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Consider the linear systemFind the eigenvalues and eigenvectors for the coefhcient matrix~3-1Find the real-valued solution the initial value problem311 2I2, 511 312...

Question

Consider the linear systemFind the eigenvalues and eigenvectors for the coefhcient matrix~3-1Find the real-valued solution the initial value problem311 2I2, 511 312'7i(0) ~7, I2(0) = 15Use as the independent variable in your answcrs:I (t) =Iz(t) =

Consider the linear system Find the eigenvalues and eigenvectors for the coefhcient matrix ~3-1 Find the real-valued solution the initial value problem 311 2I2, 511 312' 7i(0) ~7, I2(0) = 15 Use as the independent variable in your answcrs: I (t) = Iz(t) =



Answers

Determine the general solution to the linear system $\mathbf{x}^{\prime}=A \mathbf{x}$ for the given matrix $A$.
$$\left[\begin{array}{rrr}
2 & -2 & 1 \\
1 & -4 & 1 \\
2 & 2 & -3
\end{array}\right]$$
[Hint: The eigenvalues of $A \text { are } \lambda=2,-2,-5 .]$

And this problem we're calculating the Eigen vectors and the Eigen values for a given matrix. And we have our matrix A equal to the following. Uh huh, wow minus 71 minus 13 and minus 11, 2. And for okay so our strategy here is going to be well set up are characteristic equation where we take our matrix A. Subtract the identity matrix, uh find the determined determinant of that and saw that setting it equal to zero. And so that will give us a cubic equation. Uh So we just need to solve that cubic equation to find our three Eigen values. And then for each again value we check with our matrix A two and solve a set of three questions to find the I in vectors. Okay, so let's set up our characteristic equation. So are a minus lambda identity matrix. Going to look like the following. We're gonna have a minus lambda minus 21 minus seven minus one minus lambda three minus 11 two and minus four. Mine is lambda. Okay. And we want to set up determinant of our a minus, I am the I equal to and south or equal to zero and so are determinant. So first we'll do cross a first row first column. So we'll get ah minus lambda chimes to have a minus one minus and uh whoa for mine, islam and minus six, wow. Then next so we go move on to second column, first row. So crossing out the corresponding ah second column and first row. Oh and so that's gonna give us a plus two. We're gonna get a minus seven. Mhm. Four minus land. Ah Plus thirties three and then going to the third column and crossing out first row third column and we'll get plus a minus 14 minus one minus land. Uh It's a minus one minus lambda and a minus 11, horrible. Mhm. Okay And we've got adding all this up. All right, we got a minus lambda uh and a minus four minus three lambda. Yeah, plus lambda squared minus six. Uh Plus two. The minus 28 plus seven seven. Land plus 33 wow. Plus A When I was 14 minus 11 plus 11 lambda, wow, mm. All right. And if we sum all that up, we get a ah minus lambda, cute. Plus three. Land is squared plus 13 and ah minus 15 and we want to set that equal to zero. Ah So we have this cubic equation to solve and we could solve it ah numerically or graphically or whatever. If we need to solve it by hand, it's actually not too bad to solve. Ah We could see that ah If we're given that the Eigen values are all integers, so we're gonna have three roots and it's gonna be one, it has to be a one, a three and a five to get our landed to the zero term of 15. So we have to have something like ah one, A three and a five. Uh So where are possible routes are plus minus one plus minus three and plus minus five. And so we can play around with a little bit and we'll find that our ah lambda solutions for lambda are one minus three and five. So those are our Eigen values. Okay, so now we'll need to uh for each Eigen value will need to go through and find the corresponding Eigen vectors. And so that is according to are a so each a ah times are vector equals two uh lambda vector. Right? So ah well so we'll go through here for our first one, lambda one equals to one. And so we want to solve that. So taking are a uh so a times V. One equals two lambda one times V. One. So that's gonna be our our Eigen vector V. One. Uh And the corresponding with its corresponding Eigen value lambda one and so are a. Right? So we've got our a vector ah zero minus 21 minus seven minus 13 minus 11 2. For. And it's criminal applying that times are vector. And so I'll give those an X. And a. Y. And Z. And that equals to. Ah So our value here is just a. One, so X. Y. Z. Okay, so that's going to set us up with a system of three equations to solve. And so we get a minus two. Uh Of course the equals X. We get a minus seven, X minus Y. Plus three Z equals two Y. And a minus 11 X plus two. Y plus four, Z equals two Z. Okay. Ah We'll start here uh with the we've got a substitution for X from the first equation X minus two Y. Posey. So plug that into the second equation and we'll get a minus seven times a minus two. Ah Plus Z minus Y plus three, Z equals two. Why? And I'm gonna skip a few intermediate rearranging of equations here and go to that, giving us Z equals 23 Y. And then we can we can just plug that back into our first equation here and now we'll go X equals two minus two. I plus Z. When Z equals +23 Y. It's two Y plus three wise. So we get X equal to Y. Okay, so now we've got our solution so we've got uh X equals X. Yeah X equals X. Why equals X and z equals 23? Y. Okay, so that gives us our the one. So Eigen vector for our first agon value of one? Going to be 11 and three? Okay, so that will be our first solution. So we got lambda. Uh oh uh so we've got our lambda one equals to one and are corresponding Eigen vector of 11 and three. All right, next step, let's check. I mean vector for I can value of lambda too equal to minus three. Okay, so all right on our a at r zero minus 21 minus seven minus one. Three minus 11 minus to four, wow. And are again times are Eigen vector equals or I can value minus three times are again vector? Yeah. Okay. And so we'll set up those three equations, we get minus two plus C equals minus three X minus seven X minus y plus. Crazy. It's minus three Y minus 11 X. Uh Plus sweeps. Um This should be a positive too. Are they negative too? So we've got minus 11 X plus two. Uh uh Plus for Z equals two minus three. Z. Okay for this one I'll just take that first equation. Uh and use our Z as a substitution. Uh So take C equals to two y minus three X. And we'll plug that into our second equation here, which will give us a minus seven X minus Y plus three times to I minus three X. Uh huh. And skipping a few intermediate steps that leaves us with Y equals to two X. And then we can just put that back into our Z expression. So we've got Z equals to two times two X minus three X. Uh So Z equals two X. So we've got for this one, ah We had our wow, X. X equals X, Y equals two X. And Z equals two X. So are RV two. I can vector going to be a one 21 All right. So that so for our again our second Eigen value of minus three, the corresponding Eigen vector V two will be 1 to 1. Oh all right. Now we just got our third Eigen value to do. Okay, so lambda three equals 25 And our matrix a zero minus 21 minus seven minus 13 minus 11 to 4. And I in vector X Y. Z equals I can value five times Eigen vector X, Y. The animal. Mhm. And that gives us the equations a minus two plus Z equals five, X a minus seven, X minus Y plus three, Z equals five Y and a minus 11 X plus two. Y plus for z equals 25 Z. And I'll take that first equation and go Z equals five X plus two. Why? And plug that into our second equation, which will give us a minus seven, X minus Y plus 35 X plus two, Y equals 25 Y. And that gives us an X equal to zero. And then we can plug that into our first equation here, we get Z equals to two Y. And so we've got A. X equals zero, Why equals Y. And the equals to two Y. And so our V. Three, I am our third Eigen vector is going to be a zero 12 and that's going with our the corresponding Eigen vector for our, I can value lambda three equals 25 mm. So we've got our three Eigen uh three Eigen values, so we've got our lambda one equals to one corresponding Eigen vector of 11 and three. We've got our second agon value lambda two equals minus three, corresponding Eigen vector 1 to 1 and our third agon value. Lambda three equals 25 and corresponding Eigen vector 012

