Answer
a) $Rank A=2$
b) $dim.Nul.A=2$
c) $Basis\,for\, Col.A\,=\begin{bmatrix}1\\-1\\5\end{bmatrix},\begin{bmatrix}-4\\2\\-6\end{bmatrix}$
d) $Basis\,for\,Row.A=\begin{pmatrix}[1&0& -1&5] , [0&-2&5&-6]\end{pmatrix}$
e) $Basis\,for\,Nul.A=\begin{bmatrix}1\\\frac{5}{2}\\1\\0\end{bmatrix},\begin{bmatrix}-5\\-3\\0\\1\end{bmatrix}$
Work Step by Step
Taking the matrix A as row equivalent to matrix B
$\mathbf{A}=\begin{bmatrix}1&{ - 4}&9&{ - 7}\\{ - 1}&2&{ - 4}&1\\5&{ - 6}&{10}&7\end{bmatrix}$
$\mathbf{B} = \begin{bmatrix}1&0&{ - 1}&5\\0&{ - 2}&5&{ - 6}\\0&0&0&0\end{bmatrix}$
a). $Rank A=2, $It is he number of pivot columns in B and A
b) $.dim.Nul A=2$
$dim. Nul A=n-Rank A$
but , $n=4$
$dim.Nul A=4-2=2$
c) $Basis\,for\, Col.A\,=\begin{bmatrix}1\\-1\\5\end{bmatrix},\begin{bmatrix}-4\\2\\-6\end{bmatrix}$
Basis for Col.A=the pivot columns of matrix A as seen in Matrix B.
the pivotal columns in B are column 1 and Column 2.
d) $Basis\,for\,Row A=\begin{pmatrix}[1&0& -1&5] , [0&-2&5&-6]\end{pmatrix}$
Basis for Row A=The non zero rows as given by Matrix B
e) Writing matrix B in equation form as
$\begin{bmatrix}{x_1} - {x_3} + 5{x_4} = 0\\{x_2} - \frac{5}{2}{x_3} + 3{x_4} = 0\end{bmatrix}$
we can write the solution in parametric form with respect to the free variables $x_{3},x_{4}$:
$x_{3}\begin{bmatrix}1\\\frac{5}{2}\\1\\0\end{bmatrix}+x_{4}\begin{bmatrix}-5\\-3\\0\\1\end{bmatrix}$
Basis for Nul A=$\begin{bmatrix}1\\\frac{5}{2}\\1\\0\end{bmatrix},\begin{bmatrix}-5\\-3\\0\\1\end{bmatrix}$