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Chapter 2: Q40E (page 93)

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

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

Suppose \(AD = {I_m}\) (the \(m \times m\) identity matrix). Show that for any b in \({\mathbb{R}^m}\), the equation \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \) has a solution. (Hint: Think about the equation \(AD{\mathop{\rm b}\nolimits} = {\mathop{\rm b}\nolimits} \).) Explain why Acannot have more rows than columns.

Let Ube the \({\bf{3}} \times {\bf{2}}\) cost matrix described in Example 6 of Section 1.8. The first column of Ulists the costs per dollar of output for manufacturing product B, and the second column lists the costs per dollar of output for product C. (The costs are categorized as materials, labor, and overhead.) Let \({q_1}\) be a vector in \({\mathbb{R}^{\bf{2}}}\) that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year, and let \({q_{\bf{2}}}\), \({q_{\bf{3}}}\) and \({q_{\bf{4}}}\) be the analogous vectors that list the amounts of products B and C manufactured in the second, third, and fourth quarters, respectively. Give an economic description of the data in the matrix UQ, where \(Q = \left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}&{{{\bf{q}}_3}}&{{{\bf{q}}_4}}\end{aligned}} \right)\).

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5鈥�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. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

Solve the equation \(AB = BC\) for A, assuming that A, B, and C are square and Bis invertible.

Assume \(A - s{I_n}\) is invertible and view (8) as a system of two matrix equations. Solve the top equation for \({\bf{x}}\) and substitute into the bottom equation. The result is an equation of the form \(W\left( s \right){\bf{u}} = {\bf{y}}\), where \(W\left( s \right)\) is a matrix that depends upon \(s\). \(W\left( s \right)\) is called the transfer function of the system because it transforms the input \({\bf{u}}\) into the output \({\bf{y}}\). Find \(W\left( s \right)\) and describe how it is related to the partitioned system matrix on the left side of (8). See Exercise 15.
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