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Let 0 : Zxz _ S9 be group homomorphism such that 0((1,0)) (3,5,7)(2,4) and 0((0,1)) = (1,6) (8,1,9) . Find the kernel of ahd o((9,10))....

Question

Let 0 : Zxz _ S9 be group homomorphism such that 0((1,0)) (3,5,7)(2,4) and 0((0,1)) = (1,6) (8,1,9) . Find the kernel of ahd o((9,10)).

Let 0 : Zxz _ S9 be group homomorphism such that 0((1,0)) (3,5,7)(2,4) and 0((0,1)) = (1,6) (8,1,9) . Find the kernel of ahd o((9,10)).



Answers

Calculate the eigenvalues and the corresponding eigenvectors of the given matrix. All matrices have integer eigenvalues
{{0,-2,1}, {-7,-1,3}, {-11,2,4}}

So to show that this is not a hormone dwarfism. Um We essentially just have to show that one of these statements here is not true because these are the two criteria for something to be a home abort is um for some function and for this, I think the bottom one will be the easiest to actually go about doing. So How I'll do this is let's first plug in. Just half of zero or a half of the zero factor, I should say. So 00. And if we were to plug that in, so this is going to be zero, this is going to be zero and then that's going to be negative too. So we get that this outputs negative too. Well, if this is a home of dwarfism, if I were to multiply the zero vector by anything, then it should give me the same output as if I were to just multiplied on the inside and then go about doing So Let's say I were to just multiply the outside of this by two, two, two. That's going to give -4. So let's see if we do the same thing. If we could out um negative 44 and you'll end up saying that we won't because F of two times 00. Well, that's still just F of zero. And we just showed that is negative two. So essentially we just showed that two times F of 0, 0 is not equal to F of two times 00. Which would then tell us that is not a home amorphous. Um So implies not homo morph ism Um There is something a little bit more streets or we can do depending on if you have talked about it or not. Um And the thing is is that your identities should get taken to another identity. So for addition, remember this here is the additive identity. Because if we add it to anything, nothing changes. But in just the real numbers, -2 is not the additive identity, it would just be zero. So if you have where an identity does not get taken to another identity, then you can say that it's not a home a morph ISM by that, but uh that's more of a just like theorem that people often have to be taught um or go about proving I should say. But as long as you just show something like this, that would also be a valid way of showing that is not a homo dwarfism.

And this problem we have been given to force factors represented by vector de and victor G respectively. It's given that record is equal to To like a Bob -4 Jacob plus kick up Rector gee. That is given us three icap plus four Jacob Plus 10-K Cup. And here we need to check whether these two vectors are orthogonal. Artisanal means the angle between these two vectors are 90°. That we have to check. So for that we just take the dot product of those two vectors. So taking the dot product. We observed that to icap minus four Jacob plus K cap. We multiplied with another Rector Gee, that's three I capitalists for Jacob plus 10-K cap. So when we take the dot product we get two times three minus four times four. So we just multiply their corresponding coefficients together. That's the rule we apply to figure out the dot product That comes out to the 6 -16 plus 10. So when we simplify, we get the dot product at zero. When the dark product is zero, that means if we see another expression of dark product, DRG that is diggy costly to. So the only possibility for this cost to to to be zero is that when angle is 90° between these two vectors, which definitely proves that the given two vectors are orthogonal to each other

Hello. So uh this is the solution to uh the question. So this is for the uh a park with N. G. D. B. In 441 respectively respectively sorry. And then for the B. We also have that uh the does the N. G. D. Four B. 652 in the C. Part? We got uh they're supposed to be It's supposed to be a. three. So you got a straight three. Okay. So that is not uh for the scholarship question, hopefully you understand it. Thank you very much.

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


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