Answered: 3.5 1.5 A = 1.5 -0.5 k = 6

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

See how th e Rayleigh quotient method approximates the dominant eigenvalue more rapidly than the ordinary power method, compute the successive Rayleigh quotients R(x;) for i = 1, ... , k for the matrix A

Expert Solution

Step 1: Given:

See how the Rayleigh quotient method approximates the dominant eigenvalue more rapidly than the ordinary power method,

compute the successive Rayleigh quotients R(xi) for i = 1, ... , k for the matrix A.

Here, the given matrix is,

$A = [\begin{matrix} 3.5 & 1.5 \\ 1.5 & - 0.5 \end{matrix}], x_{0} = [\begin{matrix} 1 \\ 0 \end{matrix}], k = 6$

Step 2: Calculation for the value of x1 and y1:

Here, the given matrix is,

$A = [\begin{matrix} 3.5 & 1.5 \\ 1.5 & - 0.5 \end{matrix}], x_{0} = [\begin{matrix} 1 \\ 0 \end{matrix}], k = 6$

Now, by using the Rayleigh quotients to approximates the dominant eigenvalue in the form,

$R (x) = \frac{((A x) \times x)}{x \times x} = A y_{k - 1}$

whereas, $x_{0} = [\begin{matrix} 1 \\ 0 \end{matrix}]$ , therefore the first step is $y_{0} = [\begin{matrix} 1 \\ 0 \end{matrix}]$

Now,

$x_{1} = A y_{0} x_{1} = [\begin{matrix} 3.5 & 1.5 \\ 1.5 & - 0.5 \end{matrix}] [\begin{matrix} 1 \\ 0 \end{matrix}] x_{1} = [\begin{matrix} 3.5 \\ 1.5 \end{matrix}] A n d, y_{1} = \frac{1}{3.5} x_{1} y_{1} = [\begin{matrix} 1 \\ 0.429 \end{matrix}]$

Step 3: Calculation for the value of R(x1):

Here, $R (x_{1}) = \frac{((A x_{1}) \times x_{1})}{x_{1} \times x_{1}}$

$A x_{1} = [\begin{matrix} 3.5 & 1.5 \\ 1.5 & - 0.5 \end{matrix}] [\begin{matrix} 3.5 \\ 1.5 \end{matrix}] A x_{1} = [\begin{matrix} 14.5 \\ 4.5 \end{matrix}] H e r e, (A x_{1}) \times x_{1} = [\begin{matrix} 14.5 \\ 4.5 \end{matrix}] \times [\begin{matrix} 3.5 \\ 1.5 \end{matrix}] (A x_{1}) \times x_{1} = (14.5) (3.5) + (4.5) (1.5) (A x_{1}) \times x_{1} = 57.50 N o w, x_{1} \times x_{1} = [\begin{matrix} 3.5 \\ 1.5 \end{matrix}] \times [\begin{matrix} 3.5 \\ 1.5 \end{matrix}] x_{1} \times x_{1} = 12.25 + 2.25 x_{1} \times x_{1} = 14.5$

Therefore,

$R (x_{1}) = \frac{57.50}{14.5} R (x_{1}) = 3.966$

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