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That the complete solution matrix A_ Given for S0mepoinls) Suppose that Ax (16 this equation is M '01 ' " [8]-[H-[A}: X = and columns does it i.e , h...

Question

That the complete solution matrix A_ Given for S0mepoinls) Suppose that Ax (16 this equation is M '01 ' " [8]-[H-[A}: X = and columns does it i.e , how many rows and shape of this matrix What is the size have? free variables are there? How many is the rank of this matrix A? What Find this matrix A

that the complete solution matrix A_ Given for S0me poinls) Suppose that Ax (16 this equation is M '01 ' " [8]-[H-[A}: X = and columns does it i.e , how many rows and shape of this matrix What is the size have? free variables are there? How many is the rank of this matrix A? What Find this matrix A



Answers

Determine the general solution to the system $\mathbf{x}^{\prime}=A \mathbf{x}$ for the given matrix $A$. $A=\left[\begin{array}{rr}-4 & 1 \\ -1 & -6\end{array}\right]$.

Hey, everyone. So Ah, the question we're looking at today gives us ah, matrix A. And, uh, it wants us to solve the ah system of differential equations given by the ex prime equals a X, which will have three components. So three different possibly mixed differential equations wants us to use during canonical form. So how we do that is the whole basis of the method is that we set X equal to some in vertebral s. That's gonna be a three by three matrix times a, uh, three column vector y plan. Just change of variables there. And the reason we do that is because it allows us to puts the expand equation into this form. Right? And if you recognize this, it's because, ah, for a certain value of s that we can uniquely determine given a and we will we can have this equal J, which is the Jordan canonical form of the, uh, a matrix and say we will have That's an X. We will have a new equation. Why prime equals j times. Why, and we'll see that this is solvable. It's the analogous to the diagonal ization method, except for non Dagnall non diagonal Izabal Major sees it works as well. And so these air two equations that we need. And if we solve this second one here, then we can, uh, solve for y. And if we solve for y and we know s that we can solve this first one here. So what do we need to find? Well, let's get to work Finding J and s. That will be all we need. So how do we find the Jordan canonical form for something? Well, we take the Eigen values first, so that's gonna be your agon value equation that a minus slammed I the characteristic polynomial of a set that to zero. So what's a minus? Slammed? I diagonals take determinant of that, set it to zero. Well, the determinant of a three by three matrix, if you remember, is going to be It's gonna be a sum of three different parts. It's gonna be a some of this term times the determinant of this matrix. Plus this term times the determinant of this matrix, uh, starting here. And so you're sort of going around the back of the matrix if you want to think of it like that and Plus this term will that zero times the determinant of this matrix. But that's gonna cancel out. So we don't have to worry about that. Your let me cleanups factory in a little bit. Cool. Okay, so to start, that's ah, and then minus one. Perfect right, That's just negative. Landed times. Thea, do you buy two determinant? That's this Times this minus this times. This should be all familiar. And now the green one. Well, that's one we don't need to worry about it. Uh, times this times this minus this times this Well, that's negative. One time negative one minus Lambda Times. Here is just Syria. So it's one times one times one that's just one. We're gonna some these up plus equals Well, expanding out. It's, um, negative Lambda Times. That's going to be okay. You want to set that equal to zero Now, if you just look at this, you can see that ah, one is going to be an Eigen value. And also negative one is going Teoh he and Eigen value. And as it turns out, I'll save you the math of Destry writing this it's terribly hard. It's just some algebra, but we're going to get, uh, lambda minus one times one minus lambda squared equals zero. Let me just check my notes. Sorry. Yep, that's it. So, um, just make sure to keep these which to Which one is minus in which one is plus in ah, in your head. So we have the two agon value, slammed a one equals one, and that has a multiplicity. Ah, one. It's only repeated one time. It's not squared and land to It's gonna equal negative one. And that has a multiplicity of to. Is this squared right here? Okay, so we have the Eigen values here. Come back to that. And, um so that gives us the diagonal for Jordan Matrix Jordan canonical form of our matrix. A. But we need to find out how many Jordan blocks it's going to have. So we know that there will be one linear linearly independent Eigen vector for Lambda One. And that's the Mexican have, because only as multiplicity of one. But for Lambda two, we have multiplicity of two. So we