#x27;s go ahead and expand this out. So this would be negative. 35 plus seven Lambda minus five Lambda plus Lamela squared plus 40 0 to 0. And let&#x27;s think this up a little. So Lampa squared prostitute Lambda Spiv is equal to zero. Now, let&#x27;s go ahead and subtract five over because I&#x27;m going to complete the square of this. So I need to add one on each side and then over here on the left, this is Lambda Plus one squared. And this will be eat too negative or so when we take the square root on each side went up with Lambda plus one is equal to plus or minus two I and then we subtract one over. So Lambda is going to be equal to negative one Gloucester minus two I So we have both of our land is now. And actually from this, um, we can already tell what the trajectory is going to be. So when we look at this, what we&#x27;re really looking at is this value a here because remember, if we look at that e to the Lambda, well, if we plug he&#x27;s in, this is gonna be negative. One plus two high times t are just e to the negative tee times, um, e to the plus or minus. Yeah, plus reminds here, uh, to I Now remember, this thing is just going to be, like, co sign of something, plus sign of something. So we really are just looking as this here. So since e to the negative t here, this approach is zero. Then we know that this is going to be, um, a spiral that goes to the goes towards the origin. So it will be a spiral where it is attracted to the origin. And you could really just do that by looking at this a here and saying, Oh, it&#x27;s negative. What? That&#x27;s kind of like the reason as to why that happens. Okay, But now, with those we need to find our Aikin vectors. And remember, since these are complex, we only need to find one of the Eigen vectors because in the other one is just going to be It&#x27;s complex. Excuse me, It&#x27;s complex continent. So let&#x27;s go ahead and find it for the plus. First, it doesn&#x27;t really matter which one you choose. Now, let me scoop this down. All right, So we&#x27;re going to now do a minus, so I&#x27;ll do minus one plus two I times I and doing this, we should get six minus two i 10 and negative. Or then that would be six minus two y. And let me just make sure did that, right? So add 1 62. Yeah, looks good. And we don&#x27;t actually need to, like, try to do roll reductions or anything like that, because we should know that we will end up with only one Eigen vector here, Um, and we&#x27;ll zero out. Well, these have to be multiples of each other&#x27;s what I&#x27;m trying to say. So we could just set this equal to zero for one of them. So I&#x27;m just going to use this bottom one here, so it&#x27;s gonna be negative. Four x one plus six minus two I x two and this is equal to zero eso. I&#x27;ll add for X I X one over. So before x one is equal to six minus two I x two and then divide each side by four. So x one is equal to 146 minus two I x two. Um, I&#x27;m going to just tribute. The four doesn&#x27;t really don&#x27;t really need to do that, but it&#x27;ll just make the numbers a little bit prettier. I think so. That would end up being three minus I over to x two. Right, So let&#x27;s go ahead and plug this end out two So, x one x two so x one, is this three minus I over two times x two. And the next two is just staying x two so we can factor out the X two to get one in the bottom there and three minus side over to the top. And now remember, we can multiply this by any multiple we want, so I&#x27;m just gonna multiply by two. So I&#x27;m going to say our first Eigen vector is going to be three minus I to right? And again, Remember, we already know what our second Eigen vector is going to be because it should just be the complex conjugated this one. Um, actually, let me come back. So what was against the negative? One plus two, I So let me just kinda break this out. So this is when we had land one being able to negative one plus or minus or not plus or minus plus to I. And so then when Lambda is our second Lambda we have. The conjugate is just going to be for the second Eigen Vector here is going to be equal to three plus I to and you could go through and do the calculations, but you&#x27;ll see you&#x27;ll end up getting this as well. Um, and again, you might use a slightly different Eigen vector than what I have, but as long as it&#x27;s like a multiple of it, it doesn&#x27;t really matter. And it should be fine. Right? So now let&#x27;s go ahead and plug those into that solution to get the complex. So for the complex one, this is going to be so see one. So V one we had was three minus I thanks to and then e to the, um negative one plus two i t. And then plus C two three plus I to and then e to the negative one minus two i teeth. Right now, this is that complex one. And at this point, there&#x27;s two ways we can go about finding the next one because they actually give us a nice like solution for this, Um, where we could go ahead and right this in the following way. Um, so we could say so. Like X o X. I should be saying x z two. So it be like, see one. Why one plus c two. Why? To where? Why one is equal to and this is on page. I think, like 3. 16 or something in the book. It will be the rial portion of our Eigen vector. And so I&#x27;ll just use one in this case. But it doesn&#x27;t really matter which one you use co sign of Bt and then minus the imaginary of our factor here and then sign of Bt. And then this is going to be times e to the 80 and then why two will be pretty much the same thing. But now we have sign of Bt here, and this is going to be plus instead of minus and the hope that should be co sign here. Cosign of BT times E to the 18th. And remember, this is going to just be like Lambda One is equal to a plus B I So for the first one, what was our Aiken director again or value? So it is negative one plus two I so we&#x27;ll just go ahead and plug those in and to get the rial components of our first imaginary director for of our first complex Eigen Vector. We could rewrite this as 32 plus negative 10 I And so this is the rial and this is the imaginary vector. So we could just come down here and plug those in. So let&#x27;s do that. So the rial we had was just three to the imaginary is negative 10 Then we had R B value in this case is going to be, too, since we&#x27;re using the positive one. Um, so we&#x27;ll just go ahead and put two&#x27;s all here, so to to to to and then are a is going to be negative one. Oh, put negative t up there. Then we just go ahead and add these together. I&#x27;m so doing that we will end up with So X now is going to be so it BC One times why one and actually I&#x27;m gonna factor this e to the negative t all the way out. But I e to the negative t on the outside here. And then it would be see one times three to co sign of two T and then minus C one of negative one Syria sign of two T and then plus 32 times. See to his brother sign of two teeth and then plus C to negative +10 co sign of two t And we can actually go ahead and group these by the Eigen vectors that they have just to make it look a little bit prettier. Not really needed. But if you can clean something up, probably a good idea so we can match these two together. And so that&#x27;s going to give. So we&#x27;d have 32 times, Or then they&#x27;ll be C one co sign of two t uh, plus see to sign of two t. And then we can also match up the ones with negative 10 there. And so that would be plus negative +10 Um, see one. We actually need a parentheses here and actually all right, the C two first. Since we have a negative, there&#x27;s would be C two co sign of two t and then minus C one of sign of two t. Yeah, and then you could go ahead and, like, distribute, um, these so like here. But that&#x27;s really also not needed. So I&#x27;m going to go ahead and leave the solution like this. So this is going to be our riel valued solution for this Well and again just to kind of make sure I said this. We know that this is going to have a spiral trajectory because so when we look at our Eigen Factors, so this a plus B I form. So why is it to negative one plus minus two I we had where a is less than zero. So since we have both an A and B, this implies a spiral since a b or I should say both a b and it will be going towards the origin so towards origin since are a is less than zero. And again we could just think of that because, like this e to the negative t term, this will go to zero as T goes to an affinity. So sine and cosine just alternates between negative one and one. But this is going towards zero. So when we multiply that it would eventually just get closer and closer to the origin. Yeah, so those are going to be our answers

