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

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

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

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:

\(\left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0&1&0\\1&{ - .9}&{.81}\\1&{.5}&{.25}\end{array}} \right)\)

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

Interchange rows 1 and 2.

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - .9}&{0.81}\\0&1&0\\1&{.5}&{.25}\end{array}} \right)\)

At row 3, subtract row 1 from row 3.

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - .9}&{0.81}\\0&1&0\\0&{1.4}&{ - 0.56}\end{array}} \right)\)

At row 1, multiply row 2 by 0.9 and add it to row 1. At row 3, multiply row 2 by 1.4 and subtract it from row 3.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{0.81}\\0&1&0\\0&0&{ - 0.56}\end{array}} \right)\)

At row 3, multiply row 3 by \( - \frac{1}{{0.56}}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{0.81}\\0&1&0\\0&0&1\end{array}} \right)\)

At row 1, multiply row 3 by 0.81 and subtract it from row 1.

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

The matrix has three pivot columns, so the rank of the matrix is 3.

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 3.

Thus, the matrix pairs \(\left( {A,B} \right)\) are controllable.