Answer
a) $Rank A=3, $It is he number of pivot columns in B and A
b) $dim.Nul A=2$
c) $Basis\,for\, Col.A\,=\begin{bmatrix}1\\{ - 2}\\{ - 3}\\3 \end{bmatrix},\begin{bmatrix}4\\{ - 6}\\{ - 6}\\4\end{bmatrix},\begin{bmatrix}9\\{ - 10}\\{ - 3}\\0\end{bmatrix}$
d) $Basis\,for\,Row A=\begin{pmatrix}[1&-3&0&5&-7] , [0&0&2&-3&8],[0&0&0&0&5]\end{pmatrix}$
e) $Basis\,for\,Nul A=\begin{bmatrix}3\\1\\0\\0\\0\end{bmatrix},\begin{bmatrix}-5\\0\\\frac{3}{2}\\1\\0\end{bmatrix}$
Work Step by Step
Taking the matrix A as row equivalent to matrix B;
$A=\begin{bmatrix}1&{ - 3}&4&{ - 1}&9\\{ - 2}&6&{ - 6}&{ - 1}&{ - 10}\\{ - 3}&9&{ - 6}&{ - 6}&{ - 3}\\3&{ - 9}&4&9&0\end{bmatrix}$ $B = \begin{bmatrix}1&{ - 3}&0&5&{ - 7}\\0&0&2&{ - 3}&8\\0&0&0&0&5\\0&0&0&0&0\end{bmatrix}$
a). $Rank A=3, $It is he number of pivot columns in B.
b)$.dim.Nul A=2$ $dim. Nul A=n-Rank A$ but , $n=5$ $dim.Nul A=5-3=2$
c)$Basis\,for\, Col.A\,=\begin{bmatrix}1\\{ - 2}\\{ - 3}\\3 \end{bmatrix},\begin{bmatrix}4\\{ - 6}\\{ - 6}\\4\end{bmatrix},\begin{bmatrix}9\\{ - 10}\\{ - 3}\\0\end{bmatrix}$
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 3, and Column 5
d) $Basis\,for\,Row A=\begin{pmatrix}[1&-3&0&5&-7] , [0&0&2&-3&8],[0&0&0&0&5]\end{pmatrix}$
The nonzero rows of B form a basis for Row A
e)writing matrix B in equation form as $\begin{pmatrix}{x_1} = 3{x_2} - 5{x_4}\\{x_2} = {x_2} + 0{x_4}\\{x_3} = 0{x_2} + \frac{3}{2}{x_4}\\{x_4} = 0{x_2} + {x_4}\\{x_5} = 0{x_2} + 0{x_4}\end{pmatrix}$
we can write the solution in parametric form with respect to the free variables $x_{2},x_{4}$:
$x_{2}\begin{bmatrix}3\\1\\0\\0\\0\end{bmatrix}+x_{4}\begin{bmatrix}-5\\0\\\frac{3}{2}\\1\\0\end{bmatrix}$
Basis for Nul A=$\begin{bmatrix}3\\1\\0\\0\\0\end{bmatrix},\begin{bmatrix}-5\\0\\\frac{3}{2}\\1\\0\end{bmatrix}$