Ibrahima Barry · Answer

obstructed general solution of this Justin ex prime. Because there so value though one of the ideas that you get should be to close. I and you should get the curse upon the like in vector one plus i to no a construct the general solution, you know, solution gonna be He won. Well, you know, maybe GTO two plus I dissector v people here equal to plus I he time one plus I life too. Then we can use toilers formula to turn this into the beauty. Khost is key, I find here Director one plus I money get me. You know he I co signed t i t. And we have mice on Sunday My son in the second world we have ice to co sign he might have to find Let me just take the real and imaginary part separate them beef, tea and the real Korean My sanity My co sign of tea in the imaginary parties sign he no sign of t what is key side and that we should Jaros motion exit TV Could you he won t Kiki place this factor here first solution you either to be going second solution each to he. And now we know the real part of the again. We two. Plus I sport here really is all that is, Miss Trajectories. We&#x27;re gonna be spiralling of the Bible. Spiral inward, spiral outward.

Ibrahima Barry · Answer

Okay, so we&#x27;re given to me a goes my human. It&#x27;s 111 arrest solve assist in X equals. It&#x27;s probably because first we have to find value in values and collectors of a one of the idea though salted value that we have with you Geico to my team. Plus I already you see that the the real part of the even though he negative it implies that you will have spirals in your feet portrait and it take a note been inward to the second part of the question that we can automatically solve right now Analysis pointing the general solution. So the first point are you vector two bodies keep I is one minus I want have you can construct our submission the one time e V as in Victor in my eye Im ki you. The other solution is just because you get a solution This one just minus que mind If I v one you hear and then no, you know we can write Need the money What I times t using orders formula we get I need the money. Cuchi co sign. He was fine. This is the word for it. And then we can just spoke this into imaginary in real part. First, we&#x27;ll expand this out. We get U minus Q t co sider He I signed he I defector the one defector I infected. You know, when you do that you get Was he minus I Co sci fi. What? I find p and my ice We&#x27;re fine tea. The second rotation A coast I he off site then when you split into the imaginary in real part, you got so sorry people sign t over co sci fi and you have the legend report 70 my co sign T psyche. Really? We use the one that of this port years because you get solution that air con you There&#x27;s, you know, we have a real and imaginary part. I mean, our solution Just the one times Mike cuchi coastline api flying you, he and you, you can like see, I&#x27;m signed he my coast he and you know before, Like I said, the this is the spirals because we have complex like in value and they get a spiral towards or you because the idea like the real part, I think like value value, negative

