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Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

Short Answer

Expert verified

As the power increases, the matrix becomes more like \(\left( {\begin{aligned}{*{20}{c}}{1/3}&{1/3}&{1/3}\\{1/3}&{1/3}&{1/3}\\{1/3}&{1/3}&{1/3}\end{aligned}} \right)\).

Step by step solution

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01

Create the matrix

Consider the matrix\(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

Write the above matrix in MATLAB command.

\( > > {\rm{A}} = \left( {{\rm{1/6 1/2 1/3; 1/2 1/4 1/4; 1/3 1/4 5/12}}} \right){\rm{;}}\)

02

Obtain the matrix by using the MATLAB command

Obtain the matrix\({A^5}\)by using the MATLAB command shown below:

\( > > A\^{\bf{5}}\)

The output matrix is shown below:

\({A^5} = \left( {\begin{aligned}{*{20}{c}}{0.331822273662551}&{0.334587191368025}&{0.333590534979424}\\{0.334587191358025}&{0.332320601851852}&{0.333092206790123}\\{0.333590534979424}&{0.333092206790123}&{0.333317258230453}\end{aligned}} \right)\)

Thus, \({A^5} = \left( {\begin{aligned}{*{20}{c}}{0.331822273662551}&{0.334587191368025}&{0.333590534979424}\\{0.334587191358025}&{0.332320601851852}&{0.333092206790123}\\{0.333590534979424}&{0.333092206790123}&{0.333317258230453}\end{aligned}} \right)\).

03

Obtain the matrix by using the MATLAB command

Obtain the matrix\({A^{10}}\)by using the MATLAB command shown below:

\( > > A\^10\)

The output matrix is shown below:

\({A^{10}} = \left( {\begin{aligned}{*{20}{c}}{0.333337254997295}&{0.333330106839401}&{0.333332638213305}\\{0.333330106839401}&{0.333335989260343}&{0.333333903900257}\\{0.333332638213305}&{0.333333903900257}&{0.333333457886439}\end{aligned}} \right)\)

Thus, \({A^{10}} = \left( {\begin{aligned}{*{20}{c}}{0.333337254997295}&{0.333330106839401}&{0.333332638213305}\\{0.333330106839401}&{0.333335989260343}&{0.333333903900257}\\{0.333332638213305}&{0.333333903900257}&{0.333333457886439}\end{aligned}} \right)\).

04

Obtain the matrix by using the MATLAB command

Obtain the matrix\({A^{20}}\)by using the MATLAB command shown below:

\( > > A\^{\bf{20}}\)

The output matrix is shown below:

\({A^{20}} = \left( {\begin{aligned}{*{20}{c}}{0.33333}&{0.33333}&{0.33333}\\{0.33333}&{0.33333}&{0.33333}\\{0.33333}&{0.33333}&{0.33333}\end{aligned}} \right)\)

Thus, \({A^{20}} = \left( {\begin{aligned}{*{20}{c}}{0.33333}&{0.33333}&{0.33333}\\{0.33333}&{0.33333}&{0.33333}\\{0.33333}&{0.33333}&{0.33333}\end{aligned}} \right)\).

05

Obtain the matrix by using the MATLAB command

Obtain the matrix\({A^{30}}\)by using the MATLAB command shown below:

\( > > A\^{\bf{30}}\)

The output matrix is shown below:

\({A^{30}} \approx \left( {\begin{aligned}{*{20}{c}}{0.33333}&{0.33333}&{0.33333}\\{0.33333}&{0.33333}&{0.33333}\\{0.33333}&{0.33333}&{0.33333}\end{aligned}} \right)\)

Thus,\({A^{30}} = \left( {\begin{aligned}{*{20}{c}}{0.33333}&{0.33333}&{0.33333}\\{0.33333}&{0.33333}&{0.33333}\\{0.33333}&{0.33333}&{0.33333}\end{aligned}} \right)\).

It is observed that as the power increases, the matrix becomes more like \(\left( {\begin{aligned}{*{20}{c}}{0.3333}&{0.3333}&{0.3333}\\{0.3333}&{0.3333}&{0.

3333}\\{0.3333}&{0.3333}&{0.3333}\end{aligned}} \right)\)or \(\left( {\begin{aligned}{*{20}{c}}{1/3}&{1/3}&{1/3}\\{1/3}&{1/3}&{1/3}\\{1/3}&{1/3}&{1/3}\end{aligned}} \right)\).

All the entries in\({A^{20}}\)and\({A^{30}}\)approach to 0.3333.

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Most popular questions from this chapter

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

5. \[\left[ {\begin{array}{*{20}{c}}A&B\\C&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&Y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\Z&{\bf{0}}\end{array}} \right]\]

Prove the Theorem 3(d) i.e., \({\left( {AB} \right)^T} = {B^T}{A^T}\).

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

6. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

Prove Theorem 2(b) and 2(c). Use the row-column rule. The \(\left( {i,j} \right)\)- entry in \(A\left( {B + C} \right)\) can be written as \({a_{i1}}\left( {{b_{1j}} + {c_{1j}}} \right) + ... + {a_{in}}\left( {{b_{nj}} + {c_{nj}}} \right)\) or \(\sum\limits_{k = 1}^n {{a_{ik}}\left( {{b_{kj}} + {c_{kj}}} \right)} \).

Use the inverse found in Exercise 1 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)

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