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.

\[A = \left[ {\begin{array}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}&{\bf{6}}&{\bf{2}}&{\bf{5}}\\{ - {\bf{2}}}&{\bf{3}}&{ - {\bf{3}}}&{ - {\bf{3}}}&{ - {\bf{4}}}\\{\bf{4}}&{ - {\bf{6}}}&{\bf{9}}&{\bf{5}}&{\bf{9}}\\{ - {\bf{2}}}&{\bf{3}}&{\bf{3}}&{ - {\bf{4}}}&{\bf{1}}\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}&{\bf{6}}&{\bf{2}}&{\bf{5}}\\{\bf{0}}&{\bf{0}}&{\bf{3}}&{ - {\bf{1}}}&{\bf{1}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{3}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\end{array}} \right]\]

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:

\(\begin{array}{c}{\rm{rank}}\,A + \dim \,{\rm{Nul}}A = n\\3 + \dim \;{\rm{Nul}}\,A = 5\\\dim \;{\rm{Nul}}\,A = 5 - 3\\ = 2\end{array}\)

From matrix B:

\(B = \left[ {\begin{array}{*{20}{c}}2&{ - 3}&6&2&5\\0&0&3&{ - 1}&1\\0&0&0&1&3\\0&0&0&0&0\end{array}} \right]\)

The basis of Col A can be written as:

\({\rm{Basis}}\left( {{\rm{Col}}\,A} \right) = \left\{ {\left[ {\begin{array}{*{20}{c}}2\\{ - 2}\\4\\{ - 2}\end{array}} \right],\;\;\left[ {\begin{array}{*{20}{c}}6\\{ - 3}\\9\\3\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}2\\{ - 3}\\5\\{ - 4}\end{array}} \right]} \right\}\)

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

\({\rm{basis}}\left( {{\rm{Row}}\,A} \right) = \left\{ {\left[ {\begin{array}{*{20}{c}}2&{ - 3}&6&2&5\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}0&0&3&{ - 1}&{ - 1}\end{array}} \right],\,\left[ {\begin{array}{*{20}{c}}0&0&0&1&3\end{array}} \right]} \right\}\)

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

\(M = \left[ {\begin{array}{*{20}{c}}2&{ - 3}&6&2&5&0\\0&0&3&{ - 1}&1&0\\0&0&0&1&3&0\\0&0&0&0&0&0\end{array}} \right]\)

Write the row reduced echelon form of matrix M.

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

Write the system of equations.

\(\begin{array}{c}{x_1} = \frac{3}{2}{x_2} + \frac{9}{2}{x_5}\\{x_3} = - \frac{4}{3}{x_5}\\{x_4} = - 3{x_5}\end{array}\)

Consider \({x_2}\) and \({x_5}\) as free variables.

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{3}{2}{x_2} + \frac{9}{2}{x_5}}\\{{x_2}}\\{ - \frac{4}{3}{x_5}}\\{ - 3{x_5}}\\{{x_5}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{\frac{3}{2}}\\1\\0\\0\\0\end{array}} \right]{x_2} + \left[ {\begin{array}{*{20}{c}}{\frac{9}{2}}\\0\\{ - \frac{4}{3}}\\{ - 3}\\1\end{array}} \right]{x_5}\end{array}\]

So, the null space of A is \[\left\{ {\left[ {\begin{array}{*{20}{c}}{\frac{3}{2}}\\1\\0\\0\\0\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{\frac{9}{2}}\\0\\{ - \frac{4}{3}}\\{ - 3}\\1\end{array}} \right]} \right\}\].