James Chok · Answer

okay in this question, you want to solve this differential equation Where a is this metrics? So in this question, we can use and make use Capital of the Sun&#x27;s gonna skip a love the cap elections. But I&#x27;ll explain what we do along the way. The first thing that we have to do is off course computer ideal balance and Ivan function. I&#x27;m 1000 island vectors. So these are your values and your ID in our values here. And I&#x27;m in Victor&#x27;s every year. Now get general solution will then look like this up here. So then substituting B one b two B three, as as the&#x27;s eyes were here and land or two. And then the three up in Holland one, then two. Then the three. You get this solution right here? Yes. Get this solution they got Okay, So now this is your complex general solution. But we want it in really values. So to look at it from real values, all you really need to do is to look at just one of these. So this is already really so the question is why do anymore at oneself to Well, if you look at both of these two, you kind of just get in multiple of each other. But these multiples off each other cancels out later as you&#x27;ll see. So just looking at this one, expanding the reach of life I e five plus I t get five tea cozy on T I 70. So then splitting up into its real and imaginary parts. The first real part that you get from this first entry is six times curse NT. Next next play is to talk about I 70 which is minus two scientists. So the next part for this one is just nine coast auntie and three I type of ice. I t So that&#x27;s this entry and this is simply really so It&#x27;s just 10 Christ, aunty. So to get the imaginary part, the first thing that you do is you get this six any. Multiply it by the AI scientist. So got to this one and then you get this to my entire by the co sign. So that&#x27;s why we&#x27;re on the same thing for this night time tete with this one and this one and three course anti this one and this one. So that&#x27;s how you get this. So If you do this for minus, I t. Or you&#x27;ll simply get easier. Simply get a negative here and you probably get a plus plus plus plus year on dhe. Really? DC are going to differ by constant, but that doesn&#x27;t really matter. Because if you put it back into your general solution, which is 68 Martucci, that&#x27;s everything here. When you put it back, you already have Constance here anyways. And so the constants will just really readjust when you put the initial condition. So this is your final solution.

