Answer
(a) The nullspace of $A $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2}\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned}3s-t\\-2t\\s\\t \end{aligned}\right] =s\left[\begin{aligned}3 \\0\\1\\0 \end{aligned}\right]+t\left[\begin{aligned}-1\\-2 \\0\\1 \end{aligned}\right].$$
(b) The nullity is $2$.
(c) The rank of $A$ is $2$.
Since $A$ has $4$ columns, one can see that
$$\text{rank}(A)+\text{nullity}(A)=2+2=4.$$
Work Step by Step
Given the matrix
$$
A=\left[ \begin {array}{cccc} 2&-3&-6&-4\\ 1&5&-3&11
\\ 2&7&-6&16\end {array} \right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{cccc} 1&0&-3&1\\ 0&1&0&2
\\ 0&0&0&0\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned} x_{1} -3x_3+x_4 &=0\\
x_{2} +2x_4 &=0
\end{aligned}.
$$
The solution of the above system is $x_1= 3s-t$, $x_2=-2t$,$x_3=s$, $x_4=t$.
(a) The nullspace of $A $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2}\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned}3s-t\\-2t\\s\\t \end{aligned}\right] =s\left[\begin{aligned}3 \\0\\1\\0 \end{aligned}\right]+t\left[\begin{aligned}-1\\-2 \\0\\1 \end{aligned}\right].$$
(b) The nullity is $2$.
(c) The rank of $A$ is $2$.
Since $A$ has $4$ columns, one can see that
$$\text{rank}(A)+\text{nullity}(A)=2+2=4.$$