could either have one or two linearly. Independent Eigen values dragon vectors. We know that the number of Jordan blocks is going to equal the number of linearly independent Eigen vectors. So we need to determine Ah, the Agon vectors for lamb to Okay, So how do we do that? Well, the Agon vector equation a minus slammed it to which we can sub in as negative one. So a plus one i a plus r times it's Eigen vector and we want that equal to zero. So what's a plus? I Well, that's just going to be one one shoe 0110 110 011 Is that what Crites Yes, it does. Okay. And who didn't mean to draw that? And we're going to solve the system of equations for components of e gives us This is zero. So that's going to go to well, we can quickly see that the top cancels with the bottom and we can get that if we subtract this from that, then we can get and that's our role reduced form. There's there's nothing else we can do there. So, uh, that gives a solution that gives us a V. So we see there is one, um, one completely zero. So that means there will be one free variable. And if there's only one free variable. There's only one, uh, Eigen vector that we can yet that satisfies this, right, Because, uh, if you had to, then you could have, uh, the linear some. So let's write it like this. V equals Alfa Times some the one. And if it had to linearly independent Eigen vectors than we could, right? This right where v one is not in the span of v two, but, uh, it's only one free variable. So we can't have a beta which only have an Alfa, and we have that The dimension of the wagons face for Lambda two is one. So that means what we have. One Eigen vector and one Eigen vector. So we have to Eigen vectors to Jordan blocks so we can write out Jay now, Sir, J equals negative one negative one one and zeros down here as usual. Well, this is already a Jordan block, this one, so we can't do anything there, so that's a zero. And we need one more Jordan block. So we have to combine these two. Otherwise they'd be three, and that's just gonna be a zero. And there, that's our Jordan canonical form. Okay, So that's half the battle we have now. Found this, Jay. Now we need This s okay. So what is the s? Well, it's the Matrix. That, uh, sort of gives you j from a right. We have found it before in the chapter satisfies that equation. Um, So it will be basically the equivalent of, ah, diagonal ization matrix with the Eigen vectors for a fully diagonal A, um, but where a is defective, where it doesn't fill the full possibility. So the dim of some Agon space does not equal the multiplicity of its agon value, which we have for Ah, Lambda too. So our section for Lambda Two is going to be instead of the Eigen vectors of Lambda two. It's gonna be a cycle of generalized Eigen vectors. So what does that mean? Well, a cycle is of the form. Um, we need a cycle of length to so it's a minus. Slammed it to ah, I times v prime, which is the generalized Eigen vector, which is different from the item vector, right, Because for an Eigen vector, this V zero v prime and those are gonna be our first two columns and then we're going to have the Eigen vector for Ah, the, uh, land equals one because that's just it's it's non defective in that I'd in space So it's right that as w so these are gonna be your columns in column one column to going three for s So let's start by finding the cycles. So basically, what this means is that this has to be non zero. So view problem can't be an Eigen value. So Eigen vector Sorry. And, ah, the term that would come here, which is a minus landed to you. I squared times of the prime That s t equals zero. So that's where that's our constraint. That's what we need to solve for right now. Coming there. It's not multiplied. So let's go ahead and calculate a minus. Lambda two I squared. So that's going to be well, we already did a minus lambda Too high. That was 110 Perfect, right? That's what we were reduced earlier. We need that squared so we just multiply it by itself equals one to on. Okay, so we need to find some V that brings that to zero. So we're just gonna do the same augmented matrix, and we're going to reduce. You'll see. This is a pretty easy, rare reduction. We know the it's already. And that's just that we, uh all the roads are the same. So we can just nickel the zeros one big zero down there. And so this is our equation. We have to free variables when we have that The prime, the first element plus two times a second element. Plus one times the third element is going to zero. So we can write. Um Well, it's just 211 negative to write. That works. Sorry. That's not quite what we need to dio. All right, so this is this is what we get and we they receive that their to free variables. So we can separate this to, um, two different vectors, right in Alfa one in a beta. Once we get Alfa Times. Negative 210 Uh, plus data times, negative. 