Given Information
We are given with following matrix\[A = \left[ {\begin{array}{*{20}{r}}1&5&{ - 4}&{ - 3}&1\\0&1&{ - 2}&1&0\\0&0&0&0&0\end{array}} \right]\]We have to find spanning set for Nul A. In order to do so, we will find the general solution of $Ax=0$.

Step 1: The augmented matrix
\[\left[ {\begin{array}{*{20}{l}}A&{\bf{0}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&5&{ - 4}&{ - 3}&1&0\\0&1&{ - 2}&1&0&0\\0&0&0&0&0&0\end{array}} \right]\]

Step 2: Row reduce the Augmented matrix
\[\begin{array}{l}\left[ {\begin{array}{*{20}{l}}A&{\bf{0}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&5&{ - 4}&{ - 3}&1&0\\0&1&{ - 2}&1&0&0\\0&0&0&0&0&0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&0&6&{ - 8}&1&0\\0&1&{ - 2}&1&0&0\\0&0&0&0&0&0\end{array}} \right]\,\,::\,\,\left\{ {{R_1} = {R_1} - 5{R_2}} \right\}\end{array}\]

Step 3: System of Equations
\[\begin{array}{l}{x_1} = - 6{x_3} + 8{x_4} - {x_5}\\{x_2} = 2{x_3} - {x_4}\end{array}\]Since there are five variables and two equations, consider $x_3, x_4$ and $x_5$ as free;

Step 4: Solution in Parametric form
\[\vec x = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right] = {x_3}\left[ {\begin{array}{*{20}{c}}{ - 6}\\2\\1\\0\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}8\\{ - 1}\\0\\1\\0\end{array}} \right] + {x_5}\left[ {\begin{array}{*{20}{c}}{ - 1}\\0\\0\\0\\1\end{array}} \right]\]So, the Nul A is given by

\[{\rm{Nul}}\,A = \left\{ {\left[ {\begin{array}{*{20}{c}}{ - 6}\\2\\1\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}8\\{ - 1}\\0\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 1}\\0\\0\\0\\1\end{array}} \right]} \right\}\]