Answer
(a) A basis for the solution space is
$$ \left[\begin{aligned}-2\\1 \\0 \end{aligned}\right] .$$
(b) The dimension of the solution space is $1$.
Work Step by Step
The coefficient matrix is given by
$$
\left[ \begin {array}{ccc} 1&2&-4\\ -3&-6&-12
\end {array} \right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{ccc} 1&2&0\\ 0&0&1
\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned}
x +2y&=0\\
z&=0\\
\end{aligned}.
$$
The solution of the above system is $x=-2t$,$y=t$, $z=0$. This means that the solution space of $Ax = 0 $ consists of the vectors on the form
$$x= \left[\begin{aligned} x\\ y\\z \end{aligned}\right]= \left[\begin{aligned}-2t\\t \\0 \end{aligned}\right] =t\left[\begin{aligned}-2\\1 \\0 \end{aligned}\right] .$$
(a) A basis for the solution space is
$$ \left[\begin{aligned}-2\\1 \\0 \end{aligned}\right] .$$
(b) The dimension of the solution space is $1$.