10 one. So, uh, the second requirement is that neither of these air Eigen vectors So we need this to be non zero, so we can check these. So, uh, well, let's just do it out. So 110011 110 The times. Well, second one is gonna be a little easier. So to start with that medical on 01 it's getting just be some matrix multiplication equals. Ah, negative one one, negative one. So you see, that's non zero. So that's gonna work for us. Okay, so let's just choose that one. Doesn't matter. We just need one. And we need to put it in a cycle when we already calculated a minus lambda to be prime. You've already done that. We know v prime. And so we just need w. So we just need to find the I the actual Eigen vector for, um land war so that we do that well, the same way we found Eigen vectors for lead to. So I get vector equation. And since Linda one is one, it's just a minus. Sorry times Let's just say if you want equals zero. So a minus slammed I it's going to be perfect. Now we can ah go about reducing this. I'll save you guys Thea the tedium of that. Okay, so if we reduced it, we see that there's one Eigen vector as expected, this is one of three variable and V one equals well, some free variable fines. Well, negative. One of the first plus one of the last needs to equal negative one of the second plus one of the last. And that needs to be zero. So we can see that That hits that requirement pretty nicely. Um, the simple vector there. That's nice. So we have our w. Oh, I said that was w earlier. Why did I write If you want Sorry for the confusing notation. Everyone, That was very strange. I knew I had something written. Okay, so we have our columns of s You're ready to Seoul for us. Not even solves. Put. Put us together. Okay, so it's going to be to go back up here. We need it to be a minus. Landed to I, Tom's V prime. Then be prime then w used to be in that order. Okay. And that's just to do with, uh, because we put the one at the bottom. Thea actual Eigen vector for lambda. One needs to come last. And then for any Jordan block, you need to go the, uh, the order of this cycle. And that's how you determine that the order needs to be this. We had put the one first. If you put the one up here and then the Jordan block second, then the W would go first. But we did it. It's tradition to do it this way with the largest toward more. So it's gonna be this its first. This is second, and this is going to be third. OK, so step there for a second. S o s equals. Well, that's negative. 11 negative one, then Negative on 01 Finally. 111 That's that's golden. That's all we need. Okay, so that's that's there s lovely Circle that and green. And now we go way, way back up to the top to what we were originally solving for, um Seems like a long time ago. And we have this equation now, so we just need to plug everything in and we will solve the equations. Okay, So by Prime Nicholls, J times Why? Well, what does that give us that gives us a system of ah, differential equations running out of room here? I'm gonna raise some stuff. Sorry, guys. Only give us so much room in these, okay? And we're back here up at the top of this lean ice. Taking this girl appear real quick. Oof! All right, well, I messed something up there with overlay, but good thing is, we don't need that started a pet again. Uh, so this is our right prime in RJ. That's gonna be times y in resolving for the components of why, based on this equation, So remember, why is going to just be three components that were selling for Why one y two y three? My prime is gonna be the derivatives of these, so we can get that three equations. White three. Sorry. Why one prime equals negative y one us. Why, too, right? That's just this topper multiplied by why? And then why? To prime equals negative wide to and why three prime equals y three. Okay, so you'll see that unlike a diagonal ized form that we would use to solve this equation if they were diagonal Izabal. I'm like that We have this mixed term up here, but it's OK because we have these two that are unmixed, that we can just solve right off the bat and look into this and you'll see that it's pretty easy to solve actually. So let's do. Ah, this one first. So, uh, this one first? Why three Prime Michel's wife three. Well, what equals its own derivative? So we get that. Why three equals some constant of integration. C three for why, three times e to the t Perfect. That's good. Now this one. Well, it's just negative of that. So saving breath. Ah, you can just check. But this is just si two e to the negative t. All right. And now So we plugged it in up here for this one, and we get why one prime Plus why one equals C two e to the negative t. Okay. And now you might not realize that right now, but we have a chick to solve this. It's We're gonna have an integrating factor called eye of tea. You've learned this earlier in the book. Not sure how long ago, but you can go back and read this sections if you want. It's not too complicated. You'll see it works out and let that equal e to the T. So we cancel out this and we multiply through on both sides. So why one prime you do the teeth, Plus why