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.

\(A = \left[ {\begin{array}{*{20}{c}}{\bf{1}}&{ - {\bf{4}}}&{\bf{9}}&{ - {\bf{7}}}\\{ - {\bf{1}}}&{\bf{2}}&{ - {\bf{4}}}&{\bf{1}}\\{\bf{5}}&{ - {\bf{6}}}&{{\bf{10}}}&{\bf{7}}\end{array}} \right]\)

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

In the row equivalent matrix B, there are two non-zero rows. Therefore, the rank of A is 2.

Using the rank theorem, you get:

\(\begin{array}{c}{\rm{rank}}\,A + \dim \,{\rm{Nul}}A = n\\2 + \dim \;{\rm{Nul}}\,A = 4\\\dim \;{\rm{Nul}}\,A = 4 - 2\\ = 2\end{array}\)

From matrix B:

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

The basis of Col A can be written as:

\({\rm{Basis}}\left( {{\rm{Col}}\,A} \right) = \left\{ {\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\\5\end{array}} \right],\;\;\left[ {\begin{array}{*{20}{c}}{ - 4}\\2\\{ - 6}\end{array}} \right]} \right\}\)

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

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

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

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

Write the row reduced echelon form of matrix M.

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

Write the system of equations to get:

\(\begin{array}{c}{x_1} - {x_3} + 5{x_4} = 0\\{x_2} - \frac{5}{2}{x_3} + 3{x_4} = 0\end{array}\)

Consider \({x_3}\) and \({x_4}\) as free variables.

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{x_3} - 5{x_4}}\\{\frac{5}{2}{x_3} - 3{x_4}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1\\{\frac{5}{2}}\\1\\0\end{array}} \right]{x_3} + \left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - 3}\\0\\1\end{array}} \right]{x_4}\end{array}\)

So, the null space of A is \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\{\frac{5}{2}}\\1\\0\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - 3}\\0\\1\end{array}} \right]} \right\}\).