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Determine all eigenvalues and corresponding eigen values for the given matrix:3 )...

Question

Determine all eigenvalues and corresponding eigen values for the given matrix:3 )

Determine all eigenvalues and corresponding eigen values for the given matrix: 3 )



Answers

Determine all eigenvalues and corresponding eigenvectors of the given matrix. $$\left[\begin{array}{rr}2 & 3 \\-3 & 2\end{array}\right]$$.

This problem gives us in the tricks and asked us to find its Aiken values and Eigen vectors. We does by first finding characteristic polynomial, which is found by taking the determinant Uh, a minus landed times I, which will give us the determinant of the matrix to minus lambda like that 33 to minus Linda Taking the determinant, we get to minus one in the square minus three times negative three which will simplify out to land a squared gladness for Lambda plus 13. Then using a quadratic equation, excel for lambda. So negative. Negative fours for closer minus the square root Negative four squared, which is 16 minus four times a wishes one time. See, which is 13 Then we divide by to a so two times one. And this comes four plus or minus the square root of negative 36 divided by two, which is equivalent to two plus or minus three. I S o. R. Two Eigen values are not really and their two plus three out and two minus three toe find the, uh, corresponding Eigen vectors. We have to find a vector X such that a minus land I times X equals zero. So first for land equal to two plus three, I will plug that any and that gives us the Matrix Tu minus two months. Three. I negative 33 and two minus two minus three I times the Eigen vector X Y just equivalent to negative three. I negative three. Three Negative three I times a one B one equal to 00 Then solving out for this isn't equation. There's a few ways to do it. I am first going Teoh plugging a 101 And by solving out the system, you'll find that be one should be equal to, um I and just showing that calculation. So if you take anyone, you go to one just the system. We get his night of three i plus three be one equal to zero, which shelves out to give. The one is equal to I. And I'm plugging into the second equation getting negative three times one which since I was equal to one minus three I times he wants, which we found in the 1st 1 people. I am actually cool. Zero solving that out. We get negative three months, three squared plus three, and that does equal to zero. So that gives us our Eigen vector equal to one. All right, then, for land equal to two months. Three. I was ready the same process. So we get the Matrix by plugging in Lambda we get the Matrix to minus to plus three. I negative three, three tu minus to plus three I times the Eigen vector X two is equal to three. I negative 33 three I times A to B two is equal to 00 So doing the same process is above, uh, a two was gonna equal one. And then through the same calculation as I should before you'll find that B two clear the equal to negative I giving us our second Eigen vector to be one negative. Odd. And those are final answers for the Eigen values and Eigen vectors of this matrix.

This problem gives is a matrix and asks us to find its I and values and Eigen vectors. We do this by finding the characteristic polynomial which is found by taking the determinant uh, a minus land I that'll give us the determinant of the matrix. Two months, Lambda three to negative 11 might have slammed up negative 1303 months. Lambda, which will give us the polynomial to minus slammed at times one minus slammed at times three minus lambda minus nine minus six times one month's Landau plus three times three minus Linda That will simplify out to be Lambda Times two minus lambda times left a minus four set equal to zero. Giving us our Eigen values to be equal to zero two and four, then taken taking the ag value of zero. We're going to plug this into the, um a martyr slammed I equation in order to fund Eigen Vector X, that solves for the zero doctor or it is for all the aggro al use. So by plugging in zero, we get the matrix to 32 native 11 negative. 1303 times the Eigen vector. I'm gonna call X one, then we can perform gashing elimination. Teoh, simplify this matrix order Make salting easier by subtracting the first stroke to the bottom row we get to 91 3 310 and then zero all zeros on the bottom row times x one Then we can set this equal to the zero vector. So this gives us the system of equations two times a one minus B one plus three C one people zero and three a one minus B one equals zero. Which one you saw will give us the components for a one B one C one which will be Negus 39 and five making this our final answer for this first I connector, we do the same then for land equal to two. So we plug this into a Muslim I and we get the matrix 032 negative one negative one negative. One 301 times the Director x two And I find that this one can be so easily with, uh, this form no need for gashing elimination necessarily. So that equals zero vector. Then we get that B two is equal to three c two and then the U two was also eat 2382 and then we have to A to minus B two plus C two is equal to zero. Solving for this system will give us the Eigen vector X two with the components. Uh, being 131 Lastly, we're gonna do this process for Lambda equal, for we put this into the Matrix. That gives us the Matrix need to 32 negative one negative three negative one and 30 like that. One times x three this than from a gash In elimination, we can add the first row to the bottom row. So getting I give to have one three and we could also divide by the second row by three giving us one leg 10 and that bottom row music zero related to to times X three is equal to the zero vector and simplifying this we get that a two is equal to be to and that C to Z could be too and negative to a to minus b two plus plus three c two equals zero. Solving the system will give us our final Eigen vector which would be equal to 111 And those are our final answers

