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Chapter 4: Q 20SE (page 191)

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

20. \(A = \left( {\begin{array}{*{20}{c}}{.8}&{ - .3}&0\\{.2}&{.5}&1\\0&0&{ - .5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\1\\0\end{array}} \right)\).

Short Answer

Expert verified

The matrix pairs \(\left( {A,B} \right)\) are not controllable.

Step by step solution

01

State the rank of a matrix

Therank of matrix\(A\), denoted by rank\(A\), is the dimension of the column spaceof \(A\).

02

Determine the rank of the matrix

Calculate the rank of the matrix \(\left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}\end{array}} \right)\) to determine whether the matrix pair \(\left( {A,B} \right)\) is controllable.

Write the augmented matrix as shown below:

\(\left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{.5}&{.19}\\1&{.7}&{.45}\\0&0&0\end{array}} \right)\)

The matrix has 2 pivot columns, so the rank of the matrix must be less than 3.

03

Determine whether the matrix pairs are controllable

The pair \(\left( {A,B} \right)\) is said to becontrollable if rank\(\left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}& \cdots &{{A^{n - 1}}B}\end{array}} \right) = n\).

The rank of the matrix is less than 3.

Thus, the matrix pairs \(\left( {A,B} \right)\) are not controllable.

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Most popular questions from this chapter

In Exercise 1, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

1. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{6}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{array}} \right)\)

In Exercise 18, Ais an \(m \times n\) matrix. Mark each statement True or False. Justify each answer.

18. a. If B is any echelon form of A, then the pivot columns of B form a basis for the column space of A.

b. Row operations preserve the linear dependence relations among the rows of A.

c. The dimension of the null space of A is the number of columns of A that are not pivot columns.

d. The row space of \({A^T}\) is the same as the column space of A.

e. If A and B are row equivalent, then their row spaces are the same.

Let \(A\) be any \(2 \times 3\) matrix such that \({\mathop{\rm rank}\nolimits} A = 1\), let u be the first column of \(A\), and suppose \({\mathop{\rm u}\nolimits} \ne 0\). Explain why there is a vector v in \({\mathbb{R}^3}\) such that \(A = {{\mathop{\rm uv}\nolimits} ^T}\). How could this construction be modified if the first column of \(A\) were zero?

Verify that rank \({{\mathop{\rm uv}\nolimits} ^T} \le 1\) if \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 3}\\5\end{array}} \right]\) and \({\mathop{\rm v}\nolimits} = \left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right]\).

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show that a subset \(\left\{ {{{\bf{u}}_1},...,{{\bf{u}}_p}} \right\}\) in V is linearly independent if and only if the set of coordinate vectors \(\left\{ {{{\left( {{{\bf{u}}_{\bf{1}}}} \right)}_B},.....,{{\left( {{{\bf{u}}_p}} \right)}_B}} \right\}\) is linearly independent in \({\mathbb{R}^n}\)(Hint: Since the coordinate mapping is one-to-one, the following equations have the same solutions, \({c_{\bf{1}}}\),….,\({c_p}\))

\({c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p} = {\bf{0}}\) The zero vector V

\({\left( {{c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p}} \right)_B} = {\left( {\bf{0}} \right)_B}\) The zero vector in \({\mathbb{R}^n}\)a
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