Answer
a) $Rank A=3$
b) $dim.Nul A=2$
c) $Basis\,for\, Col.A\,=\begin{bmatrix}2\\{ - 2}\\4\\{ - 2}\end{bmatrix},\begin{bmatrix}6\\{ - 3}\\9\\3\end{bmatrix},\begin{bmatrix}2\\{ - 3}\\5\\{ - 4}\end{bmatrix}$
d) $Basis\,for\,Row A=\begin{pmatrix}[2&-3&6&2&5] , [0&0&3&-1&1],[0&0&0&1&3]\end{pmatrix}$
e) $Basis\, for\, Nul A=\begin{bmatrix}\frac{3}{2}\\1\\0\\0\\0\end{bmatrix},\begin{bmatrix}\frac{9}{2}\\0\\-\frac{4}{3}\\-3\\1\end{bmatrix}$
Work Step by Step
Taking the matrix A as row equivalent to matrix B;
$A=\begin{bmatrix}2&{ - 3}&6&2&5\\{ - 2}&3&{ - 3}&{ - 3}&{ - 4}\\4&{ - 6}&9&5&9\\{ - 2}&3&3&{ - 4}&1\end{bmatrix}$
$B =\begin{bmatrix}2&{ - 3}&6&2&5\\0&0&3&{ - 1}&1\\0&0&0&1&3\\0&0&0&0&0\end{bmatrix}$
a) $Rank A=3, $It is he number of pivot columns in B and A
b) $dim.Nul A=2$ $dim. Nul A=n-Rank A$ but , $n=5$ $dim.Nul A=5-3=2$
c) $Basis\,for\, Col.A\,=\begin{bmatrix}2\\{ - 2}\\4\\{ - 2}\end{bmatrix},\begin{bmatrix}6\\{ - 3}\\9\\3\end{bmatrix},\begin{bmatrix}2\\{ - 3}\\5\\{ - 4}\end{bmatrix}$
The basis for Col.A=the pivot columns of matrix A as seen in Matrix B. the pivotal columns in B are Column 1, Column 3, and Column 4.
d) $Basis\,for\,Row A=\begin{pmatrix}[2&-3&6&2&5] , [0&0&3&-1&1],[0&0&0&1&3]\end{pmatrix}$
Basis for Row A=The non zero rows as given by Matrix B
e)writing matrix B in equation form as $\begin{pmatrix}{x_1} = \dfrac{3}{2}{x_2} + \dfrac{9}{2}{x_5}\\{x_3} = - \dfrac{4}{3}{x_5}\\{x_4} = - 3{x_5}\end{pmatrix}$ we can write the solution in parametric form with respect to the free variables $x_{2},x_{5}$: $x_{2}\begin{bmatrix}\frac{3}{2}\\1\\0\\0\\0 \end{bmatrix}+x_{5}\begin{bmatrix}\frac{9}{2}\\0\\-\frac{4}{3}\\-3\\1\end{bmatrix}$ $Basis\, for\, Nul A=\begin{bmatrix}\frac{3}{2}\\1\\0\\0\\0\end{bmatrix},\begin{bmatrix}\frac{9}{2}\\0\\-\frac{4}{3}\\-3\\1\end{bmatrix}$