Bobby Barnes · Answer

so for us to find the solution to this differential pressure, remember, the first thing we&#x27;re going to do is we want to write this in the form of our solution being so some constancy one times our first vector or Eigen vector and then e to the Lambda won t where Lambda One is our first, um, Aiken value. And then we just repeat. But just with tubes everywhere. So this is the form gonna end up wanting to get in, and they tell us our solution, they&#x27;re gonna be complex. And then we could talk about how to turn them into the real ones after we get the complex solution. Right. So let&#x27;s go ahead and write that down. Eso we first need for what are Eigen? Values are going to be. So we do a my slammed I, which is going to be negative two minus lambda One negative eight to minus lambda. And remember, if we take the determinant of this, that should be equal to zero. So we get negative two minus lambda tu minus lambda. And then I would be plus eight zero. And if we were to expand this out now, we would get? Um, actually, let&#x27;s pull that negative out. So would be negative here. And then this is just gonna be the difference of squares. So this would be four minus lamb to square plus eight. And that&#x27;s gonna be zero. And now we can just go ahead and distribute the negative. So that would be Lambda Square to B minus eight plus eight to the B plus. Or is he go to zero? And then we could go ahead and move everything over. So Lambda Square, it is going to be made before. And then we take the square root on each side. And so then that gives us Lambda is going to be able to plus or minus two I and actually from that not x lambda. And actually, from this here, we can already go ahead and figure out what the trajectory off these were going to be. Because if we have something in the form of a plus B, I so if a is zero so that if a zero, then we know we will have ellipsis of Lipsky&#x27;s of round the origin. So this is going to be the trajectory of it. Since over here are a is zero. So that&#x27;s the trajectory. But now let&#x27;s go ahead and find one of our Eigen vectors of and go ahead and pick this up. And remember, we only really need to find one of the icon vectors since for the complex ones thes second Eigen vector is just going to be its complex conjugate. So if we were to go ahead and do lambda or no, not Lambda but a minus, uh, to I I this is going to give minus two minus two. I negative 81 to minus to I and we could go ahead and do like the road reduction to make sure that the bottom row is going to be zero. But since these are complex, we know it will be the case. So we could just be a little bit lazy. And I&#x27;m just going to use that last one there. Eso I&#x27;m going to rewrite this as negative eight x one and you can also use the first one. Doesn&#x27;t really matter. But I just think the numbers down there, uh, well, solve for X one a little bit easier, since that&#x27;s normally will be solved for okay, next to me said that he was zero. And now I&#x27;m gonna add X one over. So be eight x one. There&#x27;s a two to minus two I next to and then we divide the eight over. So x one is going to be equal to So the one minus one I over four, uh, to. So when we come over here and create our Eigen vector, this is going to be so one minus I over four, uh, X two and then we have X two here. Then we could pull the excuse out so we&#x27;d be left with that. And remember, now we could multiply this by anything we want. So I&#x27;m just gonna multiply this by four. And so our first Eigen vector is going to be one minus I for all right. And so, actually, let&#x27;s write this down here. So we have when Lambda is equal to two I we get B one to B one, minus I for and that means for our complex conjugate or second hike in vector is just gonna be one plus I or because we just make it the complex conjugation. So those are going to be our sectors now. And what we can do is go ahead and plug it into like, that equation. We had a top there, so are complex. Um, valued one is going to be X is equal to So it was C one and then be one. So one minus side for and then e to the one or not 12 i t. And then plus C two one plus I or and then eat negative to I teeth. And so this is going to be the complex solution. And then we can go ahead and get the rial valued solution. Um, by expanding this and then rewriting everything. But they actually give us a nice way in the book to do this. And they say that we can write this as so the X is equal to C one y one plus C two y two, where why one is equal to e riel of one of the Eigen vectors. So it doesn&#x27;t matter which one we use, plus cosine Bt and then minus the imaginary portion of one of our Eigen vectors and the sign of Bt. And then all of this is multiplied by E to the 80 and then for y two we like, switch the signs and co signs we keep the vector the same And then the subtraction is going to become addition. You could get this just by, like, using the E to the I data is co signed data, plus I sine data. But I think doing it this way is a little bit, uh, cool quicker, because then we don&#x27;t have to, like, think about, like, combinations of things. But you could do it that way as well. You&#x27;ll get a similar answer. It might just be a little different due to, like, combinations and stuff, but it&#x27;ll be the same thing in, like, the grand scheme of things. But let&#x27;s go ahead and plug everything in, although so I&#x27;m just gonna use Lambda One and we want So in this case, slammed of one is going to be. And remember, this is supposed to be like a plus B I. So that&#x27;s where our A&#x27;s and the&#x27;s air coming from. So that was just zero plus route to I, and now we can get our riel portion of the vector and imagining by breaking that up. So the real portion of this is just one or and then we add this to would be negative 10 Then we have either, So we could just come down here. So this is the rule sourcing. This is the real and this is the imaginary. So let&#x27;s go ahead and plug those in. So the reels are going to be 14 and depending on if you started with a different Eigen vector, um, you would get something slightly different here, but again, it should all give you the same thing as long as you&#x27;re consistent with everything else. And then the imaginary should be negative. 10 negative 10 for our B that&#x27;s going to be not route to just to. So let&#x27;s go ahead and replace all these bees here with twos. So to to to to And then a was supposed to be zero. So actually, this is just e 20 which that&#x27;s just one. So we can kind of just ignore that on. Then we just need to multiply the stop one. I see one c one, c two, C two, and we can go ahead and, um, collect all of the collect all the coastlines together and the signs so that gives us X is going to be so let&#x27;s see. Um, so this would be the cosign here along with this and doing that, we will end up with, um, are very I&#x27;ll just add these together. I&#x27;ll keep the coastlines and signs the same. So this is just gonna be C one. And so this is going to be host sign to t and then minus minus, actually, let&#x27;s just go ahead and make that a plus there and there will be plus sign two teeth and then this would be just four co sign to teak and then plus to see, And then it would be so sign of two t and then minus co sign of two teeth and then down below it before sign of two teams. So this would be our solution with the rial Once, um, I guess you could try toe, make this in tow. One bigger one. But I&#x27;m just gonna leave it like this here and so again, you might have slightly different numbers, um, in front of them. But as long as you can, like, pull all those factors out, you should get exactly what we have here.

Problem

In Exercises $9-18,$ construct the general soluti…

Problem 11 Easy Difficulty

In Exercises $9-18$ , construct the general solution of $\mathbf{x}^{\prime}=A \mathbf{x}$ involving complex eigenfunctions and then obtain the general real solution. Describe the shapes of typical trajectories.
$$
A=\left[\begin{array}{rr}{-3} & {-9} \\ {2} & {3}\end{array}\right]
$$

Answer

Here, the Eigen values are complex and the real parts of the Eigen values are zero.
Therefore, the trajectories are ellipses about the origin.

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

Linear Algebra and Its Applications

Lily A.

Caleb E.

Samuel H.

Joseph L.

Vectors Intro

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Here, the Eigen values are complex and the real parts of the Eigen values are zero.Therefore, the trajectories are ellipses about the origin.

Linear Algebra and Its Applications

Lily A.

Caleb E.

Samuel H.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Caleb E.

Samuel H.

Joseph L.

Here, the Eigen values are complex and the real parts of the Eigen values are zero.
Therefore, the trajectories are ellipses about the origin.

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

Linear Algebra and Its Applications