Step 1
We are given with following Matrices $A = \left[ {\begin{array}{*{20}{c}}{ - 2}&4&{ - 2}&{ - 4}\\2&{ - 6}&{ - 3}&1\\{ - 3}&8&2&{ - 3}\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}1&0&6&5\\0&2&5&3\\0&0&0&0\end{array}} \right]$

Step 2
Since the matrix $A$ is row equivalent to $B$. \[\begin{array}{l}A \approx B\\\\\left[ {\begin{array}{*{20}{c}}{ - 2}&4&{ - 2}&{ - 4}\\2&{ - 6}&{ - 3}&1\\{ - 3}&8&2&{ - 3}\end{array}} \right] \approx \left[ {\begin{array}{*{20}{c}}1&0&6&5\\0&2&5&3\\0&0&0&0\end{array}} \right]\end{array}\]Step 3
In matrix B, apply the operations
In Column 3: ${C_3} = {C_3} - 6{C_1} - \dfrac{5}{2}{C_2}$ and
In Column 4: ${C_4} = {C_4} - 5{C_1} - \dfrac{3}{2}{C_2}$ \[\begin{array}{l}B = \left[ {\begin{array}{*{20}{c}}1&0&6&5\\0&2&5&3\\0&0&0&0\end{array}} \right]\\\\B = \left[ {\begin{array}{*{20}{c}}1&0&{6 - 6\left( 1 \right) + \dfrac{5}{2}\left( 0 \right)}&{5 - 5\left( 1 \right) + \dfrac{3}{2}\left( 0 \right)}\\0&2&{5 - 6\left( 0 \right) + \dfrac{5}{2}\left( 2 \right)}&{3 - 5\left( 0 \right) + \dfrac{3}{2}\left( 2 \right)}\\0&0&{0 - 6\left( 0 \right) + \dfrac{5}{2}\left( 0 \right)}&{0 - 5\left( 0 \right) + \dfrac{3}{2}\left( 0 \right)}\end{array}} \right]\\\\B = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\0&2&0&0\\0&0&0&0\end{array}} \right]\end{array}\]So, the columns $A$ corresponding to nonzero columns of B will form the bases for Col $A$ \[Col\,A = \left\{ {\left[ {\begin{array}{*{20}{c}}{ - 2}\\2\\{ - 3}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}4\\{ - 6}\\8\end{array}} \right]} \right\}\]Step 4
Consider the null set for the matrix A$$Null\,A = \left\{ {x \in {R^4}|\,A{\bf{x}} = 0} \right\}$$As the matrix B is row equivalent to A, $$Null\,A = \left\{ {x \in {R^4}|\,B{\bf{x}} = 0} \right\}$$Solve the system of equations, $B{\bf{x}} = 0$\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}1&0&6&5\\0&2&5&3\\0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{{x_1} + 6{x_2} + 5{x_4}}\\{2{x_2} + 5{x_3} + 3{x_4}}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\end{array}\]Step 5
As there are more variables than number of equations, we assume two variables as free variables: (${x_3}\,,\,{x_4}$). With (${x_3}\,,\,{x_4}$) as free variables, we have the solution vector as\[\begin{array}{l}{x_1} = - 6{x_3} - 5{x_4}\\\\{x_2} = - \dfrac{5}{2}{x_3} - \dfrac{3}{2}{x_4}\end{array}\]Step 6
The final solution in vector form is given by\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 6{x_3} - 5{x_4}}\\{ - \dfrac{5}{2}{x_3} - \dfrac{3}{2}{x_4}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\\\= \left[ {\begin{array}{*{20}{c}}{ - 6{x_3}}\\{ - \dfrac{5}{2}{x_3}}\\{{x_3}}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 5{x_4}}\\{ - \dfrac{3}{2}{x_4}}\\0\\{{x_4}}\end{array}} \right]\\\\= \left[ {\begin{array}{*{20}{c}}{ - 6}\\{ - \dfrac{5}{2}}\\1\\0\end{array}} \right]{x_3} + \left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - \dfrac{3}{2}}\\0\\1\end{array}} \right]{x_4}\end{array}\]Step 7
As we can see fromt he above results that every column in A is a linear combination of the set of vectors $\left\{ {\left[ {\begin{array}{*{20}{c}}{ - 6}\\{ - \dfrac{5}{2}}\\{{x_3}}\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - \dfrac{3}{2}}\\0\\1\end{array}} \right]} \right\}$ and they also form a linearly independent set. Thus, the basis of Null A is given by: \[NullA = \left\{ {\left[ {\begin{array}{*{20}{c}}{ - 6}\\{ - \dfrac{5}{2}}\\{{x_3}}\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - \dfrac{3}{2}}\\0\\1\end{array}} \right]} \right\}\]