won t to the t equals C two. Now, this is where the trick comes in. Well, what is that? That's just why one either The tea, that's the product. You love it. So just that whole thing derivative versus time. It's gonna equal C to right. And that's something we can solve. Just integrate both sides. Why? I want you to the t equals C two t for a C one. So why one is gonna equal perfect, And that is going to be our wise right. But we want to solve for X. So we have to plug them back into this equation where R. S, uh, matrix that we found is gonna come into play. All right, So, uh, we have this equation for X based on why X equals s times y remember, that's a column vector. That's a components of why so X equals what's R. S salt for it down here? It will check that. That's correct. And then plug in. Otherwise, why one that's right there. All right, Perfect. Now, this is just some easy matrix multiplication to get our final X and we see that that is going to be bit of a messy some, but you can simplify it out. Perfect. That's our X Feel free to send fight out. But for all intents and purposes, this is the solution. And every problem's gonna take this form. Um, we'll do all the other words. Don't make a video, but, uh, yeah, that's that's the process. It's just these guiding equations right up here. This is the most important part. Um, okay, I hope that is helpful. And everyone have a good day.

First thing we're gonna do here. Let's find the Eigen. Values are matrix A and that means we have to look at the determinant of our major expanding minus the Agon values times your identity matrix, which will be the determinant of negative one minus lambda 001 five minus Lambda Negative 116 and negative tumor and slander. And since this is a three by three matrix, we can choose which row to take the determinant with and you can use the top row since it has two zeros in it together. This determinant is just negative one minus Lambda Times The determinant of the bottom, right for entries which is five minus lambda. I'm staying here to minus Lambda plus six minus +00 is all equals zero. We can expand the terms inside of the second Brent cease to get that it is negative. Turn minus three Lambda plus Lambda Square, our six bringing the terms together. Well, give us land squared minus three, Lambda minus four. And we can factor that into end of minus four and Lambda plus one. And you want to know when this is equal to zero. But first we can take out it factor out of negative one out of the first term to get that. This is Lambda plus one squared times Lambda minus four being with zero. And now we know what our Eigen values are is if Lambda is negative one. This whole thing is zero. And if Lambda is for this whole thing zero. But we just have a multiplicity of to in our first Eigen value. Now we're going to try to find the Eigen vectors of this matrix. And first, we're going to look for when Lambda is equal to four, which will give us looking in four for the Eigen value here. Well, give us negative. Five 00 one one negative. One 16 Negative. Six 000 And we can divide the first row by negative five to get 100 in our first round. And we can use that to zero together to entries in the first column and weaken. Leave the second room. How? How it is one a negative one. And then divide the third row by negative six. Get negative. 11 And now we can read off our Aiken Vector. So in the first row. The only term is the first entry in our Eigen Vector Wish means that it has to be zero for the whole thing to be zero. And from the second round, third row. We know that the second and third injuries, both after you won. So there's our first Eigen vector. And now we can look at Are there Eigen vector with a multiplicity Eigen value with the multiplicity of to Which means we're going to be looking at 000 zero here, zero 16 negative one and 16 negative one. And we can add the second and third columns leaving the 2nd 1 If we do, we're subtracting them. We'll get third row, you get all zeros. And from this, if we think about it and the components of the Eigen vector the first component, the one will be negative. Six times a second component plus one. I'm the third component. Where do you two is equal to? R and B three is equal to s two. Therefore, our second and third Aiken vectors will be negative. Six are plus S R and s. Now we can separate that these into a vector with bar and just s so if you look at the one that just are we're getting negative 610 plus the vector for s, which will be 101 So therefore these air are too high in vectors. And now we can construct a general solution should be some constant turned e to the first wagon value, which was four times team times the Eigen vector, which was 011 plus some other constant times e to the Eigen value, which was negative one times t times the first Eigen vector, which was negative 610 and finally, plus some third constant times et to the Eigen value. Just making one times team times the Agon vector, which was 10 What?

