91Ó°ÊÓ

From the echelon form, it can be observed that columns 1 and 3 are thepivot columns.

The basis of Col A is

\(\left[ {\begin{array}{*{20}{c}}{ - 3}\\2\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\4\\{ - 2}\end{array}} \right]\).

Nul A is given by the equation \(Ax = 0\). Then,

\[\left[ {\begin{array}{*{20}{c}}1&{ - 3}&6&9\\0&0&4&5\\0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\].

So, the equations are

\({x_1} - 3{x_2} + 6{x_3} + 9{x_4} = 0\),

\({x_3} + \frac{5}{4}{x_4} = 0\).

For the equations, \({x_3}\) and \({x_4}\) are the free variables.

\(\begin{array}{c}{x_1} = 3{x_2} - 6{x_3} - 9{x_4}\\ = 3{x_2} - 6\left( { - \frac{5}{4}{x_4}} \right) - 9{x_4}\\ = 3{x_2} - \frac{3}{2}{x_4}\end{array}\)

The solution set is obtained as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3{x_2} - \frac{3}{2}{x_4}}\\{{x_2}}\\{ - \frac{5}{4}{x_4}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3{x_2}}\\{{x_2}}\\0\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - \frac{3}{2}{x_4}}\\0\\{ - \frac{5}{4}{x_4}}\\{{x_4}}\end{array}} \right]\\ = {x_2}\left[ {\begin{array}{*{20}{c}}3\\1\\0\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}{ - \frac{3}{2}}\\0\\{ - \frac{5}{4}}\\1\end{array}} \right]\end{array}\)

So, the basis of Col A is \(\left\{ {\left[ {\begin{array}{*{20}{c}}{ - 3}\\2\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\4\\{ - 2}\end{array}} \right]} \right\}\) and the basis of Nul A is \(\left\{ {\left[ {\begin{array}{*{20}{c}}3\\1\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - \frac{3}{2}}\\0\\{ - \frac{5}{4}}\\1\end{array}} \right]} \right\}\).