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In Exercises $1-4,$ the matrix $A$ is followed by a sequence $\left\{\mathbf{x}_{k}\right\}$ produced by the power method. Use these data to estimate the largest eigenvalue of $A,$ and give a corresponding eigenvector.$$\begin{array}{l}{A=\left[\begin{array}{cc}{.5} & {.2} \\ {.4} & {.7}\end{array}\right]} \\ {\left[\begin{array}{l}{1} \\ {0}\end{array}\right],\left[\begin{array}{l}{1} \\ {.8}\end{array}\right],\left[\begin{array}{c}{.6875} \\ {1}\end{array}\right],\left[\begin{array}{c}{.5577} \\ {1}\end{array}\right],\left[\begin{array}{c}{.5188} \\ {1}\end{array}\right]}\end{array}$$

Eigenvector $x_{4}=[.5188]$ or $A x_{4}\left[\begin{array}{c}{.4594} \\ {.90752}\end{array}\right], \mu_{4}=.90752$Eigenvalue $\lambda \approx .90752$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 8

Iterative Estimates for Eigenvalues

Vectors

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Okay, So in this question, we want to basically estimate the lives item value and Becca off this aim Agents. Right here. So what I have right here is a heightened Smoot where we basically do the calculations. So a is 1.818 minus 0.8 minus trip 12 and 412 and then these out out X dollars. Right here. So what we do in this step is we perform a and multiply it by X Y. So for x zero, we have 01 and x one. We have minus 0.5 and one l can t to do that. And then for the same, I will also print out the maximum value in the entry so the results can be seen here, where a eight times zero is split up. This vector and the large sentry being forward to sending for this as any for base. Yes, as a night. You had to be careful when interpreting this because I took the transfers here. So this is a one robot to calm. So one by two matrix. But you should actually be a two by one. But I just put the transfer there because it's easy to be so. What you see is that mule zero is for one to new 16 Well, two is 5.1 and so on. So on, so on. And so I. Invicta is Expo, which is negative. 0.2. Factor zero and one, or a X four is equal to minus 1.2536 and 50064 with Reagan Value 5.64

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