Everyone. So the problem we're looking at today it gives us a matrix alien. It s is to solve the three mixed differential equations been by ex prime equals a X where x the next prime are the column vectors with three elements. So how we're gonna do that? It's similar to are the diagonal ization trick. We land for diagonal eyes ablaze. But it's more general using Jordan canonical forms, and it's based off the coordinate transformation. X equals s times. Why some in veritable s you'll see that that gives us the equation. Why prime? He calls s inverse times a times s times why, and for a suitable choice and s which we can calculate based on a we know, that s two the s inverse times. A times s um we can get that equal to the Jordan canonical form of, uh, A so we can write why Prime equals J Why? And you'll see. Uh, this call that too. Question two will give us ah, differential equations. We can solve in terms of why and then Equation one will give us X in terms of white, and so we can combine one and two and we will solve for X. Okay, So the first things first, Ah, you'll see their two unknowns we need to solve for there is J. And there is s So let's start with JJ's, the Jordan canonical form of A. So the diagonals are going to be Aydin values. So we solved the Eigen value equation. It's a characteristic polynomial set that equal Dozier the determinant of this matrix. So that's going to be right. Perfect. And, uh, we want that equal to zero. Well, uh, you need to find the determinant. And if you remove the deter, the determinant three by three major sees is going to be the some of this element times the determinant of this matrix plus to some of this element tons the determinant of this matrix, which is gonna be zero, because it's time zero. Then lastly, this element plus the determinant of this matrix, which you can see, it's also gonna be zero. So for our intensive purposes, it is just ah equals negative two minus lambda times thinking three minus slammed, uh, times negative one minus lambda. And that's going to be ah, minus finest one. So plus one right. And so that's going to give us well, expanding out. I'll save you some of the algebra of rearranging the, uh yeah, so you don't need to do the algebra. Um, you don't need to see me do it, okay? And now you can check that That indeed is Lambda Plus to keep um yes, I'll leave that up to you guys as part of this. This is an algebra course, though. This is Ah, differential equations, Salem. Anyways, we get that. There's only one Eigen value. Just call it Lambda. In that case, slammed equals negative too. And it has a multiplicity of three. Right, Because this is cubed. Say that that's gonna be the diagonals of joining canonical form. And we know that this has at least one linearly independent Eigen vector because it's nagging value, but it can have anywhere between one and three. And we have a theory that says that that number, the dimension of the Agon space, is going to determine the number of Jordan blocks we have. So we need to find the Agon vectors. So we have the Eigen Vector equation 80 minus lambda high. So a plus two I time some Eigen vector V that we're looking solid for. And we want that equal to zero. It's a bad I OK, so a pills to I You can set up a augments in matrix. Um, it's gonna be zero negative one and one on the diagonals. Syria. So here, is that okay. And we want to roll reduce this to solve for the components of E. So we see that this top row was already old zeros and theme. Next heroes are multiples of each other times negative ones. So you just add 1 to 1 000 and move this guy up. So it's just all right. And we find that there is too rose that are all zero. So their to free variables. In the US, there are two linearly independent Eigen victims. So we can write that as dim. Yeah, your wagons face is too. Okay, um, if this road was old zeroes, then, uh, that would be a different story, but it's not the dimensions to, and so we're going to have to Jordan blocks. All right, so we can write out of Jordan canonical form now Jay equals Well, it's What is it? Negative two is our Eigen value. They get to along the diagonal, tear everything below, and we have to Jordan blocks. So we need to link, um, two of these up with the one and ah, custom says that we do the largest on top. So this Jordan block size two, and then this last one is a Jordan Block it size one. That's a Jordan canonical form. Okay, All right. So if we scroll back up, we see we, uh Now I have all the components of, ah, equation two. And we need to find s for equation