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6. (20 points) Let3 A = | -1 5]Find the eigenvalues and an eigenvector for each eigenvalue Write the general solution to x' = Ax in terms of real-valued functi...

Question

6. (20 points) Let3 A = | -1 5]Find the eigenvalues and an eigenvector for each eigenvalue Write the general solution to x' = Ax in terms of real-valued functions_ expect imaginary eigenvalues for this problem. If you get real eigenvalues, please double check your work to avoid a disaster.

6. (20 points) Let 3 A = | -1 5] Find the eigenvalues and an eigenvector for each eigenvalue Write the general solution to x' = Ax in terms of real-valued functions_ expect imaginary eigenvalues for this problem. If you get real eigenvalues, please double check your work to avoid a disaster.



Answers

4. Find the linear homogeneous differential equation with constant coefficients which has an auxiliary equation whose roots are:
m = 0, 0, 0, -2, -2, 3i, -3i

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

In this question we have to find a linear homogeneous differential equations with constant coffee. Since whose rules haven't given us M equals 000 minus two minus 23 out and minus three. Health will first consider the first three zeros that have given us routes since Amy called zero. You can see the M plus zero zero again and plus zero zero again and plus zero. So you can see that Michelangelo and will multiply all these three. We'll get em kiwi called zero. Now let's just see the second one that is equal to -2. And again in cold minus. So if M equals minus that, we can see that M plus triple zero. And again, because the rules are minus and minus. So we multiply these two. That is plus 1 10 plus two. We'll get down to +00 plus two will be Now for the hard part that is m equal to three out and again, simple to minus three out. So for chemical three out and Ministry out of Visual and for me called minus helpless three out of So we'll multiple at this represents that is -3 out into the first three out and we'll get 0-00. So we'll get when you multiply these two will get them square -9. I spell it and we know that I ought to swear you called-. So we'll get em effortless 90 collagen. Now again will multiple all these 3% that is this one this one and this one and we'll get zero that is thank you And two members to elsewhere. And um start last night. Now we'll just open the brackets plus two. Open this square square one. That is m plus two holes. Whether it will get M squared plus four plus four. Now again, we'll open this bracket, keeping them to we'll get em fourth plus nine M squared plus four M. Two. Statistics name plus 26.0. Now again. Well, open up the record for the final time. That this will get empty about seven plus four and six. They're starting them to fight the strategic 4 36 M two Q. Is equal to zero. No, just try to remember this. We know that when if this is the given different allocation, we try to replace by double death by M square and wider by him. And why simply were one to get out of the equation and then solve for the roots and get the differential equations. Also the different different person. Similarly, we'll go and solve this one, but we'll go reverse this time. But if it's uh auxiliary question is this one, what will we tell modernist different person. This will put And to those seven years. What? You were seven and into a success right? Or 6 25 25 24 hours prior to the fourth and M two or three years? And this will be there. Our homelessness differential equations with Western conference whose rules will be this that is mhm 000 -2 -2, 3 out of zero, and this will be the auxiliary person. Yeah. Mhm.


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