In Exercises 1-4, assume that the matrix A is row equivalent to B. Without calculations, list rank A and dim Nul A. Then find bases for Col A, Row A, and Nul A.

\[A = \left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{1}}&{ - {\bf{3}}}&{\bf{7}}&{\bf{9}}&{ - {\bf{9}}}\\{\bf{1}}&{\bf{2}}&{ - {\bf{4}}}&{{\bf{10}}}&{{\bf{13}}}&{ - {\bf{12}}}\\{\bf{1}}&{ - {\bf{1}}}&{ - {\bf{1}}}&{\bf{1}}&{\bf{1}}&{ - {\bf{3}}}\\{\bf{1}}&{ - {\bf{3}}}&{\bf{1}}&{ - {\bf{5}}}&{ - {\bf{7}}}&{\bf{3}}\\{\bf{1}}&{ - {\bf{2}}}&{\bf{0}}&{\bf{0}}&{ - {\bf{5}}}&{ - {\bf{4}}}\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{1}}&{ - {\bf{3}}}&{\bf{7}}&{\bf{9}}&{ - {\bf{9}}}\\{\bf{0}}&{\bf{1}}&{ - {\bf{1}}}&{\bf{3}}&{\bf{4}}&{ - {\bf{3}}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{ - {\bf{1}}}&{ - {\bf{2}}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\end{array}} \right]\]

In the row equivalent matrix B, there are three non-zero rows. Therefore, the rank of A is 3.

Using the rank theorem, you get:

\(\begin{array}{c}{\rm{rank}}\,A + \dim \,{\rm{Nul}}A = n\\3 + \dim \;{\rm{Nul}}\,A = 6\\\dim \;{\rm{Nul}}\,A = 6 - 3\\ = 3\end{array}\)

From matrix B:

\(B = \left[ {\begin{array}{*{20}{c}}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}} \right]\)

The basis of Col A can be written as:

\({\rm{Basis}}\left( {{\rm{Col}}\,A} \right) = \left\{ {\left[ {\begin{array}{*{20}{c}}1\\1\\1\\1\\1\end{array}} \right],\;\;\left[ {\begin{array}{*{20}{c}}1\\2\\{ - 1}\\{ - 3}\\{ - 2}\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}7\\{10}\\1\\{ - 5}\\0\end{array}} \right]} \right\}\)

The basis of the row space of A can be written as:

\[{\rm{Basis}}\left( {{\rm{Row}}\,A} \right) = \left\{ {\left[ {\begin{array}{*{20}{c}}1&1&{ - 3}&7&9&{ - 9}\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}0&1&{ - 1}&3&4&{ - 3}\end{array}} \right],\,\left[ {\begin{array}{*{20}{c}}0&0&0&1&{ - 1}&{ - 2}\end{array}} \right]} \right\}\]

Write the augmented matrix for the system \(B{\bf{x}} = 0\).

\(M = \left[ {\begin{array}{*{20}{c}}1&1&{ - 3}&7&9&{ - 9}&0\\0&1&{ - 1}&3&4&{ - 3}&0\\0&0&0&1&{ - 1}&{ - 2}&0\\0&0&0&0&0&0&0\\0&0&0&0&0&0&0\end{array}} \right]\)

Write the system of equations.

\(\begin{array}{c}{x_1} - 2{x_3} + 9{x_5} + 2{x_6} = 0\\{x_2} - {x_3} + 7{x_5} + 3{x_6} = 0\\{x_4} - {x_5} - 2{x_6} = 0\end{array}\)

Consider \({x_3}\), \({x_5}\), and \({x_6}\) as free variables.

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\\{{x_6}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{2{x_3} - 9{x_5} - 2{x_6}}\\{{x_3} - 7{x_5} - 3{x_6}}\\{{x_3}}\\{{x_5} + 2{x_6}}\\{{x_5}}\\{{x_6}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}2\\1\\1\\0\\0\\0\end{array}} \right]{x_3} + \left[ {\begin{array}{*{20}{c}}{ - 9}\\{ - 7}\\0\\1\\1\\0\end{array}} \right]{x_5} + \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 3}\\0\\2\\0\\1\end{array}} \right]{x_6}\end{array}\]

So, the null space of A is \[\left\{ {\left[ {\begin{array}{*{20}{c}}2\\1\\1\\0\\0\\0\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{ - 9}\\{ - 7}\\0\\1\\1\\0\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 3}\\0\\2\\0\\1\end{array}} \right]} \right\}\].