one. So finding s well, um, s is going to be ah of the form. So for its Jordan block, right each each Jordan block of size, too, we're going to have, Which is this. We're going to have a cycle of size two of generalized dagon vectors, which is of the form a minus lambda I, ah, times v prime for generalized Eigen vector and then be prime. That's gonna be our cycle where it's implied that a plus Teoh squared times v prime equals zero, right? Because the cycle ends there. And, um, these two things are non zero. So they're in the cycle cycles Air lamone zero elements. And then, uh, we'll call it W um where w is Justin Eigen vector? Because this, this is if you want to think of it as a cycle of size one where you cross this mountain. But either way, just one eye confected. Okay, so let's Ah, what Sulphur Eigen. Victor's first? Because we really have that. So we have that w is going to equal Well, we have to free variables Cola, Alphen, beta. And we know that Ah, one of this first element is don't equal. Ah, a Alfa plus beta because ah, the first element minus the second element minus the third element equals zero. So that distills into two different, um, liken vectors. W equals Alfa times 110 for Spada times 101 Right. So, uh, we can choose either of those depending on our, um so it s is not necessarily unique. We can choose either of those and we can choose different generalized Eigen vectors, right? As long as they're not in the span. So let's just save that for later. Save that for later and go about finding the general generalized Eigen vectors. So to do that. Let's calculate this. And then we need to set up a, um, augmented matrix to solve for V prime, just like we did for the, uh, Eigen vectors. So it was two y squared. Well, it was too. I That's just going to be, um, zero negative 11 All right. It's just again. Was that, uh, right? That's just what we had before. And we're gonna multiply it by itself. Perfect. That is going to be Let's move these over lovely. That's gonna be equal to zeros along the top row, right? Okay. And you can check that. That's correct. It's sort of easy to see, I guess, is that the first one doesn't matter when you multiply those out. And the second to are just essentially the identity times negative. And each of these is negative one and one. So, anyways, let's check it all out. Make sure you understand why this is zero matrix. So we see that any any V prime is going to satisfy this. So our next condition that we can look for a generalized pictures of you prime is that this has to be non zero. In other words, if you will recognize the Eigen value equation. Um, it cannot be in the span of the Eigen vectors. Right? Because all Eigen vectors, definitional, iwill have zero for this, so we can choose our favorite vector. That is not in the span of this, So, uh, All right, um, well, I can see one right off the bat. Let's let ah, the prime 111 quick and simple. You can check that. That's not in the span. If you add these two together, it will give you 211 which is obviously linearly independent from Ah, this. Um, yeah. Okay, so that works. Eso we have RV prime. It's Mark that off and we have our Ah, a plus two I Right, so we just need to multiply that out. Perfect. So that's gonna be our first elements in the cycle. That's gonna be Oh, sorry. I am an idiot. Zero negative one. What? Ok, perfect. Now we are ready to write our s, and it is important that it's in this order, right? Because, um, for the Jordan block, it has to be in the order of the cycle and them since we put our Jordan block of one last, uh, the W Ask your last If we put it first, then it go first before the cycle. But racing all this nonsense that I just true. But we didn't. And that's tradition. Um, so make sure to write it in that format just to make it easier on yourself in the future. So are we. Get that s is equal to zero negative. 11 Right. The columns are these factors in their order. 