Answer
(a) The nullspace of $A $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2}\\x_3 \end{aligned}\right]= \left[\begin{aligned}4t\\-2t\\t \end{aligned}\right] =t\left[\begin{aligned}4 \\-2\\1 \end{aligned}\right] .$$
(b) The nullity is $1$.
(c) The rank of $A$ is $2$.
Since $A$ has $3$ columns, one can see that
$$\text{rank}(A)+\text{nullity}(A)=2+1=3.$$
Work Step by Step
Given the matrix
$$
A=\left[ \begin {array}{ccc} 1&3&2\\ 4&-1&-18
\\ -1&3&10\\ 1&2&0\end {array}
\right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{ccc} 1&0&-4\\ 0&1&2
\\ 0&0&0\\ 0&0&0\end {array}
\right]
.
$$
The corresponding system is
$$
\begin{aligned} x_{1} -4x_3 &=0\\
x_{2} +2x_3 &=0\\
\end{aligned}.
$$
The solution of the above system is $x_1= 4t$, $x_2=-2t$,$x_3=t$.
(a) The nullspace of $A $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2}\\x_3 \end{aligned}\right]= \left[\begin{aligned}4t\\-2t\\t \end{aligned}\right] =t\left[\begin{aligned}4 \\-2\\1 \end{aligned}\right] .$$
(b) The nullity is $1$.
(c) The rank of $A$ is $2$.
Since $A$ has $3$ columns, one can see that
$$\text{rank}(A)+\text{nullity}(A)=2+1=3.$$
