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Question: In Exercises5-8, find the steady-state vector.

7. \(\left( {\begin{array}{*{20}{c}}{.7}&{.1}&{.1}\\{.2}&{.8}&{.2}\\{.1}&{.1}&{.7}\end{array}} \right)\)

Short Answer

Expert verified

The steady-state vector is \({\mathop{\rm q}\nolimits} = \left( {\begin{array}{*{20}{c}}{.25}\\{.5}\\{.25}\end{array}} \right)\).

Step by step solution

01

Compute \(P - I\)

The equation \(P{\mathop{\rm x}\nolimits} = {\mathop{\rm x}\nolimits} \) can be solved by rewriting it as \(\left( {P - I} \right){\mathop{\rm x}\nolimits} = 0\).

Compute \(P - I\) as shown below:

\(\begin{array}{c}P - I = \left( {\begin{array}{*{20}{c}}{.7}&{.1}&{.1}\\{.2}&{.8}&{.2}\\{.1}&{.1}&{.7}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - .3}&{.1}&{.1}\\{.2}&{ - .2}&{.2}\\{.1}&{.1}&{ - .3}\end{array}} \right)\end{array}\)

02

Write the augmented matrix

Write the augmented matrix for the homogeneous system \(\left( {P - I} \right){\mathop{\rm x}\nolimits} = 0\)as shown below:

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

Perform an elementary row operation to produce the row-reduced echelon form of the matrix.

At row 1, multiply row 1 by \( - \frac{1}{{0.3}}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - .333}&{ - .3333}&0\\{.2}&{ - .2}&{.2}&0\\{.1}&{.1}&{ - .3}&0\end{array}} \right)\)

At row 2, multiply row 1 by 0.2 and subtract it from row 2. At row 3, multiply row 1 by 0.1 and subtract it from row 3.

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - .333}&{ - .3333}&0\\0&{ - 0.1333}&{0.2666}&0\\0&{0.1333}&{0.2666}&0\end{array}} \right)\)

At row 2, multiply row 2 by \( - \frac{1}{{0.1333}}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - .333}&{ - .3333}&0\\0&1&{ - 2}&0\\0&{0.1333}&{0.2666}&0\end{array}} \right)\)

At row 1, multiply row 2 by 0.333 and add it to row 1. At row 3, multiply row 2 by 0.1333 and subtract it from row 3.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{ - 1}&0\\0&1&{ - 2}&0\\0&0&0&0\end{array}} \right)\)

03

Determine the steady-state vector

The general solution of the equation \(\left( {P - I} \right){\mathop{\rm x}\nolimits} = 0\)is shown below:

\(\begin{array}{c}{\mathop{\rm x}\nolimits} = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm x}\nolimits} _1}}\\{{{\mathop{\rm x}\nolimits} _2}}\\{{{\mathop{\rm x}\nolimits} _3}}\end{array}} \right)\\ = {{\mathop{\rm x}\nolimits} _3}\left( {\begin{array}{*{20}{c}}1\\2\\1\end{array}} \right)\end{array}\)

One solution is \(\left( {\begin{array}{*{20}{c}}1\\2\\1\end{array}} \right)\). The sum of the entries of \(\left( {\begin{array}{*{20}{c}}1\\2\\1\end{array}} \right)\) is 4.

Multiply \({\mathop{\rm x}\nolimits} \) by \(\frac{1}{4}\) as shown below:

\(\begin{array}{c}{\mathop{\rm q}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{1}{4}}\\{\frac{1}{2}}\\{\frac{1}{4}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{.25}\\{.5}\\{.25}\end{array}} \right)\end{array}\)

Thus, the steady-state vector is \({\mathop{\rm q}\nolimits} = \left( {\begin{array}{*{20}{c}}{.25}\\{.5}\\{.25}\end{array}} \right)\).

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Most popular questions from this chapter

Explain what is wrong with the following discussion: Let \({\bf{f}}\left( t \right) = {\bf{3}} + t\) and \({\bf{g}}\left( t \right) = {\bf{3}}t + {t^{\bf{2}}}\), and note that \({\bf{g}}\left( t \right) = t{\bf{f}}\left( t \right)\). Then, \(\left\{ {{\bf{f}},{\bf{g}}} \right\}\) is linearly dependent because g is a multiple of f.

Let \(V\) and \(W\) be vector spaces, and let \(T:V \to W\) be a linear transformation. Given a subspace \(U\) of \(V\), let \(T\left( U \right)\) denote the set of all images of the form \(T\left( {\mathop{\rm x}\nolimits} \right)\), where x is in \(U\). Show that \(T\left( U \right)\) is a subspace of \(W\).

Define by \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 1 \right)}\end{array}} \right)\). For instance, if \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + 5t + 7{t^2}\), then \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}3\\{15}\end{array}} \right)\).

  1. Show that \(T\) is a linear transformation. (Hint: For arbitrary polynomials p, q in \({{\mathop{\rm P}\nolimits} _2}\), compute \(T\left( {{\mathop{\rm p}\nolimits} + {\mathop{\rm q}\nolimits} } \right)\) and \(T\left( {c{\mathop{\rm p}\nolimits} } \right)\).)
  2. Find a polynomial p in \({{\mathop{\rm P}\nolimits} _2}\) that spans the kernel of \(T\), and describe the range of \(T\).

In Exercise 2, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

2. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{5}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{7}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{5}}}\end{array}} \right)\)

(M) Determine whether w is in the column space of \(A\), the null space of \(A\), or both, where

\({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\1\\{ - 1}\\{ - 3}\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1\\{ - 5}&{ - 1}&0&{ - 2}\\9&{ - 11}&7&{ - 3}\\{19}&{ - 9}&7&1\end{array}} \right)\)

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