Use the power method to approximate the dominant eigenvalue and eigenvector of A. Use th e given initial vector x0 , th e specified number of iterations k, and three-decimal-place accuracy. 3 11 3 1|,Хо1,k = 60 1 3

The given matrix is A=310131013 and initial vector is x0=111 and number of iteration k=6 The power…

Answered: 3 1 1 3 1|,Хо 1,k = 6 0 1 3

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Use the power method to approximate the dominant eigenvalue and eigenvector of A. Use th e given initial vector x₀ , th e specified number of iterations k, and three-decimal-place accuracy.

Expert Solution

Step 1

The given matrix is $A = [\begin{matrix} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{matrix}]$ and initial vector is $x_{0} = [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}]$ and number of iteration $k = 6$

The power method for approximating eigenvalues is iterative. First let assume that the matrix A has a dominant eigenvalue with corresponding dominant eigenvectors. Then choose an initial approximation of one of the dominant eigenvectors of A. This initial approximation must be a nonzero vector in Rn.
Then find

$x_{1} = A x_{0} x_{2} = A x_{1} x_{2} = A (A x_{0}) x_{2} = A^{2} x_{0} . . x_{6} = A^{6} x_{0}$

Step 2

First multiply the given matrix $A = [\begin{matrix} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{matrix}]$ with the given initial matrix $x_{0} = [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}]$ .

$A x_{0} = [\begin{matrix} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] A x_{0} = [\begin{matrix} 3 \cdot 1 + 1 \cdot 1 + 0 \cdot 1 \\ 1 \cdot 1 + 3 \cdot 1 + 1 \cdot 1 \\ 0 \cdot 1 + 1 \cdot 1 + 3 \cdot 1 \end{matrix}] A x_{0} = [\begin{matrix} 4 \\ 5 \\ 4 \end{matrix}]$

Now take $4$ common out of this matrix it becomes:

$A x_{0} = 4 [\begin{matrix} 1 \\ 1.25 \\ 1 \end{matrix}]$

Step 3

Now find $A x_{2}$

$A x_{2} = A^{2} x_{0}$

Find $A^{2}$

$A^{2} = [\begin{matrix} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{matrix}] \cdot [\begin{matrix} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{matrix}] A^{2} = [\begin{matrix} 10 & 6 & 1 \\ 6 & 11 & 6 \\ 1 & 6 & 10 \end{matrix}]$

Now multiply $A^{2}$ by $x_{0}$ gives:

$A^{2} x_{0} = [\begin{matrix} 10 & 6 & 1 \\ 6 & 11 & 6 \\ 1 & 6 & 10 \end{matrix}] \cdot [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}]$

$A^{2} x_{0} = [\begin{matrix} 17 \\ 23 \\ 17 \end{matrix}]$

Take $17$ common from the matrix gives:

$A^{2} x_{0} = 17 [\begin{matrix} 1 \\ 1.352 \\ 1 \end{matrix}]$

Now for third iteration as from the above process find $A x_{3} = A^{3} x_{0}$

$A^{3} = [\begin{matrix} 10 & 6 & 1 \\ 6 & 11 & 6 \\ 1 & 6 & 10 \end{matrix}] [\begin{matrix} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{matrix}] A^{3} = [\begin{matrix} 48 & 17 & 21 \\ 51 & 23 & 51 \\ 21 & 17 & 48 \end{matrix}]$

Now multiply $A^{3}$ with $x_{0}$

$A^{3} x_{0} = [\begin{matrix} 48 & 17 & 21 \\ 51 & 23 & 51 \\ 21 & 17 & 48 \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] A^{3} x_{0} = [\begin{matrix} 76 \\ 123 \\ 86 \end{matrix}]$

Take $86$ common from the above matrix gives:

$A^{3} x_{0} = 86 [\begin{matrix} 0.883 \\ 1.430 \\ 1 \end{matrix}]$

Step 4

Now find $A x_{4}$

$A x_{4} = A^{4} x_{0}$

$A^{4} = A^{3} \cdot A A^{4} = [\begin{matrix} 48 & 17 & 21 \\ 51 & 23 & 51 \\ 21 & 17 & 48 \end{matrix}] [\begin{matrix} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{matrix}] A^{4} = [\begin{matrix} 161 & 110 & 50 \\ 174 & 165 & 174 \\ 80 & 120 & 161 \end{matrix}]$

Now multiply $A^{4}$ with $x_{0}$

$A^{4} x_{0} = [\begin{matrix} 161 & 110 & 50 \\ 174 & 165 & 174 \\ 80 & 120 & 161 \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] A^{4} x_{0} = [\begin{matrix} 321 \\ 513 \\ 361 \end{matrix}]$

Take $361$ common from the above matrix gives:

$A^{4} x_{0} = 361 [\begin{matrix} 0.889 \\ 1.421 \\ 1 \end{matrix}]$

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