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Chapter 4: Q20SE (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 becontrollableif 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

Justify the following equalities:

a.\({\rm{dim Row }}A{\rm{ + dim Nul }}A = n{\rm{ }}\)

b.\({\rm{dim Col }}A{\rm{ + dim Nul }}{A^T} = m\)

Suppose a \({\bf{4}} \times {\bf{7}}\) matrix A has four pivot columns. Is \({\bf{Col}}\,A = {\mathbb{R}^{\bf{4}}}\)? Is \({\bf{Nul}}\,A = {\mathbb{R}^{\bf{3}}}\)? Explain your answers.

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

19. \(A = \left( {\begin{array}{*{20}{c}}{.9}&1&0\\0&{ - .9}&0\\0&0&{.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}0\\1\\1\end{array}} \right)\).

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

Let \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\). Find \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\) such that \(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\2&{ - 6}&8\end{array}} \right] = {{\mathop{\rm uv}\nolimits} ^T}\) .
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