Number 34. We need to find the identities much. Uh, you can see here Metrics is in triangle of far right Like this, right? So for growing your own metrics, liken values are diagonal and please off the make ICS Thank you. Die golden antes ar minus six and seven. So I don't really use our to my six and selling right. Thank you.

To determine the general solution to the X equals X given. To make it a find the determinant of a minus blend. I get the idea. So I skip this because I assume everyone watching this video can figure out the characteristic homo mia and salt perjury. You get east B I in vector idea values. Well, I'm the one minus five. We have a minus minus five. I zero. So yet Q plus fights. My four plus five is one. And might it be five, but one 11 So what do you reduce this species get? And again, I'm bucking a hi show All the steps that I seem anyone watching this 30 year Can I guess it's major by themselves when you reduce, this matrix should get 10 1/3 0123 and heroes your own zero three Ever fishing X. If you first we have x one equals here minus one or three three, and we have excuse evil, too. Why would be extreme? We have extra free. Their solution is equal to minus 1/3. Extreme mine is through to the extreme extreme. You can write a basis. Why it's more free money to repeat one. And you can choose extreme to be equal to Maya Green. You get our first I you vector v one to be equal to one to my history. So you think for lead here equals do you well, First, the curse falling solution to this Eigen vector V one is equal here. First watches switching each one of teams equal to even the land. The one tee times be one that minus. But he I minus one to minus three. No, it's work to solve for land itude. People too. To you're a my nephew. I on the ground zero So yet tu minus 20 We have more than one. We have one minus 61 The loss were 2 to 5. Yes, I'm gonna skip the work of reducing this matrix. So the girl reduced form The reduced rochelle on form of this matrix is 10 minus. You're one minus. Have you're owns your So we get from here. They're basis. Maybe you go to to would have Europe. We're gonna choose fever extreme. You go to here, you get a second Aiken directory to use equal to for one year second solution http Legal to eat in the land between times you equal to look acuity. I'm for 10 We do the same thing for land degree. We have a why My two times I zero, we have matrix four minus one first road the second where we have one minus 21 We have two to minus one. I'm gonna reduce matrix. We should get 1000 You're one minus 00 From here we get our solution set to be able to you're would have one reach you three to be equal. Tiu, we have 1/3 eye in victory is equal to 012 That implies that we should teach the key. You could Teoh need to the line that three of p kinds Vetri that it will. Why? Fut 012 Final submission. General solution exit Easy. Go to see one inch one of t. You have to pee c three h. We just substitute feet. All oyer each substitute when you substitute. You should get your


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