In Exercises 1–4, the matrix A is followed by a sequence \(\left\{ {{{\bf{x}}_k}} \right\}\) produced by the power method. Use these data to estimate the largest eigenvalue of A, and give a corresponding eigenvector.

1. \(A = \left( {\begin{aligned}{ {20}{c}}4&3\\1&2\end{aligned}} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}1\\0\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}1\\{.25}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}1\\{.3158}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}1\\{.3298}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}1\\{.3326}\end{aligned}} \right)\)

A matrix \(A = \left( {\begin{aligned}{ {20}{l}}4&3\\1&2\end{aligned}} \right)\). A sequence \(\left\{ {{x_k}} \right\}\).

Compute the value of\(A{x_k}\)and identify the largest entry as follows:

\(A{x_0} = \left( {\begin{aligned}{ {20}{l}}4&3\\1&2\end{aligned}} \right)\left( {\begin{aligned}{ {20}{l}}1\\0\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}4\\1\end{aligned}} \right)\). The greatest entry is\({\mu _0} = 4\).

\(A{x_1} = \left( {\begin{aligned}{ {20}{c}}4&3\\1&2\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}1\\{.25}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{4.75}\\{1.5}\end{aligned}} \right)\). The greatest entry is\({\mu _1} = 4.75\).

\(A{x_2} = \left( {\begin{aligned}{ {20}{c}}4&3\\1&2\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}1\\{.3158}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}{4.9474}\\{1.6316}\end{aligned}} \right)\). The greatest entry is\({\mu _2} = 4.9474\).

\(A{x_3} = \left( {\begin{aligned}{ {20}{c}}4&3\\1&2\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}1\\{.3298}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{4.9894}\\{1.6596}\end{aligned}} \right)\). The greatest entry is\({\mu _3} = 4.9894\).

\(A{x_4} = \left( {\begin{aligned}{ {20}{c}}4&3\\1&2\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}1\\{.3326}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{4.9978}\\{1.6652}\end{aligned}} \right)\). The greatest entry is\({\mu _4} = 4.9978\).

Hence, the largest absolute entry is 4.9978. So, the eigenvalue is equal to 4.9978. The corresponding eigenvector is \(\left( {\begin{aligned}{ {20}{c}}1\\{.3326}\end{aligned}} \right)\).