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

5. \(\left( {\begin{array}{*{20}{c}}{.1}&{.6}\\{.9}&{.4}\end{array}} \right)\)

Short Answer

Expert verified

The steady-state vector is \({\mathop{\rm q}\nolimits} = \left( {\begin{array}{*{20}{c}}{.4}\\{.6}\end{array}} \right)\).

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:

\(\begin{array}{c}P - I = \left( {\begin{array}{*{20}{c}}{.1}&{.6}\\{.9}&{.4}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - .9}&{.6}\\{.9}&{ - .6}\end{array}} \right)\end{array}\)

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:

\(\left( {\begin{array}{*{20}{c}}{ - .9}&{.6}&0\\{.9}&{ - .6}&0\end{array}} \right)\)

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

At row 1, multiply row 1 by \( - \frac{1}{{0.9}}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - \frac{2}{3}}&0\\{.9}&{ - .6}&0\end{array}} \right)\)

At row 2, multiply row 1 by 0.9 and subtract it from row 2.

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

03

Determine the steady-state vector

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

\(\begin{array}{c}{\mathop{\rm x}\nolimits} = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm x}\nolimits} _1}}\\{{{\mathop{\rm x}\nolimits} _2}}\end{array}} \right)\\ = {{\mathop{\rm x}\nolimits} _2}\left( {\begin{array}{*{20}{c}}{\frac{2}{3}}\\1\end{array}} \right)\end{array}\)

One solution is \(\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right)\). The sum of the entries in \(\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right)\) is 5.

Multiply \({\mathop{\rm x}\nolimits} \) by \(\frac{1}{5}\) to obtain the steady-state vector as shown below:

\(\begin{array}{c}{\mathop{\rm q}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{2}{5}}\\{\frac{3}{5}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{.4}\\{.6}\end{array}} \right]\end{array}\)

Thus, the steady-state vector is \({\mathop{\rm q}\nolimits} = \left( {\begin{array}{*{20}{c}}{.4}\\{.6}\end{array}} \right)\).

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Most popular questions from this chapter

Is it possible for a nonhomogeneous system of seven equations in six unknowns to have a unique solution for some right-hand side of constants? Is it possible for such a system to have a unique solution for every right-hand side? Explain.

Consider the polynomials , and \({p_{\bf{3}}}\left( t \right) = {\bf{2}}\) \({p_{\bf{1}}}\left( t \right) = {\bf{1}} + t,{p_{\bf{2}}}\left( t \right) = {\bf{1}} - t\)(for all t). By inspection, write a linear dependence relation among \({p_{\bf{1}}},{p_{\bf{2}}},\) and \({p_{\bf{3}}}\). Then find a basis for Span\(\left\{ {{p_{\bf{1}}},{p_{\bf{2}}},{p_{\bf{3}}}} \right\}\).

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

22. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 1}&{ - 13}&{ - 12.2}&{ - 1.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

(M) Show that \(\left\{ {t,sin\,t,cos\,{\bf{2}}t,sin\,t\,cos\,t} \right\}\) is a linearly independent set of functions defined on \(\mathbb{R}\). Start by assuming that

\({c_{\bf{1}}} \cdot t + {c_{\bf{2}}} \cdot sin\,t + {c_{\bf{3}}} \cdot cos\,{\bf{2}}t + {c_{\bf{4}}} \cdot sin\,t\,cos\,t = {\bf{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{5}} \right)\)

Equation (5) must hold for all real t, so choose several specific values of t (say, \(t = {\bf{0}},\,.{\bf{1}},\,.{\bf{2}}\)) until you get a system of enough equations to determine that the \({c_j}\) must be zero.

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