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Question 16:Produce the general solution of the dynamical system \({x_{k + 1}} = A{x_k}\) when \(A\) is the stochastic matrix for the Hertz Rent A Car model in Exercise 16 of Section 4.9.

Short Answer

Expert verified

The required general solution is \({\bf{x}}\left( t \right) = {c_1}{{\bf{v}}_1} + {c_2}{\left( {0.89} \right)^k}{{\bf{v}}_2} + {c_3}{\left( {0.81} \right)^k}{{\bf{v}}_3}\).

Step by step solution

01

Determine the eigenvalues and eigenvector of the matrix using MATLAB

From exercise 16, we have:

\(A = \left( {\begin{array}{*{20}{c}}{0.90}&{0.01}&{0.09}\\{0.01}&{0.90}&{0.01}\\{0.09}&{0.09}&{0.90}\end{array}} \right)\)

Enter this matrix in MATLAB as:

>> \(A = \left( {\begin{array}{*{20}{c}}{0.90\,\,0.01\,\,0.09;}&{0.01\,\,0.90\,\,0.01;}&{0.09\,\,0.09\,\,0.90}\end{array}} \right)\);

For eigenvalues, enter instruction as:

>> \(E = {\rm{eigs}}\left( A \right)\);

We get:

\(E = \left( {\begin{array}{*{20}{c}}{0.81}\\{0.89}\\{1.00}\end{array}} \right)\)

Now, for eigenvectors, enter instruction as:

>> \(\left( {\begin{array}{*{20}{c}}A&B\end{array}} \right) = {\rm{eigs}}\left( A \right)\);

So, we have:

\(\begin{array}{l}{v_1} = \left( {\begin{array}{*{20}{c}}{ - 0.6700}\\{ - 0.1399}\\{ - 0.7289}\end{array}} \right)\\{v_2} = \left( {\begin{array}{*{20}{c}}{0.7071}\\{ - 0.7071}\\{ - 0.0000}\end{array}} \right)\\{v_3} = \left( {\begin{array}{*{20}{c}}{ - 0.7071}\\{0.0000}\\{0.7071}\end{array}} \right)\end{array}\)

Thus, these are the required eigenvectors.

02

The General Solution to the system.

Now, the general solution can be given as:

\({\bf{x}}\left( t \right) = {c_1}{{\bf{v}}_1} + {c_2}{\left( {0.89} \right)^k}{{\bf{v}}_2} + {c_3}{\left( {0.81} \right)^k}{{\bf{v}}_3}\)

Hence, this is the required solution.

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Most popular questions from this chapter

19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

23. Let \(p\) be the polynomial in Exercise \({\bf{22}}\), and suppose the equation \(p\left( t \right) = {\bf{0}}\) has distinct roots \({\lambda _{\bf{1}}},{\lambda _{\bf{2}}},{\lambda _{\bf{3}}}\). Let \(V\) be the Vandermonde matrix

\(V{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{1}}&{\bf{1}}\\{{\lambda _{\bf{1}}}}&{{\lambda _{\bf{2}}}}&{{\lambda _{\bf{3}}}}\\{\lambda _{\bf{1}}^{\bf{2}}}&{\lambda _{\bf{2}}^{\bf{2}}}&{\lambda _{\bf{3}}^{\bf{2}}}\end{aligned}} \right)\)

(The transpose of \(V\) was considered in Supplementary Exercise \({\bf{11}}\) in Chapter \({\bf{2}}\).) Use Exercise \({\bf{22}}\) and a theorem from this chapter to deduce that \(V\) is invertible (but do not compute \({V^{{\bf{ - 1}}}}\)). Then explain why \({V^{{\bf{ - 1}}}}{C_p}V\) is a diagonal matrix.

For the Matrices A find real closed formulas for the trajectory x(t+1)=Ax(t)wherex(0)=[01]A=[2-332]

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

14. \(\left( {\begin{array}{*{20}{c}}4&0&{ - 2}\\2&5&4\\0&0&5\end{array}} \right)\)

Question: Is \(\lambda = 2\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\)? Why or why not?

In Exercises \({\bf{3}}\) and \({\bf{4}}\), use the factorization \(A = PD{P^{ - {\bf{1}}}}\) to compute \({A^k}\) where \(k\) represents an arbitrary positive integer.

4. \(\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\)

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