91Ó°ÊÓ

From the echelon form, it can be observed columns 1 and 2 are thepivot columns.

The basis of Col A is

\(\left[ {\begin{array}{*{20}{c}}4\\6\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}5\\5\\4\end{array}} \right]\).

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

\[\left[ {\begin{array}{*{20}{c}}1&2&6&{ - 5}\\0&1&5&{ - 6}\\0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\].

So, the equations are

\({x_2} + 5{x_3} - 6{x_4} = 0\),

\({x_1} + 2{x_2} + 6{x_3} - 5{x_4} = 0\).

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

\({x_2} = - 5{x_3} + 6{x_4}\)

And

\(\begin{array}{c}{x_1} + 2\left( { - 5{x_3} + 6{x_4}} \right) + 6{x_3} - 5{x_4} = 0\\{x_1} = 4{x_3} - 7{x_4}\end{array}\)

The solution set is obtained as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{4{x_3} - 7{x_4}}\\{ - 5{x_3} + 6{x_4}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{4{x_3}}\\{ - 5{x_3}}\\{{x_3}}\\{{x_4}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 7{x_4}}\\{6{x_4}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = {x_3}\left[ {\begin{array}{*{20}{c}}4\\{ - 5}\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}{ - 7}\\6\\0\\1\end{array}} \right]\end{array}\)

So, the basis of Col A is \(\left\{ {\left[ {\begin{array}{*{20}{c}}4\\6\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}5\\5\\4\end{array}} \right]} \right\}\) and the basis of Nul A is \(\left\{ {\left[ {\begin{array}{*{20}{c}}4\\{ - 5}\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 7}\\6\\0\\1\end{array}} \right]} \right\}\).