exercises-7-12-require-matlab-or-other-computational-aid-in-exercises-7-and-8-use-the-power-method-4

Question

Exercises $7-12$ require MATLAB or other computational aid. In Exercises 7 and $8,$ use the power method with the $\mathbf{x}_{0}$ given. List $\left\{\mathbf{x}_{k}ight\}$ and $\left\{\mu_{k}ight\}$ for $k=1, \ldots, 5 .$ In Exercises 9 and $10,$ list $\mu_{5}$ and $\mu_{6} .$
$$
A=\left[\begin{array}{rrr}{1} & {2} & {-2} \ {1} & {1} & {9} \ {0} & {1} & {9}\end{array}ight], \mathbf{x}_{0}=\left[\begin{array}{l}{1} \ {0} \ {0}\end{array}ight]
$$

James Chok · Accepted Answer

okay. In this question, we want to find that Ivan values and vectors using the power method with these initial condition of this initial Steichen&#x27;s. So once again, I&#x27;ve created a pipe. A script, which was this? So a matrix is set in here. Your initial stein value is here. Then this program performs it. Performs the palace, take 10 times on dhe it before the multiplication. Here, you find the absolute maximum value off the matrix, and you will perform the palace that rights here. Now, you print out the values as it goes along. So you see that in this case, we want to list your five in your six and you see the results right here.

James Chok · Answer

Yeah, this question. We want to use a power method to find the item, Ivan. Values and vectors. Obvious matrix. So over here I have created a python script that you could run yourself to before miscalculations. So to understand a bit about what&#x27;s going on, we have a hour. Mission matrix is 2145 and accept north is 10 So if you want to change these matrices, you have to be careful. But you have to put the rose like like this, and separate collars with every number. And then you had a comma every time you&#x27;re starting, right? So that&#x27;s why this is one comma. And what this program does is that it does. It performs the power step 10 times, and Mac tries and it performs step a times by except this big eight times by X. And then you&#x27;ll find your which is the absolute which is the maximum value but the absolute maximum value off this matrix. Then you performed the Palestinian and you&#x27;ll print out the results every time your capital is this. So the results over can be seen here. We see that except no is 10 and it was no serious full. So x one is 10.51 x to these went to and you could see the rest.

Bryan Lynn · Answer

Okay, so here, the notation will be using is if we vector a being equal to a one I plus day too. J plus a three K and B is equal to be one I plus be to J plus p three k than the dot product. A and B is going to be equal to a one times be one plus a two times be too plus a three times B three. Okay. And so in this problem are a is going to be equal to to I plus J and R B is going to equal Thio I minus j plus k. So when we take the dot product of be, it will be equal to a one times be one, which is two times one plus a two times B two, which is one times negative, one plus a three times be three, which is zero times one. So this is gonna be good too minus one plus zero, which is equal to one. So the solution here is just one

Fuzail Shakir · Answer

Okay, so here we have A one is equal to three. D one is equal to two C one is negative. One a two is for B 20 and see two is five. So we have three times 43 times for plus two times zero plus negative one times five, which is equal to seven.

Mihir Nayar · Answer

Hey, everybody. I hope you&#x27;re all having a nice day today. We&#x27;re solving problem number 10 from chapter nine section for a textbook in this problem, your ass defying what piece of paper this one would be when you have pizza. It&#x27;s where? Over Jew times. Okay. Plus one. Great. So, um, for yourself, K plus one, all used being quality. Um, you&#x27;re simply going to substitute a plus one in to each of the case on the right side of your equation. So remember, just put from disuse around whenever you stop in something a plus one squared all over two times they lost one something. We&#x27;ll see you guys see where I&#x27;m substituting the substitution is your first step for this problem. Keep. I&#x27;m not one outside closed from the seasons for all that. So this simplifies down too. Hey, boss, once where and now, the next you actually all of the stops until your final answer now will be involving simplification. We&#x27;ve substituted in and then got a simplified. So this K plus one plus one is simply a plus to use. And remember, you have to be right. All of us, both anywhere and denominator in expanded for him. It&#x27;s okay. Plus once where just to show you all would look like and it&#x27;s expanded form again would be Cape us one times and Mother Kate plus one. So this would be a time scale or case where and then plus one times K plus another one times case that plus one that&#x27;s plus one plus one cane, which is plus two Harry. And then you have a plus one times one, which is just possible. And then you divide it all by two times and again I&#x27;m gonna write this out on its expanded form. So I think plus two square just for you all to see, Can you close to times another A plus two. So this is K times. Okay, which is a case where us two times gay or cheap A plus, Another to face a plus two plus two years plus four k and then you need to add plus a two times two witches. But sport, so that is all going to be equal to for a final answer. Go into red. We&#x27;ll have a gun case squared, plus two. You think plus one can&#x27;t be simplified any further in your numerator, but in the denominator, you can distribute that to to each of the terms in your fantasies. And those terms should be case where, so to face where when you add the four K so that&#x27;s four times two is eight. You&#x27;re gonna add an a K. Now again, what I&#x27;m doing is distributing into the Serbs distribute that to the two times for Kay is a K, and then two times four is eight. So your final answer here don&#x27;t need such a big fraction line. You find an answer here is in the numerator. You&#x27;ll have a case where plus two UK plus one and in your denominator you&#x27;ll have to pay square plus 80 a plus eight. So right, here&#x27;s our fine answer If you guys enjoy this video piece of the light, but next to it, I hope you all have a nice day. And thanks for watching

Daniel Nolan · Answer

Okay, So, given a formula for Peter, the K here on formula is going to be, uh, pizza. Okay. Equals case, lad. On this is the numerator here so that you know me a fraction. So the numerator here is going to k squared all times by capers. Three on all square. Okay, then in the bottom here. Just a six. OK, I believe it&#x27;s actually to find Peter the K 1st 1 Okay, So, Pete, the cables won her on. So the way we&#x27;re gonna do this is that every time you see Kate, we&#x27;re just gonna put in Capel&#x27;s one. Instead, they got a k here on the numerator. This case squared. So we&#x27;re gonna have that. Be careful. One all squared. I&#x27;m rather than chaos. Good. Okay. And that&#x27;s all going to be times by K here. So we&#x27;ve seen a case. We&#x27;re gonna replace that with capers. One, uh, post three. And that&#x27;s gonna be all square together. That could be all over Six capers. One R squared on the top here on dhe K plus four all squared there. That&#x27;s all over sex. Um, yeah, there we go. If you have any questions about that that you can email me at, um daniel dot Shapey hitch. Don&#x27;t know at Gmail. Don&#x27;t come. Thank you very much for listening. And I wish you the best of yesterday.

Problem

Another estimate can be made for an eigenvalue wh…

Problem 10 Easy Difficulty

Answer

$\mu_{5}=9.9319$ and $\mu_{6}=9.9872 .$ The actual eigemmalue is $10 .$

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

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Caleb E.

Samuel H.

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$\mu_{5}=9.9319$ and $\mu_{6}=9.9872 .$ The actual eigemmalue is $10 .$

Linear Algebra and Its Applications

Heather Z.

Caleb E.

Samuel H.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Heather Z.

Caleb E.

Samuel H.

Joseph L.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications