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skad [1K]
3 years ago
5

find the spectral radius of A. Is this a convergent matrix? Justify your answer. Find the limit x=lim x^(k) of vector iteration

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
4 0

Answer:

The solution to this question can be defined as follows:

Step-by-step explanation:

Please find the complete question in the attached file.

A = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right]

now for given values:

\left[\begin{array}{ccc} \frac{3}{4} - \lambda & \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2} - \lambda & 0\\ -\frac{1}{4}& -\frac{1}{4} & 0 -\lambda \end{array}\right]=0 \\\\

\to  (\frac{3}{4} - \lambda ) [-\lambda (\frac{1}{2} - \lambda ) -0] - 0 - \frac{1}{4}[0- \frac{1}{2} (\frac{1}{2} - \lambda )] =0 \\\\\to  (\frac{3}{4} - \lambda ) [(\frac{\lambda}{2} + \lambda^2 )] - \frac{1}{4}[\frac{\lambda}{2} -  \frac{1}{4}] =0 \\\\\to  (\frac{3}{8}\lambda + \frac{3}{4} \lambda^2 - \frac{\lambda^2}{2} - \lambda^3 - \frac{\lambda}{8} + \frac{1}{16}=0 \\\\\to (\lambda - \frac{1}{2}) (\lambda -\frac{1}{4}) (\lambda - \frac{1}{2}) =0\\\\

\to \lambda_1=\lambda_2 =\frac{1}{2}\\\\\to \lambda_3 = \frac{1}{4} \\\\\to A = max {|\lambda_1| , |\lambda_2|, |\lambda_3|}\\\\

       = max{\frac{1}{2}, \frac{1}{2}, \frac{1}{4}}\\\\= \frac{1}{2}\\\\(A) =\frac{1}{2}

In point b:

Its  

spectral radius is less than 1 hence matrix is convergent.

In point c:

\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) =   \left(\begin{array}{c}3&1&2\end{array}\right)  , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\  \to x^{(k+1)} =  \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right]  \\\\

after solving the value the answer is

:

\lim_{k \to \infty} x^k=o  = \left[\begin{array}{c}0&0&0\end{array}\right]

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Answer:

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Step-by-step explanation:

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The following picture is a square pyramid where DE=8 cm and m∠ADE=60°.
Goryan [66]

<u>Given</u>:

The length of DE is 8 cm and the measure of ∠ADE is 60°.

We need to determine the surface area of the pyramid.

<u>Length of AD:</u>

The length of AD is given by

cos 60^{\circ}=\frac{FD}{8}

       4=FD

Length of AD = 8 cm

<u>Slant height:</u>

The slant height EF can be determined using the trigonometric ratio.

Thus, we have;

     sin \ 60^{\circ}=\frac{EF}{8}

sin \ 60^{\circ} \times 8=EF

      \frac{\sqrt{3}}{2} \times 8=EF

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Thus, the slant height EF is 4√3

<u>Surface area of the square pyramid:</u>

The surface area of the square pyramid can be determined using the formula,

SA=Area \ of \ square + \frac{1}{2} (Perimeter \ of \ base ) (slant \ height)

Substituting the values, we have;

SA=8^2+\frac{1}{2}(8+8+8+8)(4 \sqrt{3})

SA=64+\frac{1}{2}(32)(4 \sqrt{3})

SA=64+(16)(4 \sqrt{3})

SA=64+64 \sqrt{3}

The exact form of the area of the square pyramid is 64+64 \sqrt{3}

Substituting √3 = 1.732 in the above expression, we have;

SA = 64 + 110.848

SA = 174.848

Rounding off to one decimal place, we get;

SA = 174.8

Thus, the area of the square pyramid is 174.8 cm²

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2 years ago
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