111 and choose their favorite, but it's not already repeated. Um, let's just go with 110 110 It's a nice looking s. Um, it's not unique, like I said before, but that will work. That's all we need. Okay, so now you'll see, we have found all the components to ah equation one. So we can start solving. Let's solve equation to So why prime equals J y? Why prime he cools. Remember, RJ, it's gonna be negative to negative to negative two. And why is gonna be? Why one why 23? And that gives us three different differential equations. I one prime equals negative two y one plus fly to what to prime equals make it 22 And finally, why three prime equals negative to why three. Okay. And, uh, you'll see the difference here is if if, uh, J was diagonal, then we would not have this mixed term, and we could just immediately solve everything. But we do have two equations that are not mixed. And so we can sold for those and then use that salt for this. So this is completely solvable and we know how to do this. Um, well, let's start with Let's work our way up. So why three prime equals Negative two y three. What is happening for a wife? Three. You divide both sides and integrate. You can know where you can just see that this is going to be why three equals. Eat the negative to t and always, always, always remember your integration factor. This two close together remember, that's multiplied because just the way that intervals workout, I suggest that you do it. If you don't know what I'm talking about, just do the integral ab but sides by Why three and salt. Okay, then why too? It's just the same thing. The different integration constant and why one is going to be. Well, let's set this up. What's right in the form and still become clear why I'm doing this second. Because, well, you might not see immediately how to solve this. We actually do. You have a method. It's got using an integration factor. I have t we want i of tea to cancel it this side of the equation. So I have t is going to be That is awful tea. If he is going to be each, the two t gonna multiply both sides by it. So why one crime either the to t plus two. Why one needed the to t Afghan equals C two. Uh, and if you don't know, do you see it? This is the general form of integration factor questions. This is going to equal if we product rule. I want each of the two teeth. If we take that derivative with respect to T, we're actually going to get this left hand side up here. And so this is an equation we know to solve. Okay, So we ever Why one and we can read out are Why now? Because we have y one y two and white three Just mark he's off for later use. Okay, so we have our why. Now we need our X. So we're done with this. You need our X X equals s times. Why Will weaken is too simple matrix multiplication to figure that out. So X equals what is our s? Well, it's this right then times why it's going to be a bit of a hassle to write. Well, actually, let's just think about it. Ah, formulaic Lee for now. So that's gonna be a vector. That's gonna be two. Plus why three negative y one plus why two plus why three. And finally, why one us? Why, too? And that's going to be ah are different x one x two and x three Maybe these rays. So let's expand this hold on a second while I clear things up. All right, there we go. So getting with giving us a little more room and we can just see the definitions. So finally we get that X. Remember, this is ah column vector. So just write that. I know I haven't been doing that, but just remind you is, um what two plus y three. All right, the negative way. One right which is plugging in these equations that we solved right here in the red into Ah, this on And well, uh, and it does not ask us. It does not give us any initial conditions. So we do not have to solve for these integration constants. That's where the initial conditions would come in. Um, so we are completely done. This is our answer. And remember, X hat is expect er's x one x two x three So this is X one. This is 62 This is X three, but this is that's a solution. And all of these problems take very similar forms. Pretty much the same. Just different Eigen vectors. Different mentions of the dragon space Daddy yada yada. But yeah, this was informative. And, uh, one Have a good day.

The question asked to find a solution off the system. First, we need to change the system off equation into a semantic metrics for remember, we need to find the reducer or I shall inform to find an answer for road to we can do to times roll one minus wrote to for Rose three We can do no one. Why does the rosary copier one do the elementary rule operation for row two and a rosary switch? Your rosary was a row two for rosary We can do five times wrote to minus a rosary cover You know what in a row to for grocery we can do 1/4 times a role street Help your other Rose Furrow to begin Do road to minus two times Siro Serie for no one we can do Roll one. Why this three times rosary for row one, we can do roll one minus two times, bro. To now we have the reducer or echelon form so we can get X one equals zero acts two equals zero X ray equals zero and this is the final answer


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