This problem gives them matrix and asks us to find it again. Values and Eigen vectors. We do this by finding the characteristic polynomial, which is given by the determinant of a minus lander times I and that is equal to determine up three months lambda for negative to native one minus lambda, which is equal to three minus lambda times State of one minus lambda minus night of two times four, which will solve out Teoh Equal Lander squared minus two Lander, 1st 5 So then take, uh, taking the cord attic formula saw for land. But we get negative, negative to which is to cluster a minus the square root of negative two squared which is four minus four times one times five over to a just to which is equal to two plus or minus the square root of negative 16 all over to which is equivalent to one plus or minus two. I being that both are Eigen values are not really so solve for the Eigen vectors. We need to find vectors at such that a minus slammed I times. Eigen vector is equal to zero. So in order do this. First we take Landy equal to one plus two. I will plug that in. That will give us three minus one minus two I four negative to negative one months. I'm honest to i times the Eigen vector X one, which is equal to two months to I for negative too. And they have two minutes to i times a one B one equal to 00 Now, this could be solved as, ah, system of equations. Um, if you solve in many different ways what I like to use, I'm going to set a one equal to one. And then if you sell out the system of equations, which I'll do just for this examples, we have a one to minus two I minus to be one, since a one is one that we don't have any a one there is equal to zero, and that sells out to give us be one is equal to one minus I. If you plugged the one to the second equation, we get four minus Ah, Earl do plus one months, I times negative two minus two. I Z equals zero. And if you saw about this equation yet negative too minus two I plus two I minus two. I squared people. Um, sorry, which is equal to minus two. Sneak one night before. And that gives us that negative four was leg before. So it solves the system you need. The X one is equal to one one minus side. And for Lambda equals, one wants to. I do the same thing. We plug in Warren martyrs to our gives us three minus one plus two I four native to night of one minus one plus two I times next to is equal to two plus two I for native to a negative two plus two i times to be to should equal 00 solving the system the same way as we did for one plus two I you'll find that a one again is equal to one and hurts our should be a two and B two is equal to one. Plus I giving us the second i director to be 11 plus I and those are our final answers

This problem asked us to find the alien bodies and Eigen vectors for a given matrix. We do this first by finding the characteristic polynomial, which is found by taking the determinant of the matrix a minus lander times the identity matrix. So that will give us the polynomial negative two minus Lambda Times one minus Landau Times Negative three months Lambda minus one plus deeply turns negative. Two minus lambda minus negative. Three Minus lambda which will solve out to be five plus four lander plus lander squared times Negative. Lambda minus two equals zero. Then solving out for Landau, we find our Ivan values Teoh equal Negative too negative. Two plus I and negative two minus side, then self Eigen vectors We want to solve for vectors X such that a minus land. I times that I correct X is equal to this year. A vector. So first will do this woodland equal negative too. We plug us into a months Land I, which will give us the matrix 011 113 and zero Negative one negative one times Eigen vector X one And this should equal this year a vector. So as we can see since Ah, the top row 010 That means that our component B one has to equal zero. As a result, we see for the second or third row that we have one a negative feeling for the components at a one and C one. Meaning that a 11 and see what a legal one are. Sorry. Yes, well, even one. And that gives us our first Eigen vector to equal 101 Then we have land equal to native to Plus I were you the same thing. We plug this into the matrix that gives a snake that I won one 11 less eyes three and zero negative. One native one minus I times x two equals which well, then equal the zero vector, then solving out for this system of equations, which is going to be a little bit tedious. But by some things out, you should find that X two is equal to two months. I one plus two are and one. Lastly, we do the same for Lambda. Equal to negative two minus side. Putting this into the matrix. We get I born one 11 plus I three zero negative one native one plus I times x three equal to zero. And that gives us our third again vector to equal two. Plus I one minus two. I won and those are final answers.


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