Find bases for the column space, the row space, and the null space of matrix A. You should verify that the Rank-Nullity Theorem holds. An equivalent echelon form of matrix A is given to make your work easier.

$A = \left[\begin{array}{ccc} 4 &16 &3\cr 3 &1 &0\cr -4 &2 &5\cr 2 &7 &-3\cr -4 &10 &6 \end{array}\right]$ $\sim$ $\left[\begin{array}{ccc} 1 &0 &0\cr 0 &1 &0\cr 0 &0 &1\cr 0 &0 &0\cr 0 &0 &0 \end{array}\right]$
Basis for the column space of $A$ is
$\Bigg \lbrace$ $\left[\Rule{0pt}{6em}{0pt}\right.$$\left]\Rule{0pt}{6em}{0pt}\right.$, $\left[\Rule{0pt}{6em}{0pt}\right.$$\left]\Rule{0pt}{6em}{0pt}\right.$, $\left[\Rule{0pt}{6em}{0pt}\right.$$\left]\Rule{0pt}{6em}{0pt}\right.$$\Bigg \rbrace$

Basis for the row space of $A$ is
$\Big \lbrace$ $\left<\Rule{0pt}{1.2em}{0pt}\right.$,,$\left>\Rule{0pt}{1.2em}{0pt}\right.$,
$\left<\Rule{0pt}{1.2em}{0pt}\right.$,,$\left>\Rule{0pt}{1.2em}{0pt}\right.$,
$\left<\Rule{0pt}{1.2em}{0pt}\right.$,,$\left>\Rule{0pt}{1.2em}{0pt}\right.$$\Big \rbrace$

Note that since the only solution to $A{\bf x} = {\bf 0}$ is the zero vector, there is no basis for the null space of $A$.